On Planar Rational Cuspidal Curves
by
Tiankai Liu
A.B., Harvard University (2008)
C.A.S.M., University of Cambridge (2009)
Submitted to the Department of Mathematics
in partial fulfillment of the requirements for the degree of
Doctor of Philosophy in Mathematics
I
OF TECHNOLOGY
~~~
-
MASSACHUSETTS INSTITUTE OF TECHNOLOGY
@ Massachusetts Institute of Technology 2014. All rights reserved.
Signature redacted
Department of Mathematics
April 24, 2014
redacted
,-.v .........
James McKernan, FRS
Norbert Wiener Professor of Mathematics
Thesis Supervisor
Accepted
JUN 1 7 2014
L! BRARIES
June 2014
-Signature
Certified b7
cL
-
MASSACHUSETTSiINS
at the
Author ..
-
by.Signature redacted ........................
William P. Minicozzi II
Chairman, Depa fentCommittee on Graduate Theses
E
On Planar Rational Cuspidal Curves
by
Tiankai Liu
Submitted to the Department of Mathematics
on April 24, 2014, in partial fulfillment of the
requirements for the degree of
Doctor of Philosophy in Mathematics
Abstract
This thesis studies rational curves in the complex projective plane that are homeomorphic to their normalizations. We derive some combinatorial constraints on such
curves from a result of Borodzik-Livingston in Heegaard-Floer homology. Using these
constraints and other tools from algebraic geometry, we offer a solution to certain
cases of the Coolidge-Nagata problem on the rectifiability of planar rational cuspidal curves, that is, their equivalence to lines under the Cremona group of birational
automorphisms of the plane.
Thesis Supervisor: James McKernan, FRS
Title: Norbert Wiener Professor of Mathematics
3
4
Acknowledgments
I am deeply grateful to my advisor, James McKernan, for mentoring me throughout the past five years, for sharing some part of his boundless knowledge, and for
repeatedly encouraging me during moments and months of despair. Thanks go to
Samuel Grushevsky, Vivek Shende, and Chenyang Xu for helpful conversations, and
to my thesis committee, Bjorn Poonen and Frangois Charles. I am indebted to my fellow students Joshua Batson, Lucas Culler, Shashank Dwivedi, David Jackson-Hanen,
Ailsa Keating, John Lesieutre, Gregory Minton, Jennifer Park, and Roberto Svaldi,
for answering my silly questions over the years, and to many other friends, without
whom I could not have survived graduate school. Thanks, finally, to my parents for
their constant support, and for first instilling in me a love of mathematics.
This research was partially supported by NSF grants DMS-0943787, 0701101 and
1200656.
5
Contents
1
Introduction
2
The topology of cusps
2.1 Topological invariants of cusps . . . . . . . . . . . . . . . . . . . . . .
2.2 N otation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.3 The semigroup distribution property . . . . . . . . . . . . . . . . . . .
9
15
15
18
19
3 The Coolidge-Nagata problem
3.1
3.2
3.3
3.4
3.5
4
Conditions for rectifiability . . . . . . . . . .
An effective divisor and a volume computation
Three or more cusps . . . . . . . . . . . . . .
O ne cusp . . . . . . . . . . . . . . . . . . . .
Two cusps, each with only one Newton pair .
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Triangulating sequences
4.1 Strongly triangulating and pseudo-triangulating sequences of length
4.2 Simplest fractions, fair approximations, and continued fractions . .
4.3 Perturbations of strongly triangulating sequences . . . . . . . . . .
4.4 Weakly triangulating sequences of length 1 . . . . . . . . . . . . . .
4.4.1 Weakly triangulating sequences (6; a) with 1 ' a < 94 . . .
9 . . .
4.4.2 Weakly triangulating sequences (6; a) with se4 < a
4.4.3 Classification of weakly triangulating sequences of length 1
4.5 Weakly triangulating sequences of length 2 . . . . . . . . . . . . . .
4.5.1
4.5.2
4.6
Cases of the form
Cases where n
-'=
mi
+F
_F2
k 21
F
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1+Fk-3
.+F
.
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23
28
30
34
43
1.
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+1Fk-1+2 . . . . . . . . . . . . . . .
+2
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46
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52
60
60
67
72
78
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. . . . . . . . 121
Assorted results on triangulating sequences of length 3 . . . . . . . . . . . . . . . . . 138
7
8
Chapter 1
Introduction
As its title indicates, this thesis studies rational cuspidal curves in the complex projective plane IP2
Before we begin the formal discussions, perhaps a few words for the lay reader are in order, since
the objects of study are sufficiently concrete (by the standards of modern mathematics) as to be
accessible to many people with a scientific background, even quite far from algebraic geometry. The
complex projective plane P2 may be thought of as the ordinary complex plane C 2 completed with
a "line at infinity" that contains one point corresponding to every direction in C2. It was realized
centuries ago that many geometrical problems can be stated more elegantly in P 2 as opposed to C2,
and that is the case, in particular, for the problems we will be considering. That being said, readers
who are not familiar with this notion will do reasonably well, for the time being, just to think in
terms of the more familiar complex plane C2, or indeed even the real plane R2.
An algebraic curve (which is the only kind of curve we will be considering) is, more or less,
the locus of points (x, y) in the plane that satisfy an equation of the form f(x, y) = 0, for a
specified polynomial f in two variables with complex coefficients. Such a curve is rational if it
admits a parametrization by rational functions (ratios of polynomials in x and y). Thus, the line
y = mx + b (or y - mx - b = 0), which may be so parametrized as (x, y) = (t, mt + b), is rational.
Perhaps more surprisingly, the circle x 2 + y 2 = 1 is also rational, because it may be parametrized
t,
). Indeed, we can quickly verify that
as (x, y) = (t
2t
t2 + I
2
_12
4t2
t2 +1
2
t4 + 2t
+1
t4 - 2t 2 +1
t4 + 22 +1
It turns out that not all curves are rational (and in some sense "most" curves are not), though
perhaps this is far from obvious at first.
As is often the case, just by looking at the R-valued points on a curve, it may not be so clear
geometrically what rationality means. But if one is able to imagine the complex-valued points as
well, then a complex "curve" really looks to us like a real surface, and from this point of view, "rationality" has a clear topological meaning, at least for smooth "curves." The kinds of real surfaces that
arise this way are classified topologically by a non-negative integer called their genus. Informally, a
sphere (by which we always mean the surface only, not a solid ball) has genus 0, a torus (the surface
of a bagel with one hole) has genus 1, and in general a torus with g holes has genus g. Then it turns
out that a smooth complex "curve" is rational if and only if it looks like a real surface of genus 0.
At any rate, all curves, rational or not, may have smooth points (points on the curve where
the defining polynomial f has at least one non-zero partial derivative) and singular points, or
singularities (points on the curve where both partial derivatives Of/Ox and Of/y are zero). The
curve itself is called smooth if all its points are smooth points. Much research has been done on
9
classifying the singularities that can arise on curves. For our thesis, we will be concerned with one
dichotomy in particular.
At some singularities, a curve may have at least two distinct "branches." For example, the origin
on the curve y2 = x 2 (X + 1) is a singularity known as "the ordinary node"; the curve is seen to have
one branch tangent to the x-axis, and a different one tangent to the y-axis. If one zooms in close to
this singularity, it looks like two separate lines meeting at a point. If we try to visualize the complex
points, it looks like two separate real planes meeting at a point inside a four-real-dimensional space.
At other singularities, a curve may have only one "branch." For example, the origin on the
curve y2
X3 is a singularity known as "the ordinary cusp"; the curve has only one branch there.
Here the real picture may be a bit misleading, since it looks as though one could draw a distinction
between the points with positive versus negative y. But one should keep in mind that in the world of
complex numbers, such distinctions are usually not meaningful. If we zoom in on the ordinary cusp,
while trying to visualize all the complex points, we see what looks like the cone over a trefoil knot
in three-dimensional space. That is, just as in everyday language a "cone" is the result of taking
every point on a circle in the plane, and joining that point by a straight line to a fixed cone point
outside the plane in three-space, here we are asked to imagine the result of taking every point on
a trefoil knot in three-space, and joining that point by a straight line to a fixed cone point outside
three-space, in a larger, four-dimensional space. This cone point is precisely the point at the cusp
itself. From this picture, I hope it is at least somewhat plausible that this singularity has only one
branch: there is no natural way to partition the points on the trefoil knot into multiple distinct
sets, in the way that we teased apart the two real planes in the example of the ordinary node above.
A curve is called cuspidal if its singularities (and it may not have any at all) are of the second
kind, that is, they have only one branch. A rational cuspidal curve in the complex plane, then,
looks like a topological sphere that, although it may be "pinched" at certain points, cannot "cross
itself." In the language of topological, we say that the surface is homeomorphic to a sphere. Now, at
least, I hope I have conveyed some sense of what the basic objects of this thesis - rational cuspidal
curves in the plane - are.
Now let us begin the more formal introduction. All our varieties are defined over the field of
complex numbers C. A projective curve is called cuspidal if is integral and all its singularities are
unibranched (locally analytically irreducible). In other words, its normalization map is a homeomorphism; intuitively, the curve does not "cross itself" as it would at a node. A curve is rational
if it is birational to P 1 , which is to say it admits a parametrization by rational functions. Thus,
in the analytic topology, rational cuspidal curves are characterized by being homeomorphic to the
2-sphere S 2 . Here are some examples of rational cuspidal curves:
1. A straight line is rational and has no singularities at all, so it is vacuously cuspidal.
2. A smooth conic is also a rational cuspidal curve.
3. The cuspidal cubic, given by the equation y 2 = x 3 , is the most basic singular example. More
generally, the curves ya = xb, for relatively prime integers b > a > 1, are rational and each
have one cusp at the origin [0 : 0 : 1], and possibly a second cusp at [0 : 1 : 0], if b - a > 1.
4. The tricuspidal quartic: imagine the curve T traced out in R 2 by a point on the circumference
of a wheel of radius 1 rolling along the inside of a fixed unit circle [?]. This turns out to be a
cuspidal rational quartic curve. It has three A 1 cusps, namely, the real points where the curve
meets the fixed unit circle.
10
5. Let C be the image of the mapping PI
-4 p2 given by
t(s) = [3s4 : 2s5 + s 2 : 4s3
-
1.
It is a quintic with four cusps: in the ADE classification, it has one A 5 cusp at t(0) = [0 : 0 : 1],
and one A, cusp at t(C) = [:
:1 for each cubic root of unity (. This example is discussed,
for example, in [6].
Many more examples are known; see [10], example, or [3, 5, 4, 6, 12, 15].
The requirement that a cuspidal curve have no nodes is quite restrictive. Tono [16] proved
that every rational cuspidal curve in the plane has at most 8 cusps, and, more generally, a planar
cuspidal curve of geometric genus g has at most (21g + 17)/2 cusps, regardless of its degree. In
fact, no rational cuspidal curves are known with 5 or more cusps, and only one such curve (up to
PGL(3) actions) is known with 4 cusps, namely our fifth example above. See [15] for a conjectural
classification of all rational cuspidal curves with 3 or more cusps. At any rate, the situation contrasts
sharply with the abundance of nodal curves, since there is no upper bound on the number of nodes
of a planar nodal curve of any fixed genus.
By using the methods of Heegaard-Floer homology, Borodzik-Livingston [1] produced a series
of constraints on rational cuspidal curves, known as the "semigroup distribution property" (see
Fact 2.2 below). In the case of a rational cuspidal curve of degree d with exactly one cusp, these
constraints say that, for 0 < j < d-1, the number of elements of the semigroup W of the cusp in the
interval [0, jd] equals the triangular number (j+2). We define a notion of triangulatingsequences to
study such semigroups, and calculate many constraints that they must satisfy. We show that these
constraints are in fact sufficient to recover the classification in [4] of unicuspidal rational curves
whose cusp is homeomorphic to one of the form ya = Xb. We also prove the following theorem, in
the next simplest case:
Theorem 1.1. Suppose C C P2 is a rational cuspidal curve of degree d with a single cusp locally
homeomorphic to the cusp parametrized by (x, y) - (tnM2, tnm2(1 + tn2)), for some positive integers
where gcd(m,n) = gcd(m 2 ,n2) = 1 and m,n,m2 ; 2. Then, up to interchangingthe roles of m and
n, one of the following ten statements holds. (Here Fi denotes the ith Fibonacci number; F0 = 0,
F 1 = 1, and Fk+1 = Fk + Fk_1, for all k E Z.)
1. There exist k, f E Z such that d
2), and (M 2 , n 2 ) = (F2k
= F2k-lF2k+lm, and (in,
2. There exist k, f E Z such that d = F2k+lm, and (in, n)
and (M2, n2) = (F2k-1, £F2k-1 + F2k-5).
3. d = 8n 2 + 4n+ 1, (m,n) = (4n+ 1,n), and (M 2 ,n2)
4. d
=
nM2, (m, n)
=
(n - 1, n), and (M 2 , n 2 )
= £(Fk-1, Fk+)+(F2k-3,F2k1+
F2k+1)+ (F2k3, Fik 1+2)
= f(F2k_,
=
(4n-+ 1,(2n+ 1)2).
(M 2 , nm 2 - 1).
=
5. d = n 2 + 1, (m, n) = (n - 1, n), and (M2, n2)
n)
=
(n, (n + 1)2).
6. d = 20, (m, n) = (2, 3), and (M 2 , n2) = (6, 31).
7. d
=
2mm2, (m, n)
8. d
=
17, (m, n) = (2, 7), and (M2, n2) = (4, 17).
9. d
= F4k+2,
(i,
n)
=
(m, 4m - 1), and (M2, n2)
= (±,
F4k+4) ,
= (M2, MM2
and (M2, n2) = (3, 1).
11
1).
10. d = 2F4k+2, (m, n) =
(F4,,
F4k+4)
, and (m2, n2) = (6, 1).
It would be interesting to determine which of these invariants belong to actually existing curves.
We know that at least the curves in case (4) exist by [3].
Let us turn now to a specific question that can be asked of planar rational curves. A rational
curve C c p 2 is called rectifiable if there is a birational automorphism of p 2 that is defined at the
generic point of C and maps C onto a line. To my knowledge, Matsuoka and Sakai [9] first stated
the following conjecture:
Conjecture 1.2 (Coolidge-Nagata problem). Every rational cuspidal curve C
C p2
is rectifiable.
The problem is named after Coolidge, who studied the images of algebraic curves under Cremona
transformations in his treatise [2], and Nagata, who reinterpreted classical problems in a modern
language [11].
The conjecture has been verified for all known rational cuspidal curves. Palka [13] proved it for
curves with 5 or more cusps, and later [14] for curves with 3 or more cusps, and in some other cases
that we will describe shortly. He mentioned in private correspondence that he and Mariusz Koras
have proved the full conjecture, but that paper seems not to be available at the time of this writing.
Let us show how to rectify the examples we listed above.
1. A straight line is by definition rational.
2. A smooth conic C is rectified by a Cremona transformation centered at any three points of
C. That is, blows up the three chosen points, and then contracts the proper transforms of the
lines between them.
3. For gcd(a, b) = 1, we may choose integers s, t such that as + bt = 1. Define the birational
map p 2 -- + p 2 sending (x, y) to (u, v) = (XbY-a, X8Yt). (Its inverse sends (u, v) to (x, y) =
(Utva, u-svb).) Then the curve ya = xb becomes the line u = 1.
4. The tricuspidal quartic can be transformed into a smooth conic by a Cremona transformation
centered at the three cusps. This reduces the problem to item 2 above.
5. The quintic with four cusps is a bit more difficult to rectify explicitly. Consider the conics
Go = {y 2 = xz} and G( = {4 x 2 + 4(xy + 3y 2 + xz = 0}. Note Go passes through all four
cusps of C, and GC passes through all but t(C). Form a surface S -4 p 2 by blowing up three
times along C over t(0) with exceptional divisors E 1 , E 2 , E 3 , and once more at each t(()
with corresponding exceptional divisor F(: see Figure 1.0.1, in which W = -'+=. In S, the
proper transforms of Go and the three curves G( are pairwise disjoint (-1)-curves. Contract
them and the two curves E 2 and Ei to obtain a morphism S -+ p 2 whose composition with
2
f -1 : p 2 -- + S is a birational automorphism of p rectifying C.
In this thesis, we prove the following:
Theorem 1.3. Let C C p 2 be a rational cuspidal curve of degree d and s cusps, such that the ith
cusp has ki Newton pairs. Then C is rectifiable if s >, 3, or if E'_1 ki < 2.
(See section 2.1 for the definition of Newton pairs.) The proof is carried out in Lemmas 3.15,
3.16, 3.17, 3.18, and 3.20. We also prove a partial result, Lemma 3.19, toward the case of one cusp
with at least 3 Newton pairs.
12
G1
Gw
El.G
E,2
GF
1
E3j
F,
F,
Fw
Figure 1.0.1: Diagram in S showing rectification of quintic rational curve with four cusps
13
14
Chapter 2
The topology of cusps
2.1
Topological invariants of cusps
We first recall the basic topological invariants of cusps. These facts are all standard; see [10, 5], for
example. Let P be a cusp of a curve C in a smooth surface X. On a small analytic neighborhood
of P, we can choose holomorphic coordinates (x, y) centered at P such that C is described by the
parametric equations
x = ta
y = citbi
+ c 2 tb2 +
as t ranges over a small neighborhood of 0 in C. Here 1 < a < b, < b2 < ... are integers such that
a { bi and gcd(a, bi, b 2 ,...) = 1. The multiplicity of the cusp is a. The sequence of bi's may be
finite or infinite, but we require all the ci to be non-zero constants. Such a choice of coordinates and
parametrization is not uniquely determined, although the numbers a and b, are. If we momentarily
let gi denote gcd(a, bi, b2 ,.. , bi), then there is a finite sequence il < i 2 < ... < ik of indices at
which gi decreases; that is, ii = 1, and
gi1 =
'' '
=
i2-1
>
gi2
=''
=
i-1
> 9i3
''
>
gik
=1
It is convenient to set io = bo= 0 and go = a as well. Let us define
Mj = gi,_ 1
Nj = bij - bij
(1 <j < k + 1),
I<(1<j
k).
One associates a sequence of k Newton pairs to the cusp:
(mj,rnj) =
(Mj+1,
N,
Mj+1)
=
=
gii
94,
bi, -bij),
(1 <j <, k).
gii
The two numbers in any given Newton pair are relatively prime positive integers, and ml < ni. The
sequence of Newton pairs is an invariant of the cusp, independent of the choice of coordinates and
parametrization. It is not a complete analytic set of invariants, but it does determine the topology
1
The multiplicity of a point P on a planar curve C is the coefficient m such that f*C = f-1C + mE, where f is a
blow-up of the plane at P, with exceptional divisor E. In fact, the multiplicity can be defined intrinsically in terms
of the Hilbert-Samuel polynomial of the local ring Op,c, but we will not need to use such a definition.
15
of cusp, in the classical topology. Some exponents of the parametrization can be recovered from the
sequence of Newton pairs:
Mj = mmj+1 -mk,
Nj = nmj+1 - -nk,
a = M1,
bi, =N
1
+N
2
+---+Nj.
The advantage of Newton pairs is that they are bound by no a priori relations. To facilitate
comparison with other papers, however, let us mention some equivalent ways of presenting the same
data. The k Puiseux pairs of the cusp are
( aI~
9i1
gi
bi,
1 /
9i 2
(9-1'__
2\
9i2 /i
k
gik
and the characteristicsequence is
(a; bi, ...
b ik).
The semigroup of the cusp is the set
W := {dimc Opc/(f) 10 =A f E
p,}C} C N = {0, 1, 2, ... },
where OpC is the local ring of C at the cusp P. In other words, W is the set of all possible
local intersection multiplicities at P of C with other curves. The semigroup W has k + 1 minimal
generators W1, ... , Wk+1, given by
W1
=
MI,
W2
=
Ni,
Wj
=
mj-2Wj-1+Nfti,
(2.1.1)
(3 ( j ( k + 1).
The semigroup encodes the same information as the sequence of Newton pairs, so it is also a complete
topological invariant of the cusp.
The 6-invariant of the cusp is
k
J=
(MI-
1)(NJ
-
1) +
k(M
- 1)Nj)
(2.1.2)
j=2
and it is the contribution of P to the difference between the arithmetic and geometric genera of C.
It equals the number of elements of N\W, and 26 - 1 is the maximum element of Z\W.
When we blow up the ambient surface X at the point P, the strict transform of C has either a
smooth point or another cusp lying over P. We may, by repeated blow-ups, pass to a minimal log
resolution 14 X of the cusp P, for which the r-exceptional curves along with the strict transform
o = r,-1(C) form a simple normal crossings divisor D on X. Then two different irreducible
components of D intersect each other either once transversely, or not all all. The dual graph for this
resolution is a graph whose vertices are components of D and whose edges correspond to intersection
points between two different components. Often the dual graph is considered along with the data
of the self-intersection numbers of its various components.
The dual graph for a cusp is a tree whose structure that can be computed from the Newton
pairs. To avoid confusion with the algebro-geometric notion of degrees of planar curves, we will say
that the branching number of a component of D is the number of other components that it meets,
or, as we will sometimes say, is adjacent to. Every r-exceptional curve has branching number at
16
3
3
2 2 2 2
2 5
2 3
Ei2
4
2
2 2 6
Figure 2.1.1: Dual graph of minimum resolution for Example 2.1. The labels are the negatives of
the self-intersection numbers.
most 3. The curve C is adjacent to exactly one of the exceptional curves; that curve has branching
number 3, and is the only exceptional curve with self-intersection number -1.
Exactly k + 1 exceptional curves are tips (i.e., they have branching number 1); let us number
them 0,... , k in order of their appearance (when we think of constructing X -+ X by a sequence
of blow-ups). If, starting at one of these tips, we walk along components of branching number at
most 2 as far as possible, then we trace out a maximal twig of D.
We now explain how to draw the dual graph for a cusp. To avoid a morass of notation, we trust
that the algorithm will be clear enough from one example:
Example 2.1. Let P be a cusp with three Newton pairs: (mi, ni) = (37, 53),
and (M 3 , n 3 ) = (3, 11). Compute the continued-fraction expansions:
ni1
M1ml=1+22+g
1
TM2
1 ,
=0+-,
4
n3
1
-=3+
M3
+
(m2, n2) =
(4, 1),
1.
Let us agree to abbreviate this as
ni = [1,2,3,5],
MI
n2 =[0,4],
M2
n = [31, 2].
M3
The log resolution of C then has (1 + 2 + 3 + 5) + (0 + 4) + (3 + 1 + 2) = 21 exceptional curves
lying over P. In Figure 2.1.1, the corresponding vertices of the dual graph are arranged, from left
to right, in the order of their appearance when we resolve C by repeated monoidal transformations.
Since the cusp has three Newton pairs, the resolution graph has three branching divisors, labeled
El, E 2 , and E 3 in the figure. We may think of each Ei as corresponding to the ith Newton pair
(mi, ni). Reading from left to right, we see 1 + 2 + 3 + 5 = 11 vertices up to and including El, then
0+4 = 4 vertices up to and including E 2 , and finally 3+1 + 2 = 6 vertices up to and including E3.
Let us first understand how the 11 vertices up to and including El are arranged. Each one of
them appears on one of two branches, which are drawn as the "lower" and "upper" branches; the two
branches join together at El. Looking closely, we see that these 11 vertices are arranged into units
of 1, 2, 3, and 5 - as in continued-fraction expansion of ni/mi = [1, 2, 3, 5] - which appear on
alternating branches. That is, there is a unit comprising 1 vertex (the one labeled "4") on the lower
branch, then a unit comprising 2 vertices (the ones labeled "2" and "5") on the upper branch, then a
unit comprising 3 vertices (the ones labeled "2, 2, 6") on the lower branch again, and finally a unit
comprising 5 vertices (labeled "2, 2, 2, 2, 5," including E 1 ) on the upper branch again. This process
culminates in El, which is joined by an edge to the lower branch also. (The numerical labels in the
figure are self-intersection numbers, which can be ignored for the time being.)
Thus, the rule is to take the four numbers 1, 2, 3, 5 from the continued-fraction expansion of
ni/mi = [1, 2,3, 5], make four units with the corresponding numbers of vertices in each, distribute
17
the units alternately among the lower and upper branches (starting with the lower branch), and
finally connect the two branches together at El.
Now let us understand how the next 4 vertices, up to and including E2, are arranged. They, too,
can be thought of as appearing on two branches: a "lower" one connected to El, and an "upper" one
created anew from a tip of the tree. Recall the continued fraction expansion n 2 /m2 = [0, 4]. The
rules above suggest that we should take the two numbers 0, 4 from this continued fraction expansion,
make two units with the corresponding numbers of vertices in each, distribute the units alternately
between the lower and upper branches (starting with the lower branch), and finally connect the two
branches together at E2. This is indeed the procedure. We have purposely drawn an extra-wide
space to the right of El to remind ourselves there is a unit of 0 vertices there.
Finally, let us how understand the last 6 vertices, up to and including E3, are arranged, but
there should be no surprises here. These 6 vertices also appear on two branches: a "lower" one
connected to E2, and an "upper" one created anew from a tip of the tree. In accordance with the
continued fraction expansion n 3 /m 3 = [3, 1, 2], they occur in units of 3 vertices on the lower branch,
then 1 vertex on the upper branch, and finally 2 vertices on the lower branch, ending on E3, which
is joined to the upper branch as well as the strict transform C of the curve.
In sum, the tree is structured so that the 21 vertices naturally fall into units of 1, 2, 3, 5; then
0, 4; then 3, 1, and 2 vertices. The three branching divisors have been labeled El, E2, and E3, and
each naturally belongs to the unit immediately to its left. 2 We have drawn an extra-wide space
to the right of El to illustrate the "0" occurring in the continued-fraction expansion of 2. The
graph-theoretic significance of this "0" is that the exceptional curve appearing immediately after
E occurs on a maximal twig, and is not adjacent to El. Compare this to the case of the third
Newton pair: there the continued-fraction expansion of ' begins with a positive number, so the
exceptional curve appearing immediately after E2 is adjacent to it, and not on a maximal twig.
The self-intersection numbers of these exceptional curves are all negative, and their absolute
values are shown in the figure. To compute this value for any exceptional curve, except the (-1)curve Ek at the very end, count the number of vertices between it and its right-side neighbor,
inclusive. We also have
k
k
0
2
=
2
C 2 - 26 +1 - M, -
M N
-
2.2
N.
(2.1.3)
j=1
j=1
Notation
Let us fix the notation for the rest of this chapter and the next. Let C C p2 be a rational cuspidal
curve of degree d with s cusps P1,..., Ps, and let the cusp Pi have ki Newton pairs, namely
(mni,,
ni,1), ... , (Mi,ki, ni,i).
m. and Nij = nijMij+1, so in particular Mil is the multiplicity of P on C. Let
Let Mij =
W be the semigroup of the cusp Pi, so in particular the lowest two elements of Wi are 0 and Mil.
20f course, we may choose to write the continued-fraction expansions as
-mi
= 1+
1
2+
n2 =0+
1
m2
4+T
1
3+1'
-=3+
m3
1
1+1
1
instead, in which case we may think of the three branching divisors as constituting their own units. It is hardly
coincidental that there are two ways of writing the end of the continued-fraction expansion, just as the vertex for the
branching divisor is adjacent to two vertices on its left.
18
Recall that the 6-invariant for the cusp Pi is
1/
k
(Mi - 1)(Ni -1)
= 2
+ Z(Mj - 1)Nj,
j=2
so
SoS
(d - 1)(d - 2)
(2.2.1)
i=1
Following [1], let
Ri(a)= #Wi n [0, a),
and for functions
f, g
: N
-
a E N,
N let * denote the convolution product in the min-plus semiring:
(f o g)(c) = min (f(a) + g(b)).
a+b=c
Let V -7+ P2 be the minimal log resolution of (p2, C). Let C be the proper transform 7r.'(C),
and let D be the sum of all irreducible components of r-'(C). Let the ki branching divisors lying
over the Pi be Ej, 1 , ... , Ei,ki, so that Ei,ki is a (-1)-curve meeting C.
Let , be the Kodaira dimension of KV + D, or in other words the logarithmic Kodaira dimension
of the open surface V\D = IP2 \C. If n = 2, let Kv + D = (Kv + D)+ + (Kv + D)- denote the
Zariski decomposition of KV
2.3
+ D.
The semigroup distribution property
By using the methods of Heegaard-Floer homology, Borodzik-Livingston [1] produced a series of
constraints on rational cuspidal curves, known as the "semigroup distribution property":
Fact 2.2 ([1], Theorem 6.5). For natural numbers j < d - 1, we have
(Ri
,) (d + ) -(j + 1)(j + 2)
2
We should note that, in the paper they state this for j ( d - 2, but in fact the same equation
holds follows for j ( d - 1 as well, simply by the genus equation:
S
(d - 2)d +1 > (d - 1)(d - 2) =Z (max(N\Wi) + 1),
so for all
j > d - 2 we have
(R1 o ...
o R,)(jd + 1) = (Ri o ...o R,)((j - 1)d + 1) + d.
follows from the "semigroup density property":
-o Rq)(jd+ 1) > (j+)(j+2)
2
+ x, < (+ j+2, there exists a curve in P 2 of degree j
for any x1,..., x E Nsuch that x, +whose local intersection multiplicity with C at the ith cusp is at least the xi-th non-zero element
of the semigroup Wi, and Bezout's theorem ensures that these local intersection multiplicities sum
to at most jd. The deeper result of Fact 2.2 is the < direction. The proof involves calculating the
"d-invariant" or "correction term" for the chain complex CF1 associated to the boundary ON of a
The inequality (R i o
19
tubular neighborhood N of C in CP 2 , together with Spinc structures on ON that extend over the
complement CP 2 \N, which is a rational homology ball.
At any rate, the combinatorial implications of Fact 2.2 are quite intricate, and we have relegated
the details of these calculations to Chapter 4. In Definition 4.1, we introduce a notion of a weakly
triangulating sequence (4;
, - - - 'k), which is a necessary condition for a rational cuspidal
curve of degree d and a single cusp with Newton pairs (mi, ni),..., (Mk, nk) to satisfy the purely
combinatorial constraints imposed by Fact 2.2. This notion is useful in cases of more than one cusp,
also: if a rational cuspidal curve C of degree d has one cusp with Newton pairs (mi, ni),... , (mk, nk),
and mi+1 ..
-Mk exceeds the multiplicity all other cusps of C, then (mdmk; . -- , ) must also
be weakly triangulating.
As a direct consequence of Proposition 4.25, we recover the classification in [4] of unicuspidal
rational curves whose cusp has one Newton pair:
Theorem 2.3. Let C C P2 be a unicuspidal rational curve of degree d whose cusp has only one
Newton pair (m, n). Then one of six cases holds. (Here F denotes the ith Fibonacci number; F = 0,
F 1 = 1, and Fk+1 = Fk + Fk_1, for all k E Z.)
1. For some k > 3, we have d = F2k-1 and (m,n) = (F2k-3, F2k+l).
consecutive odd-indexed Fibonacci numbers with 2 ; m < d < n.
That is to say, m,d,n are
2. For some k > 2, we have d = F2k-1F2k+1 = Fk + 1 and (m,n) = (F2,
3. For some n
4.
F2k+1)-
3, we have d = n and (m, n) = (n - 1, n).
For some m >2, we have d = 2m and (m,n) = (m,4m - 1).
5. We have d = 8 and (m, n) = (3, 22).
6. We have d = 16 and (m,n) = (6,43).
We also derive constraints on unicuspidal rational curves whose cusp has two Newton pair,
namely Theorem 1.1 from the Introduction, which is a direct consequence of Theorem 4.26.
20
Chapter 3
The Coolidge-Nagata problem
In this chapter we discuss some results on the Coolidge-Nagata problem (Conjecture 1.2). As we
mentioned in the Introduction, Karol Palka has mentioned in private correspondence that he and
Mariusz Koras have proved the full conjecture, but that paper seems not to be available at the time
of this writing. However, Palka [13, 14] has released some partial results, see Fact 3.5 below.
3.1
Conditions for rectifiability
In this section, we survey the previously known results on rational cuspidal curves that will be useful
to us. The main tool for proving rectifiability of plane rational curves is the following theorem:
Fact 3.1 (Coolidge [2], Kumar-Murthy [8]). A rational cuspidal curve C C p 2 is rectifiable if and
only if
ho(V, 2Kv +
)
0.
Coolidge [2, Book IV, page 396] defined a "special adjoint" to C of index i > 1 to be a curve in the
linear series Ji K + Cj. He showed that the property of having special adjoints is preserved under
Cremona transformations, and that a singular curve with no special adjoints can be transformed into
one with milder singularities. Since the straight line has no special adjoints, it follows that a rational
curve is rectifiable if and only if it has no special adjoints. An argument with Riemann-Roch [8,
Corollary 2.4] then yields Fact 3.1.
Actually, to be slightly less historically inaccurate, we should mention that Coolidge, writing in
2
, without explicit reference to V. The point is that, given a divisor L on
the 1930's, worked on 1P
2
p of degree
degir,(iKv + C) = -3i + deg C,
one can determine whether L is the push-forward (under 7r) of some effective divisor L - iKv+O on
V: such L exists if and only if, for every point P of multiplicity m on C (possibly an "infinitely near
point," as the classical algebraic geometers called valuations), L passes through P with multiplicity
at least m - i. Strictly speaking, it is L, rather than L, that Coolidge calls the "special adjoint,"
but we may use the term more liberally.
The rational cuspidal curves C for which r, < 2 have been classified [5], and the Coolidge-Nagata
conjecture has been verified for them:
Fact 3.2 ([5], [17]). If , < 2, then C is rectifiable. Moreover, C has at most 2 cusps.
If K = 2, then the Zariski decomposition of Kv+D = (Kv+D)++(Kv+D)21
can be computed:
Fact 3.3 ([16], [13]). Call a maximal twig of D "admissible" if its components have self-intersection
at most -2.
If r, = 2, then (KV + D)+ has zero intersection with all components of admissible
maximal twigs of D, and (KV + D)- is supported on the union of these admissible maximal twigs.
Note that C forms a maximal twig of D if and only if s = 1. Since the intersection form on V is
negative-definite when restricted to the linear span of the components of admissible maximal twigs,
Fact 3.3 specifies the Zariski decomposition of KV + D exactly. Thus, we are able to compute the
volume of Ky + D. Our intention is to exploit the following:
Fact 3.4 (Logarithmic Bogomolov-Miyaoka-Yau inequality, [7]). If K = 2, then
vol(Kv + D) < 3 - e(V\D) = 3,
where e denotes the topological Euler characteristic.
By exploiting the above facts, Palka [13, 14], proved the following:
Fact 3.5 ([13]).
The curve C is rectifiable if s + k1 +
+ k, > 7, and in particularifs
3.
Matsuoka and Sakai [9] proved the following:
Fact 3.6 ([9]). We have
d < 3maxMii.
In particular, we immediately have the following:
Fact 3.7. The curve C is rectifiable if d < 6.
Proof. Suppose that, in accordance with Fact 3.1, there were a curve L
-
2Kv + C. Then
d - 6 = deg C + 2 degK2 = degTr, (0 + 2Kv) = degr*L > 0,
so d = 6. Then some cusp Pi has multiplicity Mil > d/3 = 2 by Fact 3.6. From Coolidge's original
point of view (of working on P2 ), the fact that L is an adjoint of index 2 to C implies that 7r.L must
pass through Pi with multiplicity at least Mil - 2 > 0. But deg 7rL = 0, so this is impossible. E
Orevkov [12] sharpened the Matsuoka-Sakai inequality to
d < 0 2 (1 + max Mi) +
where p is the golden ratio, 1+2.
Fact 3.8 ([12]).
1
,
Moreover, Orevkov showed the following:
If K = 2, then
d < p2(1 + max Mii) -!
22
3.2
An effective divisor and a volume computation
We begin with a simple observation.
Lemma 3.9. Fix 1 < i < s, and let x be the unique integer such that 0 < 2x < nki and
mkjx =±l
(mod nki).
Consider the longest chain of (-2)-curves in D that meets the (-1)-curve Ei,k, and lies over the
cusp Pj; let ei be the length of this chain. Then
M
if nk
- 1
IW
1,
=
if nki ) 3 and mk
nk - 2
=
1
-1
(mod nk ),
otherwise.
Proof. For simplicity, let us drop the subscripts i and ki. Express IL
m as a continued fraction
n
n-=ao
+
1
1
=[aO, .. .,a,]
with ar > 1. (We require ao, al, ... , ar E N and al, ... ,ar-1 > 0.) The algorithm described in
Example 2.1 shows that f = ar - 1. The theory of continued fractions tells us that
[ar, ... , ao] =-
n
provided we interpret 1/0 as 00 and 1/oc as 0. Let us consider the three cases described in the
lemma:
Case 1: n = 1. Then m = [0, m], so f = m - 1, which agrees with the lemma.
Case 2: n > 3 and m = np - 1. Then
-
n
m
[1, m] = [1, n - 1]
[0,p-1,1,n-1]
if p = 1,
ifp >2,
so f = n - 2, and the lemma again yields the correct result.
Case 3: In all remaining cases, it remains to prove that ar = [ar,... , ao]]. This equation
usually holds, with two exceptions (if we keep in mind that m > 2 and gcd(m, n) = 1). The first
exception occurs when r = 1 and ao E {0, 1}, for then [ai, 0] = oo and [ai, 1] = a, + 1. But then
n E {1, m + 1}, and we should be in Case 1 or 2 above. The second exception occurs when r = 3,
ao = 0, and a2 = 1, for then [a 3 , 1, a,, 0] = a3 + 1. But then
n
[0, a1,1, a3]
m-=
M
=
a3 +1
aja3 + a,+ a3l
l
so we should be in Case 2.
nk, and mkjX - ±1 (mod mk)
With Lemma 3.9 in mind, and with 0 < 2x
define
E:=
Mk
if nki
1,
nk - 1
if nk
3 and mk
-1
as above, let us
(mod nTk),
otherwise.
Then Ej > 2, and there is a length Ej - 1 chain of (-2)-curves in D that meets Ei,ki. A few values
of Ei are shown in Table 3.1.
We now have the following:
23
2
1
2
3
4
5
(mi,k , ni,ki mod mi,ki)
2
3
4
5
6
3 14 15
2
2
3
2
4
5
Table 3.1: Values of ei
Lemma 3.10. If C is not rectifiable, then in V there is an effective divisor linearly equivalent to
L := 2Kv + C-
EiEi,ki.
i=1
Proof. Fact 3.1 guarantees that there is an effective divisor
L' ~ 2Kv +
.
Our aim is to prove that Ei,ki is a component of L' appearing with coefficient at least ei. We
can consider the various cusps separately, so fix a particular i. Let F 1 , ... , F be the chain of
so that F
F_ 1 = -2 and Fe, = Ei
wr-exceptional divisors in D such that F1 = Suppose that Fj occurs with coefficient aj > 0 in L'. Let a-1 = 0, for convenience. Then L' . F 1 =
... = L' - Fi_1 = 0 and L' - Fe = 2(-1) + 1 = -1, so for each j
L'
2aj - aj-1 - aj+1
a3 - aj_1 - 1
ajFj
-
j=1
if 1
( j ('<
-
1,
if j = Ei.
Then
1
a
-
ai_i < asi1
-
aEi-2 <,
< a2
-
al < al,
so by induction on j we see that aj > j for all j. In particular,
a i > Ei,
F-1
as desired.
Suppose we had a counterexample to the Coolidge-Nagata conjecture, so there is an effective
divisor
L ~ 2KV +
-
Ei Ei,ki.
i=1
Our goal is to exploit the fact that L must have non-negative intersection product with the nef
divisor (Kv + D)+. Palka [131 used the same strategy, but without the EI eiEi,ki term.
Lemma 3.11. If a rational cuspidal curve C is not rectifiable, then
S
6
2
_4 _ 2s.
< _4 _
i=1
24
Proof. By assumption the effective divisor L exists. Note that
deg rL = d - 6 < deg 7r.C,
so C cannot be a component of L. It follows that
0 < L. C= 2(Kv + C)-
CC2
2
_
_4 _
i=1
as claimed.
0
Thus, if s = 1, then C appears with positive coefficient in (Kv + D)-, according to Fact 3.3.
Let us make some relevant calculations. First note that (Kv + D)+ has zero intersection product
with all 7r-exceptional divisors that are not branching in the dual graph D. One can check that, if
s > 2, then
(KV + D)+ . C
if j = 1,
if 2 < j < k2 .
Mi
- nil
(Kv + D)+ -
s - 2,
-
Egg
=
I
+ D)- with coefficient 1/(-C2)
When s = 1, the only difference is that C appears in (Kv
1
i
1
(Kv + D)+ - Ejj =
if 1 = j < k,
1
-_
if 1 < j < ki,
11'
-nl1l
1 -l
if 1 = j = ki,
if 1 < j = ki.
-02
-02
It follows that
k,~
(KV +D)+ -
Kv +D-
L)= (Kv + D)+
2 Ei,k + ZEij
2 +
j=1
In light of Fact 3.4, one of the most basic inequalities at our disposal is this:
Lemma 3.12. If C is not rectifiable, then
-y < vol(KV + D) + min
,1
}
, 41,
min {3+
i=1
where
+ (Kv + D)+ -
i Ei k +
E 3 j.
/
j=1
If s > 2, then, since Ei > 2, we have
3-e
-
_-I +
2+a
ki+
11
I)
2 ni
5 -2
>
J2
nil
2
-Z_
2
mil1
nil
)
ki-l
-y
L.j=l
1
2 22
mi,
1
Mi,k
if k= 1,
if ki
2,
-
61
mil
22
_
1
_
_
mni,k1
25
_
_
> 0, So
2
Mi k
2
±
3
if ki
1,
if ki
2.
If s = 1, then, since Ei > 2, we have
- 2+Ei
3
2
=
tki + 1
nj,1
ml,1
1
2
m
2
-
1
3=1
ni,1
2
5
1 +-1-)
+
(-I
2
m
if kl=1
-C2
1
2i
ifk
1
2,
-02
(Mi,kl
_2
ki + q
+
1
2
mi~j
if ki =1,
-
-21
i-
-
2
ki
2
2.
Proof. The formulas for yj follow from our previous calculations. The inequality
,1
vol(Kv + D) + min
i=1
is nothing more than a restatement of the fact that
(Kv +D)+ - Kv +D- -)E
At this point, we are ready to deal with the case of four or more cusps, so the reader may skip
ahead to Lemma 3.15 if desired.
From 2.1.3 and the genus equation, we find
-02 =Z
Mi,jNij - d,
i~j
s
ki
Mi, +
- 1
N
(3.2.1)
+ 2 - 3d.
j=1
i=1
For future reference, let us calculate the volume of KV + D.
Lemma 3.13. Assume that r,= 2. If s > 2, then
vol(Kv + D) := ((Kv + D)+) 2
If s = 1 and
U2
=
7 - 3d +
- + M+1
-
i=1
rni +
RZ A
N
-
_
j=1
< _2, then
7-3d-1+M 1 1-
vol(Kv+D) =
Ni
l- + -02.
mlL
j=l
kij
C2
s
j
mij
2. One may check that
Proof. First suppose s
KV = r*Kp2+
nil
k'
Mi+2
E
j+
i=1 j=1
i
jNi2++
-1)
Eij
Mij+1
+ (non-branching exceptional divisors).
26
So
vol(Kv+D)=(Kv+D)+-(Kv+D)-(Kv+D)+-(Kv+D)(Kv + D)+ -
-
Now
[
=
0)
(d - 3)
Ey
i~
i=1 j=1
(Kv + D)+ - (7r*Kp2 +
)
++Ni' +
Mi,1I+Ni,+
+
wr*Kp2 +
0.
(-3) + (s - 2) = 7 - 3d + s,
and
(Kv + D)+.Z
i=1 j=1
(Mi, +
Ni,1 + Ni,2 + -- - + Ni)
Mil +Ni
-mki
M il
nil
M[
iS
-1
n1
MI2
-1
-t
j
-+ Nii)
(Mi,
Mij+1
1
j=1
+ L (Mi, + Ni, 1 +
1+ M
(I
1
(I1
E
+
+
-
Mi )
Mij+1
m1)
Mij)-
(Mj+1
j=
j
The sums over
E
1
Nif
(M +1
1 <j',<j <ki
Mi
telescope, so the expression equals
[-Ml-
1 + Mil
(1
-
I
+
Nij,
)
I -
Therefore,
-M
vol(Kv + D) = 7- 3d+s +
-
ni
i=1
(Nj -
1 + Mil - 1+
-]
=1
proving the lemma when s > 2.
When s = 1, the only difference is that C appears in (Kv + D)- with coefficient 1/(-0
We should therefore subtract
(Kv+D)-O
_
(Kv+0)0+E1,ki -0
-02
_02
2
) > 0.
-2+1
_02
from the formula of the s > 2 case. That is indeed what the lemma claims.
The following Corollary is sometimes useful in conjunction with the formula (3.2.1) for -02:
Corollary 3.14. We have (under our standing assumption that r,= 2)
2+
_C2 <,1
4=
( ni,1
k1
+ L
2
27
-
if s = 1
if s > 2.
If C is not rectifiable, then in fact
Sk
(mi +
N
n,j
+ L+
ni,
1
1
L
_
if S=1
-2
<-c
if s > 2
6
= mij
(_
si
i1
E__j+E
8
<-C(
ni,1
i=1
+
)
'3
mi
(2
-2
2
Proof. The first follows from 3.4 and Lemma 3.12, which says
m
n,
i
5+
+vol(Kv + D) _
n,
+mi
-C2
j)
i,
ni,1
-1
-c2
if s8
1
if s >2.
5
If C is not rectifiable, then by Lemma 3.12, the right-hand side of the above equation is at least
S
ki
ni,
3.3
if
L,+
+NO, +
""I
Mj
_
-
2
6
if s >2.
Three or more cusps
We can immediately deal with the case of at least four cusps:
Lemma 3.15. If C has s >, 4 cusps, then C is rectifiable.
Proof. Suppose that s > 4, but C is not rectifiable. We may assume d > 7, by Fact 3.7. If s > 5,
then, in the notation of Lemma 3.12,
Z
i=1
a contradiction. Likewise, if s
=
5
25
> 4,
'Yi > 6 s > 6 >~
4 but any ki > 2, then
3
2
5
2
67 + +
>
25
=6 > 4,
a contradiction. Therefore, we must have s = 4 and k1 =
let just write (mi, ni) for (mi, 1 , ni,). Then yi> - 2
Mi
+
ni
4 = 1. For simplicity of notation,
so we have
=
.
,
>3.
On the other hand, Lemma 3.6 says max mi > 3. Without loss of generality, assume mi > 3. Since
2
+
it must be that ml = 3 and m2
1
+2
+ 2
=
m3
+
=
<
+
+3
<(I
+
+
<3,
m4 = 2. As
+
+
28
+3
+
< 3
if s1
if s
2.
we must have ni < 5 and n2 = n3 = n4 = 3. Then the sum of the 3-invariants of the four cusps is,
by 2.2.1,
(mi - 1)(ni - 1)<,4+1+1+1
2<4±+1 + 1+ 1= 7,
7
4
i=1
which contradicts the fact that C has arithmetic genus
(d-
1)(d -2)
2
2
>
- 15.
The case of three cusps is a bit more involved, and uses both Lemma 3.13 and, in one place,
Fact 2.2.
Lemma 3.16. If C has s = 3 cusps, then C is rectifiable.
Proof. Suppose that s = 3, but C is not rectifiable. Because
5
6
4
2
2
3
5
6
(
2
2
2
3
we contradict Lemma 3.12 if any cusp has ki > 4, or if there are at least two cusps with ki > 2.
Therefore, we may assume without loss of generality that ki = k2 = 1, and k 3 < 3. Our goal now
is to find an upper bound on d.
For simplicity of notation, let us write (mi, ni) for (mii, ni,i) when ki = 1, and let us write
(pj, vj) for (m3,j, n3,j).
If k3 = 1, then since
15
5
1
+264
2 4 -2-5
2
we must have mi < 23 for all i, so Fact 3.8 implies that
d < 24 2 < 63.
If k3 = 2, then since
i=1
2
1 (22 mu 3
/2
2
m2
1
1
2
7
Vi1
v
P2
6
23
it is easy to see that we must have
1
1
1
mi
ni
2
which implies
m1, m2
<3,
It2
< max{11
2,
3 - 3 , 2 - 6}=
Fact 3.8 implies that
2
d < 232 -29
1
f5
< 60.
22.
If k 3 = 3, then since
5
3
4 i=1
(2
=
5
6
2
2
n
ml
3
2
5
6
2
3
5
2
2
2
n2
m2
+
3
1
1
1
2
2
vi
Al
A2
A3
23
6
it is easy to see that we must have
(mi, n 1 )
= (M
2
, n 2 ) = (2, 3), Al = P3 = 2,
since these are lower bounds for the variables in question, and increasing any one of them would
force a contradiction, if we bear in mind that gcd(mi, ni) = 1 and vi > MI. Moreover, A2 < 3, so
Mi,1 = MI,2 = 2 < M 3 ,1 = PIP2/I3 < 12.
Fact 3.8 implies that
d < 13 p 2
V5
< 34.
So, in all three cases, we see that d < 62. Given this bound, there are finally many combinations
of Newton pairs satisfying the genus equation (2.2.1). A computer program can enumerate them
all, and, with the help of Lemma 3.13, check whether vol(Kv + D) < 3 (in accordance with Fact
3.4) and (Kv + D)+ - L > 0.
It turns out that only one combination of numerical invariants passes all these tests: d = 13,
(mi, ni) = (m2, n2) = (2,3), k 3 = 2, (pi,1vi) = (3, 4), and (A 2 , v2) = (3, 20). However, such a curve
cannot exist. Indeed, it would have semigroups
Wi = W 2={0,2, 3,4,...}
and
W3 =
{0, 9, 12, 18, 21, 24, 27, 30, 33, 36, 39, ...
so that
(RI c R 2 cR 3 )(3 - 13 + 1) = R 3 (40) = 11 > (3+1)(3+2)
2
which violates Fact 2.2 for
3.4
j
=
E
3.
One cusp
for the
Since there is only one cusp, instead of writing (mij, n 1i), we will simply write (mj, nr)
Newton pairs, and so forth.
The unicuspidal curves whose cusp has only one Newton pair were classified in [4] (see Theorem
2.3 above), and they have been verified to be rectifiable. For the sake of completeness, let us
reproduce the quick verification:
Lemma 3.17. Let C be a rational cuspidal curve of degree d with one cusp of one Newton pair
(M, n). Then C is rectifiable.
30
Proof. Formula (3.2.1) and Lemma 3.11 together imply that
m + n-3d+1
=
02>6
if C fails to be rectifiable. But if we go through the list in Theorem 2.3, we find this is never
the case. (As before, F denotes the ith Fibonacci number. We will use some identities involving
Fibonacci numbers without proof.)
1. If d = F2e-1 and (i, n) = (F2e-3,F2e+1), then we have
_02 = F2 --3 + F2e+1 -
3F 2 f- 1 + 1 = 1 < 6.
2. If d = F2e-1F2t+1 and (m, n) = (F2_ 1, F2+ 1 ), then
2 =
1 + F+
- 3F2 -F2f+1
+ 1 = -1
+ 1
0 < 6.
3. If d = n and (m, n) = (n - 1,n), then
-02 =n- 1+n-3n+1=-n<6.
4. If d = 2m and (m, n) = (m, 4m - 1), then
_02
= m + 4m - 1 - 6m +1 = -m < 6.
5. If d = 8 and (m, n) = (3, 22), then
_02=3+22-24+1=2<6.
6. If d = 16 and (m, n) = (6,43), then
_02=6+43-48+1
=2<6.
11
So in each case d is indeed rectifiable.
Let us show the same in the case of two Newton pairs:
Lemma 3.18. Let C be a rational cuspidal curve with one cusp of 2 Newton pairs. Then C is
rectifiable.
Proof. We try to follow the same strategy as in Lemma 3.17: formula (3.2.1) and Lemma 3.11
together imply that
M 1 + N 1+ N 2 - 3d + 1 = -02 ;, 6
if C fails to be rectifiable. Let us run through each of the ten cases enumerated in Theorem 1.1. As
before, Fj denotes the ith Fibonacci number, and we will use identities involving Fibonacci numbers
without proof. We will play a little fast and loose with the ordering of m, n; besides the fact that
the multiplicity of the cusp in min{m, n}M 2 rather than max{im, n}M2, the situation between m
and n is completely symmetric. So for the duration of this proof, we do not require m < n. This
allows us to avoid writing two sets of formulas in cases 1 and 2 below.
31
1. If for some i, f E Z we have d = F 2ji-F
and (m 2 , n 2 ) = (Fj_ 1 , m), then
2
±i+m, and (m, n) = e(F- 1 , FN+1 )+(F 2 -3 , F22_ 1
+2),
-02 = 0 < 6.
2. If for some i, f E Z we have d = F 2i+lm, and (m, n) = f(Fj_1 , F2+1 ) + (F 2i- 3 , Fj_1 + 2),
and (m2, n2) = (F 2i-1, fF2i- 1 + F 2 i- 5 ), then
_62 = 1 < 6.
3. If d = 8n 2 + 4n + 1, (m, n) = (4n + 1, n), and (m 2 , n 2 ) = (4n + 1, (2n + 1) 2 ), then
-02 = (4n + 1)(5n + 1) + (2n + 1)2 - 3(8n2 + 4n + 1) + 1 = n.
4. If d = nm2, (m, n) = (n - 1, n), and (m2, n2) = (m2, nm2 - 1), then
-02 = m 2 (2n - 1)
+ nm2 - 1 - 3nM2 + 1 = -M2 < 6.
5. If d = n2 + 1, (m, n) = (n - 1, n), and (M 2 , n 2 ) = (n, (n + 1)2), then
-02 = n(2n - 1) + (n + 1)2 - 3(n2 + 1) + 1 = n - 1.
6. If d = 20, (m, n) = (2, 3), and (m2, n2) = (6, 31), then
-02 = 5 - 6 + 31 - 3 - 20 + 1 = 2 < 6.
7. If d = 2mm 2 , (m, n) = (m, 4m - 1), and (M 2 , n2) = (M 2 , MM2 - 1), then
-C2 = (5m - 1)M2 + MM2 - I - 3 -2MM2 + 1 = -M2 < 6.
8. If d = 17, (m, n) = (2, 7), and (M2 , n2) = (4, 17), then
_02 = 4 .9 + 17 - 3 -17 + 1 = 3 < 6.
9. If d = F4k+2, (m,I n) =I(L,
F4+4,
and (M2,
n2)
= (3, 1), then
-02 = 3(m + n) + 1 - 3(m + n) + 1 = 2 < 6.
10. If d = 2F4k+2, (m, n)
(f4k
-02
=
F4 k+ 4 ),
and (m 2 , n 2 ) = (6,1), then
6(m + 1) + 1 - 3 - 2(m + n) + 1 = 2 < 6.
So by Lemma 3.11 we are already done in eight of the cases. In the remaining two, let us use
Corollary 3.14. Assume for the sake of contradiction that C is not rectifiable.
32
3. If d = 8n 2 + 4n + 1, (m, n) = (4n + 1, n), and (m2, n2)
MI -
~- n- +1Z J1+
_+
mj
ni
)4+,
nk
1+E
+k+
Mk
n -I
4n+ I
1
4+E
4n+
+2+
i
5;
R
6
-6
2n
33
2
3
2
+2
4(4n + 1)
2
5
4n+1 In
3
3
2
47 > n
25
25
6
2(-C2)
7
+n+-4(4n + 1)
4
3
3
=
then note that
1)2),
6
-
2
(2n+1)2 -2
~+
4+-4
5
2+e
2
(4n + 1, (2n +
=
3
n
-6
-0
contradicting 3.14.
5. If d = n2 + 1, (m, n) = (n - 1, n), and
k-i
m-1
+
ni
nj-
1
+
+=1
(n, (n + 1)2), then note that
(m2, n2) =
nk-
21+E
.k+
+
Mk
2
(n+1)2 -1
>n-22
3
n-
3
n-
2
n
3
2
3
2
n
n-1
3
n 2
2
3
7
6 =7+
5
-6
6
2
+
=1+1--+n+2+2+
-6
2(-C2)
+2+3-
n
n
=n+----
4+E
-
2(n - 1)
1
-6
2
n-1=-0,
again contradicting 3.14.
E
This completes the proof.
In the cases of k > 3 cusps, we are able to prove a partial result:
Lemma 3.19. Let C be a rational cuspidal curve with one cusp of k > 3 Newton pairs (ml, n1),..., (Mnk, n7k).
Then -a2 < m2 + 2, and if mk < 4, then C is rectifiable.
Proof. The semigroup W' generated by the k integers
W1
mk
M1
W2
= m1m
mk
mk1,
2 ...
_N
mk
1
-
= n m2
- -2mk-1,
mk
includes all but
Wk
...
mk
k-i
I
(M
)
(N,
1)Ni]
Mk
1I
(Let us not confuse W' with the semigroup W, which is generated by
non-negative integers.
w 1 , ... , Wk+1.) For every positive integer j < d - 2, the interval [0, jd] contains
of Mk, so, in light of the above observation, Fact 2.2 says
1 +
d
1
M1
Ni)
Mk
-i
+ E
i=2
33
A
M
m
1)Ni
Mk--I
[_J
<(j + 1)(j + 2)
2
multiples
Now plug in for j the integer closest to Mk
d - 2', to find that
1 [(Mi
2
mnk
-1
(Ni
)Mk
-1
)
k-i
M
-
i=2
I
1)N)
M- -
1 1'
Lk
(j + 1)(j + 2)
jd
mk
d
1
2
1
2
(j+1)(j+2)
2
Mk
d
Mk
2
2
3
2
2
3
28
my
2
-
1
d
3
2
Mk
2
2
1
--.
Multiply both sides by 2m2 to find
k-1
2
(M 1 - mk)(N1 - ink) + Z(M - mk)Ni > d - 3mnkd.
i=2
Subtract this from the genus equation, which says
k
2
(Mi- 1)(Ni - 1) + Z(Mi - 1)Nj = d - 3d+ 2,
i=2
to find that
+ Nk- mk- 1) < 3(Mk - 1)d + 2.
(mk- 1)(M + N + N2 +
Because both sides are integers, we have in fact
+Nk
(mk - 1)(M1+ Ni + N2 + -
-mk-
1) < 3(mk - 1)d+ 1,
and therefore, by equation (3.2.1), we have
C 2 = 3d -1-(M1+N1+
N 2 +---+
1
Nk) > -Mk - 2 mk-
1
If Mk E {2, 3}, then
02 > -2 -
2-
-
1
= -3 - 2 - 2 = -5,
2
so C is rectifiable, by Lemma 3.11. If mk = 4, then note that e > 3 (see Table 3.1), but 02 y
-4 - 2 - } = -6}, so again C is rectifiable, by Lemma 3.11.
3.5
Two cusps, each with only one Newton pair
Lemma 3.20. Let C be a rational cuspidal curve with two cusps, each with one Newton pair. Then
C is rectifiable.
Proof. Suppose C is such a curve, but C is not rectifiable. In light of Fact 3.2, we may assume
K = 2. Recall that d is the degree of C. To simplify the notation, let the Newton pairs be
(m, n) = (mi,1, ni,1 ) and (m', n') = (m2,1, n2,i), with m > m'. Given any combination of concrete
numbers for d, (m, n) and (m', n'), we may try to prove that C cannot exist, by using the following
tools:
34
"
The genus equation, (d - 1)(d - 2) = (m - 1)(n - 1)
+ (in' - 1)(n' - 1).
" The BMY equality (Fact 3.4) with the aid of the volume formula in Lemma 3.13.
" Fact 2.2, for 0 ( j ( d - 2.
If that fails, we may try to prove that C must be rectifiable. By Lemma 3.9, it suffices to show that
no divisor
L ~ 2K1 + C
Z
-
Ei,ki
i=1
can be effective. Here we have the following tools (assuming C is not rectifiable):
" L - C > 0, since C is not a component of L.
" L - (Kv + D)+ ) 0, since (Kv + D)+ is by definition nef.
All the above tests are mechanical. Recall, too, that d > 7 by Fact 3.7. We wrote a computer
program to test all the finitely many combinations of numerical invariants with 7 < d < 250.
(Even the genus equation alone shows that the number of combinations is finite, though it is hardly
efficient to enumerate all the combinations from that angle. In practice, Lemma 3.12 serves as a
useful first line of defense, cutting down the search space to something manageable.) Among all
those possibilities, only one combination of invariants is not automatically eliminated by the tools
we listed above:
d = 8,
(m, n) = (3, 16),
(m', n') = (2,13).
We prove in Lemma 3.21 that no curve C with these invariants can exist.
Having checked all curves of degrees up to 250, we may assume that d > 250 from now on. By
Fact 3.8, we have
m = max{m, m'} > 95.
By Lemma 3.12, we have
71+72=
(5
2
2
(2
96
97
(5
2
2'
+ (5 -+ m'
n'
2
4,
which means either n' = 2 or (M', n') E {(3, 4), (3, 5)}. But if (m', n') = (3, 4), then we can calculate
E2 = 3, so Lemma 3.12 in fact implies
5
2
2
96
2
97
1
5
1
)
<4
which is (just barely) false. So we are left with two cases: either m' = 2 , or (m', n') = (3, 5).
Before diving into a morass of technical details, let us informally describe our strategy for dealing
with the remaining cases. We have at our disposal the following formulas, from equation (2.2.1)
and Lemmas 3.12 and 3.13:
(m -1)(n -1) +(m' -1)(n' -1)
n'
M'
n
m
+m'+n'- m+nM
in
M
n
2 (-+-+
m
n
m
+35
n
=
(d-1)(d-2)
=
3d+vol(Kv+D)-5
4-vol(Kv+D).
(3.5.1)
Pretend for a moment that m', n', and vol(Kv + D) are small, in comparison to m, n, and d, which
are comparable in size. Then the first two equations above imply that
d3 - 3d
mn-m-n
m+ n
3d,
which, taken together show that
O2
m
and in particular
(3.5.2)
+- "-~~7.
m
n
This observation is the starting point of our proof. Let
2'
2'1
f := (m + n-3d) + m' +
then e must be an integer (since either m' = 2, in which case n' must be odd, or else n' =5). Then
n, and show that it is far from being an
we will try to estimate the difference between f and !n + m'
(3.5.2).
contradicts
integer, which
Now let us continue with the formal proof.
Case 1: m' = 2. Then from (3.5.1) we have
7
e =m+n-3d+ +
2
-
n
+ -+
M
+vol(Kv + D) - 7
2
n
and
Ti'
m+n+ -=3d+e-7
7
2
2
2
- 2e)
(18
(m - 3)(n - 3) = d - 9d+
2
1 Im +n
+n-2
m-2
First we claim that e
2
m
n
n
m
n
7. Indeed, Fact 3.8 says that
d < W2 (1+ m)
1
,
so if f > 8, then
Sd29d
+2
<7in-3
m- 3
the last inequality involves some tedious calculation and the fact that d > 250. But then
Ti-34
1 <m
n
2
1
2
n
m
n'
7
3'
2
contradicting the assumption that f > 8.
From the fact that d > 250 and f 7, we easily see that
_ 9d+ 18 - 2f > d - 4.6.
-d2
Let
d
n
M
d
36
If a > 2, then we would have
Vd2 - 9d+ 18-2
d-4.6
m-3
d-3a
>
m-3
and consequently
d2 -9d+18-2f
(m - 3) 2
n-3
m- 3
so that
m- 3
n-3
-3
2
2
-2
m-3
Since
n- 2
M
n
>
m-3
M
m-3
n7-3
+n -3)
and m > 95, we conclude that
m- 2
n - 2 > 0.96(a
2
m
n
+ a-2).
(3.5.3)
On the other hand, we always have
vd 2 - 9d + 18 - 2f
d - 4.6
n - 3
n- 3
4.6-3/3
n- 3
d - 3/03
n- 3
1
4.60-3
>
(1
4.6)
1
d
#'
(1
4.6 )2
d-3
3
(n - 3)
1
4.63
-
and consequently
M
n
m- 3
d2 - 9d + 18 - 2f
n - 3
(n - 3)2
d
1
o2l
so that, since d > 250,
m + n
n
m
2
/
<
(0.9816
0.9816
)2
#
< 1.04
/2
.
(3.5.4)
+ 0.97. -2.
More precisely, we could note that
n
m
- -3
n - 3
3(n - m)
n(n - 3)
m - 3
n- 3
3(0 - 1/a) >
2
1
0 d
9.2 - 3+
1/a
d
1
) /2'
so that in fact
9.2 - 30 + 1/a
m
n
- + - <
n
d
M
From Fact 2.2, applied to
/
o2+
j
E {1, 2, 5}, we find the following:
nf
min{2m, n}, min m5m
2 ' 2
In terms of oz and
/,
9.2 - 3/ + 1/a
d
I
'
mn {13m 2n
we have the following four possibilities:
37
5
' 5
<d.
; .
1
#32
(3.5.5)
* #3< 1.
In this case, inequality (3.5.4) says that
+ - < 2.01.
n
m
Since evidently
m-2
n-2
(1
n
m
(n
1)
3
> -,
2
we must have f = 2 and n' = 3. But then
m + n = 3d - 3,
which is patently absurd when m < n < d and d > 250.
* /3
< 2 < a. In this case, inequalities (3.5.3) and (3.5.4) imply that
4.08 < m -2
n -2
so we must have
e=
m+
- < 4.4025,
m
n
m
n
4 and n' < 21. But then
m + n > 3d - 10,
which is patently absurd when m < d/2, n < 2d and d > 250.
* /3
< j < a. In this case, inequality (3.5.3) says that
- 2 +n - 2
n
m
6.1536 <
so f > 6. On the other hand, inequality (3.5.5) says that
n
m
- + -
m
n
o2+
d
9.2-3,3+1/a
Note that the term (i
1
9.2 - 3/ + 1/a)
1
d
#32
and consequently
So in fact we must have f
1
9.2 - 3 + 1/a
9.2 - 3 + 1/a'
<1-
(1
d
)
02-
is decreasing in a and d, and increasing in P, so
1
9.2 - 3(5/2) + 2/5>
> 0.15866,
(5/2)2
251
(1
m
n
- + - < 6.462.
m
n
=
6 and n' < 53. But then
m + n > 3d - 24,
which is just barely absurd when m < 2d/5, n < 5d/2, and d > 250.
e
L-
a. In this case, inequality 3.5.3 says
m- 2
n
so f ; 7. Now by Fact 2.2 for
cases:
j
=
+
n- 2
->
m
6.63,
3, we have min{8m, m + n1} < 3d. Let us consider two
38
- If n' = 3, then we have
m+n =
3d+2
d 2 -9d+4
d 2 +1.
(m-3)(n-3) =
mn =
so it's of course impossible to have 3d > m
jid and n> Ll d + 2.
3d> 8m, so m
d 2 d2+1=8n(
1m
+ n. It must, consequently, be the case that
Then
3 d21 8 d
=
2)
d2+
64
4
This is impossible for d > 250.
- If n' > 5, then !
+ - > L. Then note that
m + n - 1 < 3d,
but
(m - 3)(n - 3) = d 2 - 9d + 4,
so some arithmetic shows
2(m + n)+ 17 > m
2
+ n2 - 7mn
Divide both sides by mn:
2(m + n) + 17 >rm
mn
n
1
> 107
-,
n
10
m
and
(m - 20)(n - 20)
570,
which is evidently absurd when n > m > 95.
Case 2: (m', n') = (3, 5). This case is easier:
7
247
M
n
2
1 :=m+n+7-3d= -+-+L2-3n
m
and
m+n =
mn
3d+l-7
d 2 +l-14,
=
which together imply
m
-+
n
and
-1
n
- = 7+2(1- 7) -- + - +
m
n)
(M
m- 2
n
n- 2
m
1
15
39
7
24
2
+ 51 +77
mn
m
n
n
m
d.
0
If 1 ( 6, then
2(l -8)
1
1
7M
n
-+-
-l
+
2
+51+77
mn
1
+
_l+-7+-1=1+--,
120'
24 15
or in other words
mn < (16 - 21)(m + n) + (12 - 51
12 _ 1
(7-
77).
-
Now
12
- 51 - 77 < 36 - 30 - 77 = -71,
so in fact
and
1 mn
20
7
< (16-21)(m+ n)
i 17--~-l
S
m
>
n
1120
16-21
7-4-6_
77
16120
'16-2.6
480'
which is absurd when n > m > 95.
. If I>7, then
-12+51+77
7+2(l-7)
-+
(M
n
+
>I
+ 7
mn
24
Now
-l2+51+77-49+35+77=63,
s63
- >63(1-7)
mn
1--
2 2
M
n
7
+ -
24
which is absurd when n > m > 95.
El
2
Lemma 3.21. There is no rational cuspidal curve C C p of degree 8 with two cusps, the first with
one Newton pair (3,16), and the second with one Newton pair (2,13).
Proof. Of course, an assertion of this type is expressible in the first-order theory of algebraically
closed fields of characteristic 0, so it must be decidable, given sufficient computing power. More
practically, in the notation of Fact 2.2, note that
(Ri
R 2 )(2d + 1) = R1 (15) + R 2 (2) = 5 + 1 = 6 =
_(2+±1)(2
2'
+2)
since
15, 16, 18, 19, 21, 22, 24, 25, 26, 27, 28, 30, 31, 32, ... },
W,= {0, 3, 6, 9, 12,
2, 4, 6, 8, 10, 12, 13, ... }.
W 2 = {0,
By dimension-counting, there exist divisors A', B C Ip 2 of degree 2 with the following local intersection multiplicities:
(A' C)p 2 > 2,
(B C)p2 >"0.
(A' - C)p, >, 12,
(B -C)p, > 15,
40
Let A be the unique line through P with maximal (A -C)p; that is, A is "tangent" to C at P
in the sense that its proper transform continues to intersect the proper transform of C even after
blowing up P1. We have (A - C)p,
6.
We first claim that B must be a smooth conic. If B is a sum of two lines, then since we are only
[L] = 8. But
interested in maximizing (B. C)p, we might as well take B = 2A, so that (A -C)p
8 does not belong to W1, so in fact (A - C)p, > 9. This is absurd, since deg A = 1 and deg C = 8.
Thus, B is a smooth conic. Now, since (B -C)p, > 15, the proper transforms of B meet the proper
transforms of C over P 1 even after 4 blow-ups. (The reader may wish to bear in mind the resolution
graph for P1 , which we explained how to calculate in Example 2.1.) Any other smooth conic B'
could say the same for only up to 3 blow-ups, since B' can at most osculate to C to order 3; it
follows that (B' - C)p < 9. Likewise, the line A can be at most tangent to B, so
(A - C)p1 = 6.
Note, by the way, that B cannot pass through P 2 , for otherwise (B - C)p 2 would be at least 2, and
that would contradict the fact that B - C = 16.
Now we claim that A' = 2A. Indeed, A' passes through P 2 , so A' # B. If A' were smooth, then
as we noted above, (A' . C)p, could be at most 9. Therefore, A' is a sum of two lines. Now A is the
only line with (A -C)p ) 6; all other lines have intersection multiplicity at most 3 there. So in fact
A' must be 2A. It follows that A passes through P 2 .
We have arrived at a relatively concrete geometrical picture. By performing a suitable projectivelinear transformation, we may assume P = [0 : 0 : 1], P 2 = [1 : 0 : 0], A is the x-axis (Y = 0), and
B is the parabola y = x2 (that is, X 2 = YZ). Choose a parametrization [f(t) : g(t) : h(t)] :P -+ C
mapping 0 - P and oo 4 P 2 . The fact that C has a double point at P 2 and is tangent to A at
P1 , where it has a triple point, means that
f(t) = _t3 + et4 + dt5 +ct
6
+ bt7 + at 8,
(a
0)
g(t) = _t6
h(t) = - + _ti + _t2 + _t3 + _t + _t5 + _t6
where _ stands for coefficients we have not bothered to name. We may rescale (f,g, h) and t
separately to ensure that the coefficient of t 3 in f(t) and that of t6 in g(t) are both equal to 1. Since
(B - C)p, = 15,
f(t)2 - g(t)h(t)
(mod t15)
in the polynomial ring C[t], so
h(t) = (1 + et + dt 2 + ct 3 + bt 4 + at 5 ) 2
(mod t9 ).
Because of the cusp P 2 , we know that h has no t7 or t8 term, so
2da + 2cb = 2ca + b2 = 0,
(3.5.6)
and
h(t) = 1 + 2et + (2d + e2 )t 2 + (2c + 2ed)t3 + (2b + 2ec + d 2)t4
+ (2a + 2eb + 2dc)t 5 + (2ea + 2db + c2 )t 6 .
41
Now let us look at this from the perspective of t = oo and P 2 . Let s :=
s f(t) = s8 f(1), and so on, so that
, so that F(s)
8
F(s) = a + bs + cs2 + ds3 + es4 + s,
G(s) = s2,
3
H(s) = (2ea + 2db + c2 )8 2 + (2a + 2eb + 2dc)s + (2b + 2ec + d2)S4
+ (2c + 2ed)s 5 + (2d + e2 )S6 + 2es7 +s8.
The cusp P 2 is an A 12 cusp. In particular, it is not an A 2 cusp, so the coefficient of s3 in H(s) must
vanish:
(3.5.7)
2a + 2eb + 2dc = 0.
Since a = 0, the equations (3.5.6) and (3.5.7) together imply that b = 0. Therefore, we can solve
for c, d, e in terms of a, b. The important thing is to consider the power series expansions of (
and H
, which are the local affine coordinates of points on C. A little calculation shows that
H(s)
F(s)
_
7b5
8a 4 G(s)
2
4a b
F(s)
- 3ab
(G(s)'\
2
F(s)
=
b2 (4a 4 + b5 ) -.s8 + 0(s 66.).
6
4a
Thus, the fact that P 2 is not an A 6 cusp implies that the coefficient of s5 in the above expression
5
4
must vanish. Since b 0 0, this implies 4a + b = 0. But then
H(s)
F(s)
7b' - 8a 4 G(s)
4a 2 b F(s)
3ab (C(s)
F(s)
2+
9b (G(s))
2 F(s)
2
2 .7
b
S
so P 2 is in fact an A 8 cusp, not an A 12 cusp as we supposed. This contradiction shows that the
curve C cannot exist.
42
Chapter 4
Triangulating sequences
In this Chapter, we define a notion of triangulating sequences, and prove some combinatorial results
about them. The motivation, as we discussed in section 2.3, is to prove some restrictions on
unicuspidal rational curves.
We abandon the notation from previous chapters. Let N be the additive semigroup of nonnegative integers. Let ai, ... , ak (k > 1) be a finite sequence of positive real numbers such that ai
is rational whenever i < k. We may write, in lowest terms,
gcd(mi, ni)
(mi, ni E N,
Oi = n,
Mi
=
1)
for i < k. Define 91,... gk+1 by the following formulas:
k-1
91
=
7j,
j=1
92
=
aCl im,
k-1
j=1
k-1
=
gi+1
miigi + aif
mj,
(for 2 < i < k).
j=i
(An empty product is taken to be 1.)
Note that 91, ...,9k E N, and, for 1
i ( k,
k-1
gcd(gi,.. ., gi)
= f
nm
3 ,
j=i
so let W be the semigroup
k-1
Wi =JJm
1
(gi, . . . , gi) C (N, +).
j=i
Note that W 1 = N, and for 2 ( i < k,
k-1
k-i
max(N\Wi)
=
rf m )
i-1
(mi-igi - Q mi - E
j=1
(j=i
43
i'=1
k-i
nit 1
j=i'+1
mj,
and
1 + max(N\Wi)
2
We are mostly concerned with Wk, so write W for Wk when no confusion can arise.
in lowest terms, and define
In the case that ak is also rational, we may write ak =
#(N\Wi) =
Wk+1 := mk(g,. .. , 9k+1) C (N,+);
it, too, contains all but finitely many non-negative integers.
It makes no difference to the semigroups W, ... , Wk if we replace ai by aj, that is, if we swap
m 1 and ni. So we will sometimes prefer to deal with the case ni > mi, and at other times exploit
the extra flexibility of swapping ml and ni to simplify our formulas.
Definition 4.1. Given a sequence al, ...
jth triangulatingset is
,C
as above, a real number 6 > 0, and j E N, we say the
Sj(6; a,..., ak) := {(w, z) E Wk x N I W + Z9k+1
j}.
The jth triangulating condition, or triangulating condition j, for (6; al,..., ak) is the statement
that
+2
We say (6; ai, ... , ak) is a strongly triangulating sequence of length k if it satisfies the jth triangu-
lating condition for all j E N. We say (6; al,..., ak) is weakly triangulating if ak = n E Q and
(6; a 1 , ... , ak) satisfies the jth triangulating condition for all j E N such that both of the following
conditions hold:
(i) J < ik,
(ii)
6
or (j - 1)(j - 2) ( max(N\Wk+1), or Mik (j - 3) < max(N\Wk+1) - 1.
j < gk+1, or #{(w, z) E Wk x N I w + z9k+1 < Mkgk+1}l
(J+2
We will often say a sequence Z1, ... , C is strongly or weakly triangulating if there exists 6 > 0
making (6;al,...,ak) so.
Note that, if (6; a, ... , ak) is weakly triangulating and
j
is small enough to satisfy (ii) above,
then in fact
#{(w, z) E Wk x N I w + zgk+1 < j6} = #(Wk+1 n [0, jMkJl).
Lemma 4.2. (1)
If (6; a,..., ak) is weakly triangulating, then so is (m-m-1
a1,..
.,ai)
for
;a,... , ai) satisfies the jth triangulating condition for all j
satisfying the equivalent of (ii) in the definition of "weakly triangulating," that is, 6j < gi+1 or
#{(w,z) E mi ..-mk_1Wi x N I w + zgi+1 < Mi9k+1} (J2
(2) If (6; a 1 , ... , ak) is weakly triangulating but, for some j E N, we have #S(6; 1, ... ak) >
2
(j, ) then j must fail condition (ii) in the definition of "weakly triangulating."
any 1 < i < k.
In fact, (mce1
Proof. Statement (1) holds because, for any j small enough to satisfy (ii) relative to the seai) is also small enough to satisfy (i) and (ii) relative to the sequence
o,...,
quence (m 6 1ak l;
(6; ai, ... , ak), and moreover the jth triangulating conditions for (mi.-men
(6; a,...,ak)
are equivalent.
44
;ai,...,aj) and for
As for (2), for such a j, by definition, j must fail at least one of the conditions, (i) or (ii). We
have to show that in fact j must fail condition (ii). Indeed, if j were the smallest counterexample,
satisfying (ii) but failing (i), then we would have
# S J 1(0 ; a l ,., a k) <,
2
# S i (6; Ce,
'
.. , k) >
2
'
and so
#{(w, z) E Wk x N
(j
-
1)mkO < mkw + mkgk+1z ( jmk3 } > j + 2.
The coefficients mk and mkgk+1 are both integers, so mkw + mkgk+1z E N.
1)mkO, jmk3 ] contains at least j + 2 integers, so we must have
MkO
The interval ((j -
> j + 2,
which means that (i) in fact holds, contrary to our hypothesis.
D
The triangulating conditions are readily seen to boil down to some counts of lattice points.
So, in the limit as j -+ oc, we are more or less computing areas of triangles. It follows that, if
(3; al,... , ak) is strongly triangulating, then we must have
62 =
9k+1-
We are sometimes interested in weakly triangulating sequences (6;
, .. .
) for which MO E
N. The following lemma shows that we need only consider the minimum possible 3 for such questions.
Lemma 4.3. Suppose (6; -,..
-) is a weakly triangulating sequence with 2,...,L
all in
lowest terms, n1 > 2, and all mi > 2. Let 3 min be the minimal real number for which (3';
is weakly triangulating. Then 6min E Q, and if MiO E N, then 3 = 6min.
,.
'k
Proof. Every triangulating condition for (3'; -,...,
) imposes the constraint that 3' lie inside
some half-open interval depending on (-,...,
n).
The number of triangulating conditions that
(3'; I ,.,
k ) is required to satisfy (in order to be a weakly triangulating sequence) is a weakly
increasing function of 3' that jumps up only at certain rational values of 3'. This explains why
... , g)
is
6min E Q. Suppose that, according to the definition of "weakly triangulating," (6min; -,
required to satisfy the jth triangulating condition for all j < Jmax, so in particular inkmin(jmax -2) > max(N\Wk+l) - 1 or MikminJmax > Mkgk+1. In either case, Mikminjmax > max(N\W).
For all j ( jmax, we must have Sj(6; n, . ... n) = Sj (6min; ... , n),
which is to say
there cannot exist an element of Wk+1 in the interval (mkominj, mkoj]. Every integer exceeding imaxMkomin in fact belongs to Wk+1, so there cannot be any integers at all in the interval
(MkOminjmax, Mikjmax]. But if 3 > Jmin, then by assumption jmaxmkJ would itself be such an
integer. This cannot happen, so we must have 3 = 6min.
Our motivation for studying weakly triangulating sequences comes from Fact 2.2 restricting the
topological types of cusps of cuspidal rational curves in P2 . For example, suppose we have such
a curve, of degree d, with only one cusp with Newton pairs (mi, ni),... , (Mk, nk). Then Fact 2.2
n,., ) satisfies the jth triangulating condition for
(the main result in [11) says that (d;
0 < j < d - 1. In sum, the jth triangulating condition is satisfied for all non-negative integers
L) satisfies our definition of
j < d. From this it is not hard to see that in fact (;
"weakly triangulating."
Our goal is to prove necessary conditions for a sequence to be weakly triangulating. In fact,
many of the conditions we derive will also happen to be sufficient for this combinatorial question,
.',...,
45
but as our primary purpose is to constrain the possible topological types of singularities on planar
cuspidal rational curves, the sufficiency of conditions for a sequence to be weakly triangulating is
of less interest to us.
As it turns out, we will often find it useful to relax the condition that the last element of a
triangulating sequence be a real number, and tweak it by an infinitesimal E:
Definition 4.4. Define the jth triangulatingset to be
S (; a,...,
Sj (6; al,
ak + e)
:=
E)
:=
. . .E, a
We say (3; al,
tion:
ak t
-
N w + zgk+1 < j3 or (w, zgk+1) = (j, 0)),
{(w, z) E Wk x N IW + zgk+1 < j6 or (w,Zgk+1) = (0,j)}.
{(w, z) E Wk
X
) is strongly pseudo-triangulatingif it satisfies the jth triangulating condi#SJ(;a1,..., ak±t )
=
{+2
(2)
for all j E N. (One could analogously define a notion of "weakly pseudo-triangulating," but we will
not be using such a notion.)
4.1
Strongly triangulating and pseudo-triangulating sequences of
length 1
In this section we give some examples of sequences (3; a) that are strongly triangulating. In each
case, the proof is the same: we check jth triangulating condition for 0 < j ( 2 by hand, and then
induct by 3's.
Let p := 1+'1 be the golden ratio.
Proposition 4.5. The sequence (6; a) = (p2
P4 ) is strongly triangulating.
2
4
2
Proof. The jth triangulating set Sj (6; a) consists of pairs (x, y) C N with x + ', y < (p j, or in
other words
We will show #Sj(3; a) = (i+2) by induction on j. Three base cases, for 0 < j ( 2, can easily be
checked.
Suppose we are given an arbitrary j > 3, and we know the claim holds for j - 3. The crucial
point is that
Consider any (x, y) E N 2 such that 9 2X + p 2y < j, or, what is equivalent,
97 2 (x- 1) +
2
(y-1)
j - 3.
(4.1.1)
2
The induction hypothesis says that (4.1.1) has (i-2)(-) solutions with x, y > 1. Since p is
2
solutions with y = 0. Since (0, 0) is
irrational, there are p j] solutions with x = 0 and [K
also a solution, we just have to show that
i]
(j - 2)(j - 1) +
2
+
-j
(~]
PJII+ J2
(j + 1)(j + 2)
2
or in other words
f-2j] + [p 2 j] = 3j + 1.
This is true because p- 2j and cp2 j sum to exactly 3j, but neither one is an integer.
46
E
Let F_ :=
be the Fibonacci numbers, which are integers. For example,
F_1 = 1, Fo = 0, F1 = 1, F 2 = 1, F 3 = 2, F 4 = 3, F5 = 5, F6 = 8, F 7 = 13, ....
Proposition 4.6. For every k E N, the sequence (6; a)
=
(2; (kl)2)
is strongly triangu-
lating. (Note: for k = 0, this gives (1; 1).)
Proof. For notational simplicity, let a := F2k-3, b := F2k-1, c := F2k+1, and D := bc. One can
easily verify that
ac - b2 = 1,
b2 + c2 = 3D - 1,
and
a
+ c = 3b.
The proof proceeds along the same lines as that of Proposition 4.5, but is slightly more complicated.
pairs (x, y) E N2 with
We will show by induction that, for all j E N, there are exactly (j)(j+2)
b2 x + c2 y < Dj.
(4.1.2)
Three base cases, for 0 < j < 2, can easily be checked.
Suppose we made (4.1.2) into an equation:
b2 x + c2 y = bcj.
(4.1.3)
Since b and c are relatively prime, any putative solution must have c I x and b
x = cx' and y = by', we have
bx' + cy' = j
for (x', y') E N 2 . Since ac - b2
=
I y,
so if we write
1, the solutions to this equation have the form
y' = aj - bz
x/ = cz - bj,
for z E Z. In order for both x' and y' to be non-negative, we need
a,
bj <
<z
-i
c
b
Hence, the number of solutions to (4.1.3) is
aj
b
Suppose we are now given an arbitrary j > 3, and we know the triangulating condition holds
for j - 3. Consider any (x, y) E N 2 such that b2 x + C2y < Dj. Either equality (4.1.3) holds, or else
b2 x + c 2 y < Dj - 1. In the latter case, since b2 + c 2 = 3D - 1, we have
b2 (X
-
1) + c 2 (y
-
1) < D(j - 3).
(4.1.4)
So the number of solutions to (4.1.2) is equal to the sum of the number of solutions to (4.1.3) and the
number of solutions to (4.1.4). The induction hypothesis says that (4.1.4) has (-2)(J-1) solutions
with x, y > 1. And (4.1.4) has
LDj-1
b2
Dj-11±
+
c2
solutions with xy = 0, if we remember to count (0, 0) once.
47
+
Since D
L
b
2
and D
are never integers,
=
I
- 1^
+ D2- 1 +I[D
DJ-1
1
It now suffices to show that
(j + 1)(j + 2)
2
_b _
+1
Ci
)
FDj]
+
Dj~
b2
+ (j - 1)(j - 2)
2
+
c2
(
b1
F
+
Fbj~c
b icj~
c1bj+
or in other words
3j =L
+
.
This is true because in fact
3j =
b
c,
+
since a + c = 3b.
Proposition 4.7. The sequence (6; a + e)
+ e) is strongly pseudo-triangulating.
=
Proof. The jth triangulating condition says that the number of pairs (x, y) E N 2 with
(x, y) =8
or
9x + 64y < 24j
0)
is (j+2). This is easily verified for 0 < j < 2. Now suppose the triangulating condition for j - 3
holds; we wish to prove the jth triangulating condition holds as well. To that end, let N denote
the number of pairs (x, y) E N 2 with
9x + 64y < 24j,
Then
724j
N=
64
i=0
7
- 9i
(4.1.5)
x < 8.
KLE
i=0
3j - i
8
64l
For 0 < i < 7, we have 0 < i/64 < 1/8, and the fractional part of (3j - i)/8 is a multiple of 1/8, so
indeed
3j -i
8
6I4]
8
]i'
Now as i runs from 0 to 7, the fractional part of -(3j - i)/8 runs from 0/8 to 7/8 in some order,
with an average value of 7/16, so
7
N = Ei8=
i=0
7)
16
=3j+
i-i
28
=3j.
i=0
It suffices to note that (x, y) satisfies the jth triangulating condition if and only if (x, y) satisfies
(4.1.5) or x > 8 and (x - 8, y) satisfies the (j - 3)rd triangulating condition. So we are done, by
induction.
Proposition 4.8. The sequence (6; a - E) = (3; 9 - e) is strongly pseudo-triangulating.
48
Proof. The jth triangulating condition says that the number of pairs (x, y) E N2 with
(xy)
or
x + 9y < 3j
=(O,
is (j+2). This is easily verified for 0 < j < 2. Now suppose the triangulating condition for
j - 3 holds; we wish to prove the jth triangulating condition holds as well. The number of pairs
(x, 0) satisfying the jth triangulating condition is 3j, and, if y > 1, then (x, y) satisfies the jth
triangulating condition if and only if (x, y - 1) satisfies the (j - 3)rd triangulating condition. So we
E
are done, by induction.
4.2
Simplest fractions, fair approximations, and continued fractions
Lemma 4.9. Fix real numbers a,/ such that 0 < a < / and
fraction n/d of minimal denominator in the interval (a,/).
/
-
a < 1.
Then there is a unique
Proof. If not, say the fractions n/d and N/d both fit the criteria, but n < N. Then d > 1, since
1
Let x:=
.
Then xd
n
N
<- d---
n
d~p
</3
a
a
~11
xd+(d-1),so
n
n-x
< <d-1
n+1
N
d
d
In particular,
(a,/3),
d - d1E
I
E
contradicting the minimality of d.
Lemma 4.10. Fix real numbers a,/3 such that 0 < a </3 and
a- - 0
1,
0- a
1.
Let n/D be a fraction of minimal numerator in the interval (a, /), and let N/d be a fraction of
minimal denominator in (a,/). (Here we require n, D, N, d E N.) Then we claim in fact n = N
and d = D.
Proof. By minimality, we have n < N and d < D, and so
n
n
N
D
d
d
By the previous lemma, we have n = N. Taking reciprocals, the previous lemma also shows that
d = D.
Definition. Fix distinct real numbers a,,/ of the same sign such that a < /
a--' - #--1,
and
0 - a <, 1.
Then the simplest fraction in the interval (a, /) is the unique fraction p/q in that interval whose
numerator and denominator have minimal absolute values; such a fraction exists by the above
49
lemma. We similarly define the simplest fractions in half-open and closed intervals the same way;
the only caveat is that, in the case of a closed interval [a, 3], we require both a-' - 0-1 and / - a
to be strictly less than 1, so we can avoid ties for the lowest numerator or denominator.
We say a fraction n/d is a fair lower approximation to a real number a if n/d < a and n/d is
itself the simplest fraction in the half-open interval [n/d, a). We define fair upper approximations
similarly. Note that, in particular, we require In/d - a1, In/d - a~' < 1. For example, 1/2 is a fair
approximation to 2/3, and vice versa, although in general "fair approximation" is not a symmetric
notion.
Every real number a has an increasing sequence of fair lower approximations converging to a
and a decreasing sequence of fair upper approximations converging to a.
One way of computing simplest fractions and fair approximations involves continued fractions.
Let us follow the standard notation:
[ao, al, a2, ... ] :== ao +
1
al ±
1
a2+...
.
We require ai > 0 for i > 0. Recall the following standard fact:
Fact 4.11. Fix any real number a = [ao,a1,...] > 2, and define the following two sequences
recursively:
P-1 = 0,
po = 1,
q-1 = 1,
q0 = 0,
Then piqi+1 - pi+1qi = (-l)i for all i, and pi/qi
pj+1 = aipi + pi-1,
qi+1 = aiqi + qj-1.
=
[ao, ... ,ai_ 1] for i >, 1. In particular,
p1 < P2 P< 3 < qi < q2 < q3 <
and p and qj are relatively prime for all i.
Lemma 4.12. If a = [ao, a 1 , a 2 , ... ] > 1 is irrational,then its sequence of fair lower approximations
is
ao < [ao, ai,l] < [ao, ai, 2] < ...< [ao, ai, a2] < [ao, al, a2, a3, 1] < . . < [ao, al, a2, a3, a4 < ..
a2
a4 terms
terms
(*)
and its sequence of fair upper approximations is
[ao, 1] > [ao, 2] > -
> [ao, ai] > [ao, a1, a2, 1] > [ao, a1, a2, 2] > -
> [ao, al, a 2 , a 3 , >
(t)
a3 terms
al terms
Proof. One can easily verify that the sequences of inequalities in (*) and (t) are true, and their
terms converge to a. If we can show that every fair approximation to a appears in one of those
sequences, then it follows that every term is in fact a fair approximation to a.
Suppose that / E Q is a fair lower approximation to a. Then / > ao. If / E Z, then evidently
/ = ao, so let's ignore that case and assume / V Z. Every rational number has two continuedfraction expansions, so we may choose a continued-fraction expansion of / with an odd number of
terms:
# = [bo, b, . . . , b2k].
Now if bi = ai for all i < 2k, then / appears in (*), as desired. So assume there is a smallest index
i such that bi = ai. Let's consider two cases, based on the parity of i:
50
"
If i is even, then bi < ai, since 3 < a. If i = 2k, then 3 appears in (*), as desired; note
here that the case i = 0 is a bit special, but it creates no problems. If i < 2k, then since
[bi+ 1 , . . , b2k] > 1 and bi + 1 ; ai < [ai, ai+, .. .], we have
# < [bo, ... , bi, 1] = [bo,.
, bi-1, bi + 1] < a.
Since moreover # has a numerator greater than that of [bo,..., bi, 1], we conclude that /
cannot be a fair approximation to a.
" If i is odd, then bi > ai, again because / < a. Then because [bi,..., b2k] > bi > [ai, ai+,. . .],
we have
#< [boI ...,bi] < a.
Since moreover / has a denominator greater than that of [bo, b1 , ... , bi], we conclude that
cannot be a fair approximation to a.
#
The proof for fair upper approximations is similar.
Lemma 4.13. The fair approximations to a rational number a >, 1 may be found by pretending
has a continued fraction expansion of the form
a = [ao, a,, ... , ak, 1, ool
or [ao, al.... , ak + 1, 00];
the expansion with an even number of finite terms produces the fair lower approximations, while the
expansion with an odd number of finite terms produces the fair upper approximations.
For example, the fair lower approximations to a = 1 = [2, 3, oo] are
2 < [2, 3, 1] < [2, 3, 2] < [2, 3, 3] < ...< [2, 3, k] < - and the fair upper approximations to a = 1 = [2, 2, 1, oo] are
[2, 1] > [2, 2] > [2, 2,1, 1] > [2, 2,1,2] > [2,2,1,3] > ... > [2,2, 1, k] >
We omit the proof is of Lemma 4.13, which is similar to the one for 4.12, only slightly more
tedious.
be two consecutive fair approximations to a number
Corollary 4.14. Let #1 = P'q, and /2 = L
q2
Then
a > 0, with /2 lying between /1 and a (i.e., either /1 < /2 < a or a < /2 < /1).
|p1q2 - P21 I = 1, and the simplest fraction strictly between /1 and /2 is P1
.
Proof. Write a = [ao, a 1 ... ] and
/1
= [bo, ... , bi] according to Lemma 4.12 or 4.13. If bi < ai, then
[bo,..., bi_,, bi + 1] = [bo, ... , bi, 1].
/2
=
=
[bo, . . . , bi, ai+1, 1] = [bo, ... , bi, ai+l + 1].
If bi = ai, then
/2
In either case, Fact 4.11 shows that
|p1q2 -p2qlI
= 1. By Lemma 4.13, the simplest fraction between
is either [bo,... , bi, 1, 1], if bi < ai, or [bo,.. ., bi, ai+1 + 1, 1], if bi = ai. In both cases, the
fraction in question equals P
by Fact 4.11.
l
/1
and
/2
The following lemma relates simplest fractions to the triangulating conditions.
51
Lemma 4.15. Let (Vi; a) be strongly triangulating with a > 1, and let / > 1 be a real number
with 0 < |a - 1 < 1. Let q/p be the simplest fraction in the interval (a, 0) or (/3, a), depending on
whether a or / is lesser. Let 6 > 0 and j E N.
(1)
Suppose that (6;
x
+ ay < ji
/) satisfies the jth triangulatingcondition. Then
-
> x
+
y <
whenever (x, y) E N 2 and x < q and y
j6
< p.
(4.2.1)
(2) Conversely, if
, pVa&,
j < min{
-1/0z
-, p}
6
and (4.2.1) holds, then (6; 0) satisfies the jth triangulatingcondition.
Proof. (1) Assume for the sake of contradiction that (x, y) is a counterexample, with x < q and
y < p. Assume also that
x + ay
x+ y
(The other case, with inequalities going the other way, is similar.)
implies there must then exist some other (x', y') E N 2 such that
X'+
y'
The triangulating condition
x'+ ay'
6
This means that
x'+
y'
6
and
x+/y
6
x + ay
x'+ ay'
so
a(y - y') < x' - x < 0(y - y').
(4.2.2)
is strictly between a and /. Because q/p is the simplest fraction
X''
It follows that y - y' : 0 and y-y
>, q and ly - y'I ) p. But this is impossible when x < q and
have
Ix'
xl
between a and /, we
y < p, since x' - x and y - y' have the same sign, in light of (4.2.2).
(2) is obvious.
4.3
Perturbations of strongly triangulating sequences
The following are two theorems that allow us to construct new triangulating sequences out of old
ones.
Proposition 4.16. Let Mi1 , ... , Mk, ni, ... , nk be positive integers, with gcd(mi, ni) = 1 for all i,
and let S be a rational number such that mkJ6EN. Suppose that m', n' E N and nkm' -mkn' = 1.
Then A:= (6;
n,...,
is strongly triangulatingif and only if
A/
i
nk-1 n/
Mk-1 MI) Mk
ni
M1
is strongly triangulating. (In particular, if these conditions hold, then
triangulating.)
52
6;
,..
,k1
is weakly
Proof. As a matter of notation, when we write gi or Wi in this proof, we always refer to the A
sequence. Whenever mo appears below, we take it to equal 0. Note that
mk-19k +
nk
mk
- 32,
= 9k+1
so
m mk-lgk + n = m'32
+
=-
-mkn.
mk
mk
Let
S := {(X,YZ) E Wk x N x N I Y < m'}.
The jth triangulating condition for A deals with the number of triples (X, Y, Z) E S such that
X +
2
(Y + m'Z) < j,
(This comes from the jth triangulating condition, where we have - seemingly arbitrarily - broken
up the integer Y + m'Z into its quotient Z and remainder Y upon division by m'.) If we multiply
through by Mk, we get an inequality
mkX + iMk2(Y -+ M'Z) < Mk j
in which all the coefficients, including Mk j 2 and mkc, are integers.
The jth triangulating condition for A', meanwhile, deals with the number of triples (X, Y, Z) E S
such that
m'X+ m/j2
-
Y+mi 2 j 2 Z
=
mi2mklgk + mW' +
m'X+(mn'mk-1gk+n')Y+
M/) Z < m'3j.
(Here the stipulation that Y < m' is necessary to prevent the expression iM'X + (M'iMklgk + n')Y
from double-counting elements of the relevant semigroup generated by m'W and m'm-igk + n'.)
We may multiply through by mk/im' to rewrite this inequality as
mk X + nk 2(Y + Mi'Z)
Y
-
and here all the coefficients, except 1, are integers.
Kj< 1, so, since Mik
For all (X, Y, Z) E S, we have 0 <
MkX + mk6 2 (Y + M'Z) < Mi/j 4=
, < mYoj,
2
, Mik
E N,
MkX + Mk3 2 (Y + M'Z)
-
-/
<, Mkj,
l
as desired.
Proposition 4.17. Let Mi,..., Mk, ni, ... , nk be positive integers, with gcd(mi, ni)
and let J be a rational number such that Mk 3 E N. Then the following are true:
(a) Suppose that m',n' E N and nkm' - mkn' = 1. If A-
=
ni ,
pseudo-triangulating,then
(_ mklgk + n'/i'
M= ink k + nk/m
is weakly triangulating.
53
; n1
nk-1
n
M1
Mk-1
M'
=
1 for all i,
is strongly
(b)
Suppose that m', n' E N and nkm' - mkn' = -1 instead. If A+
(6;
,.
n
+'E
is
strongly triangulating, then
B+
n,
m1
n.1,
k
,
mk-1
MI
is weakly triangulating.
Proof. Imitate the proof of Proposition 4.16. Let
T := {(X,Y) E Wk x N I X m'nk_1gk +n',
For all
with
j
Y < m}.
in the range we need to consider, the jth triangulating set for A± consists of pairs (X, Y) E T
MkX + (mkl1Mkgk + nk) Y < Mk 6 j,
subject to the additional condition that, if equality holds, then X = 0 (for A-) or Y = 0 (for A+).
With some arithmetic, we can see that the jth triangulating set for B- consists of pairs (X, Y) E
T with
Mk +
M'Mk_1gk
+ n')
X
+ (mkmklgk + nfk)Y < mk 6 j.
The jth triangulating set for B+ consists of pairs (X, Y) E T with
MkX +
Mkinklgk + nk + -7)
Y < Mk 6 j.
It is not difficult to see that the jth triangulating sets for A- and Bi
are the same.
E
Now we prove a result restricting the set of weakly triangulating sequences that are too close to
a strongly triangulating one.
i
k - 1) and a' = m be positive fractions in lowest terms.
Lemma 4.18. Let ai = g- (for 1
Let a be either a positive real ak, or an infinitesimal perturbationthereof, that is, ak ± E. Let I
p be
the simplest fraction in the "open interval" between a and a', where this "open interval" includes ak
if a is an infinitesimal perturbation of ak away from a'. Let 6,6' > 0, A = (6; al. .. , ak- 1 , a), and
A' = (6'; al, ... ,a 1, a'). Suppose j E N is small enough such that
(M-gkp + q, 0), (0, p) V Sj (A),
and both A and A' both satisfy the jth triangulatingcondition. Then Sj(A) = Sj(A').
Proof. Suppose not; then there must exist (w, z) E S(A)\Sj(A') and (w', z') E Sj(A')\Sj(A). But
then it is a simple matter to see that w - w' and z - z' are both non-zero and have opposite signs,
must lie strictly between M-igk + a and mkligk + a'. But Mk_1gk + 1 is
the simplest fraction in the open interval between mi-lgk + a and M-lgk + a', so we must have
1w - w'f ; Mkigkp + q and Iz - z'| >, p. Thus, if w > w', then w >, mk-1gkp + q, and if z > z',
then z >, p. In either case, one of (Mk-g_1p + q, 0) and (0, p) must belong to S (A), contrary to our
0
hypotheses.
and the ratio
-
54
Proposition 4.19. Suppose 6 = 2 and ci = IL, for 1 < i < k, are positive fractions in lowest
terms, and A = (3; a1,..., ak) is strongly triangulating. Suppose, too, that
Phi
P1o
for some fair lower approximation 1
that
plomk-lgk
to cek and some fair upper approximation h
+ q10 ) max(7Z\Wk) + c + 2
Suppose that a' 4 ak, but A' = ('; a 1 ,...
Then, we claim,
,
and
to Cek, such
Phi > b.
k1 a') is weakly triangulating with Sj(A) 9 Sj(A')
for all j E N.
i. We must have 3' > 3, and Sj(A)
=
S (A') for all j such that
b(m'mk-lgk
j cm'
+ n')
I
c
ii. Moreover, a' must be a fair approximation to ak.
Proof. i. To prove that 3' > 3, note that, for all sufficiently large positive integers i, we have
ic E Wk and, consequently,
(i c,0) E Sib(A) 9 Sib(A).
This implies that
c
f c
=6.
=ib
b
> Then for
j < min {,
( 'b
b(m'mk-lgk
+ n')
c
we have
j < m' < n''.
b2 and b2 9k+1 = c 2 . For any
satisfying the given inequality in the lemma statement, and for any (w, z) E Sj(A), we have
Moreover, the fact that 62 = 9k+1 = mk-lgk +-L
j
b2 w + c 2z
=
implies that mk
b2 (w + gkz) < bcj < min
=
{c 2 m', b2 (m'mk-1gk
n')},
so
m'w + (m'mk-lgk + n')z < max
/, m'm-12k + n'
. (b 2 w + c 2z)
< m'(m'mk-lgk + n'),
and
j+2
= #S (A) < #{(w, z) E Wk X N I m'w + z(m'mk-lgk + n') < mk(m'mk-lgk + n)}.
2~
2)
Thus, A' is required to satisfy the jth triangulating condition, so
#Sj(A') = (i
+2)
2
Since Sj(A) 9 Sj(A'), this implies Sj(A) = Sj(A').
55
ii. Suppose for the sake of contradiction that n'/m' is not a fair approximation to nk/mk. Let
be the simplest fraction strictly between them. Let us consider two cases, depending on which of
a or ak is larger.
Case 1: If a' < ak, then write
p
Plo
<p'
m'
p
k
where q'/p' and q/p are consecutive fair lower approximations to ak; then by Corollary 4.14,
m' >,p' + p >p10 + p
and
n' >, q' + q >, q10 + q,
so
(m'mk_1gk
+ n') - (pmk1gk + q + max(Z\Wk)) = (n' - p)mkg1k + (n' - q) - max(7Z\Wk)
> p10 mk-1gk + q10 - max(7Z\Wk) > c + 2.
Therefore, we may choose x' E N such that pmk-1gk + q + max(Z\W) < cx' < m'mklgk + n'.
Then, by (i), we know that A' is required to satisfy the jth triangulating condition, when
b, <b(m'mklgk + n') cm'
<b b
c
j := bx <
Note that
cx' - (pmk-1gk + q) > max(Z\Wk),
so we can be assured that
cx' - (pmk 1kA + q), cx' E Wk.
Thus, (cx', 0) E Sj(A), so we must have (cx', 0) E Sj(A') also. Then a little calculation shows
(cx' - (pmk-1gk + q),p) E Sj(A')\Sj(A),
so A' fails the jth triangulating condition, a contradiction.
Case 2: If, on the other hand, a' > ak, then write
>hi
Phi
q>
q1>
-n'> ak,
PI
m'
p
where q'/p' and q/p are consecutive fair upper approximations to ak; then by Corollary 4.14,
m' >, p' +p
> Phi +p > b+p.
Therefore, we may choose x' E N such that p < by' < m'. Then, by (i), we know A' is required
to satisfy the jth triangulating condition, when
<CM/
,CY'
b
b(M'Mklgk + n')
c
Now (0, by') E Sj(A), so we must have (0, by') E Sj(A') also. Note that
pMk-1gk + q > nkgk > max(Z\Wk),
so PMkl1gk + q E Wk. Then a little calculation shows
( pmk-1gA + q, by' - p) E Sj (A')\Sj (A)
so A' fails the jth triangulating condition, a contradiction.
56
I
> 1 and ai = g-,
Proposition 4.20. Let 6 =
i ( k, be positive fractions in lowest terms,
for 1
and suppose that A+ = (3; a1,..., ak + ) is strongly pseudo-triangulating.Let a' = . be a rational
,ak-1, a') is weakly triangulating, and let
number strictly within 1 of ak such that A' = (3'; a,...
- be the simplest fraction in [a', ak] if a' < ak, or in (ak, a'] if a' > ak.
p
Then, we claim,
i. A' must satisfy the jth triangulating condition for all j such that
(j -
1)(j - 2) < (n'
1)(m'mk-Igk + n') + m'max(N\ Wk)
-
and
b(m'mklgk
j < min {M,
(
+
n')
I
c
'b
If in fact
j < min{
, b(mk-1gkP + q)
c
b
then Sj(A') = Sj(A+).
ii. If p > b, then Sj (A')
iii. If ak < a' < q-
-
Sj(A+) for all j < c, 6' ;, 6, and a' > ak.
for some fair upper approximation 2
to aje such that
Ph
Phb
Phi > 2b + b 2 (c+ 1)
then a' must be a fair upper approximation to ak.
9k+1 = mk 1 g1k + 1
Proof. As in the proof of Proposition 4.19, the fact that 32
2
2
Mk = b2 and b 9k+1 = c
i. The semigroup W' generated by M'Wk U {m'Mklgk + n'} is such that
implies that
max(N\W') = (M' - 1)(m'mklgk + n') + m'max(N\Wk).
For any
j
< min
-m , b(m'mk19k+n)
and any (w, z) E Sj (A+), we have, just as in Proposition
4.19,
b2 w + c2 Z = b2 (w + gkz) < bcj < min
{c 2 M',
b2 (M'Mklgk + n') ,
so
m'w + (M'mklgk + n')z < max
Mnmklgk
+ n/
(b2 w + c2 Z) <in'm'nm-1g + n,
C2\~l~l
J
and
+ 2)
C
= #Sj(A+)
<
#{(w, z) E W x N I m'w + z(m'inklgk + n') < Mk(M'Mklgk + n')}.
Thus, A' is required to satisfy the jth triangulating condition if j satisfies the first two inequalities
listed in (i). If moreover
Sj (A').
j
< min
{,
b(mkj1kp-q)
57
,then by Lemma 4.18, we must have Sj(A+) =
ii. Note that m' > p ,> b+ 1 and
mk-19k +
q + 1C2
ak =
,
P
so
pmk_1gk+q
To show that S 3 (A')
(M'
-
=
j
S4(A+) for all
c2
b
< c, it suffices to note that
> (p - 1)(pmk-lgk + q) - m'
1)(m'mk-lgk + n') + m'max(N\Wk)
2 (b +1)_
/C
) - P>b 1
> (p-i)
b - 1 = c2 +
-
- 2b - 1
> C2 - 3c + 2 > (j - 1)(j - 2)
and
cp
b(mk-lgkp+ q)
2p - 1
b >C >c
> C
(b2-2(b
) 1
1)b
c+
-
c>
and then use part (i).
; note that bx < c. Thus, Sbx(A')= Sb,(A+), and in particular, as
Let x = [
C2
> max(N\Wk),
cx >-
we have
(cx,0) E Sbx (A+) = Sx(A'),
which shows that 6' >
=
6. Likewise, we have Sc(A') = Sc(A+), and in particular
(0,b) $ S (A+) = Sc(A').
Thus,
<
b - (mk-lgk + a')
bx
C
and in particular
a >
C2
- mk-19k =
ak,
as claimed.
iii. Suppose for the sake of contradiction that a' is not a fair upper approximation to ak. Then
write
<
ak
for consecutive fair upper approximations
so
c(p+ p')
<
IV
p
to ak, with p'
,
'
= cm
> Phi > 2b+
b(m'mk-lgk
b(c+T1.
Then m' > p+p',
+ n')
b
b
and recall from (ii) that 6 < 6'. Thus, A' must satisfy the jth triangulating condition for all
c(p+ p')
b
58
Let x
3
]3c~
1, and note that
-
Mk-lgkp+q
-
M
S
[1(c
1=
-=
2
-i
-~~~
(c2
<
1(
p-9+
IC
\)1
[c
11~
First,
cx >
3
4
c(x + 1)>
3
3pc
3(mk-gkp
2
2
C
max(N\W),
+ )>4 b2 > T2 >
so cx E W. Second,
bx+c< b
/cp
+C
b2
=
\b
c(p+ 2b)
c(p+ p')
b
b
b)
(P
so A' must satisfy the jth triangulating condition for j = bx + c. Now (cx, b) just barely fails to
belong to Sbx+c(A+); we claim that (cx, b) V Sbx+c(A') also. If not, there would have to exist some
(w, z) E Sbx+c(A+)\Sx+c(A'). Since a' > ak, we must have w < cx and z > b, and
Cx - w
mk-19k + ak < z - b < mk1g1
a
+ C,
,
so
CX > cx - w > mk-19kp
+ q,
which is evidently false, by the way we defined x. So indeed (cx, b) V SbX+c(A').
Now let
mk-1_gk(p + 1) + q
y :=
C
First,
cy - (mk-lgkp + q) > mkgk > max(N\W),
so cy - (mk-lgkp
+ q) E W. Second,
by < b mk_1gk(p+ 1) +q + I) < C
so
b2
by + 2c < - p + b (
<c(p+ p)
+ -b2 + 2b)<<
b
C
)
- +1
cm'
b +
+b,
C
b(m'mklgk
+ n')
c
b
so by (i), A' must satisfy the jth triangulating condition for j = by + 2c. Note that (cy, 2b) just
barely fails to belong to Sby+2c(A+). We claim that (cy, 2b) E Sby+2c(A'), however. Otherwise,
since
mk-19k
+ ak < mk<kP + q <
p
mk-19k
+ a',
we would have
(cy - (mk19kp
+ q), 2b + p) E Sby+2c(A+)\Sby+2c(A'),
so there must exist (W', z') E Sby+2c(A )\Sby+2c(A+).
z' < 2b, and
Since a' > ak, we must have w' > cy and
w1 - cx
mk_1gk
+ ak < 2b - z' < mTgk19 + a ,
so that
2b > 2b - z' >, p,
59
which is evidently false. So indeed (cx, 2b) E Sbx+2c(A').
It follows that
cx + b(mk _gk ± a')
cy + 2b(mk _gk + a')
6
bx + c
by + 2c
Both bounds are weighted averages of 'b and
heavily, because
>
2x-y
mnk1lgkP - q
C
2
mk-lgk(pC
so it must be the case that
The right-hand side weights C more
b(mklgk±+)
q
(mk_1k(P+1)+
+1)
C
)
1) + q -3
>
(p - 1) - 3 > 0,
b(mklgk + a')
C
C
b
which is to say
2
a <
C
-~
Mk-19k =k,
Li
a contradiction.
4.4
Weakly triangulating sequences of length 1
In this section, we classify all weakly triangulating sequences of the form (6; a
4.4.1
=
-), with m, n > 2.
Weakly triangulating sequences (6; a) with 1 ; a <so 4
In this section, we will use Proposition 4.6 to find constraints on weakly triangulating sequences of
the form (6; a) with 1 ; a < s
4.
When dealing with Fibonacci numbers, it is helpful to introduce a bit of notation for continued
fractions: when we write "a, b, c "k within a continued fraction, we mean imagine the sequence
"a, b, c" is written out k times. If the superscript k is missing, the sequence is meant to be repeated
ad infinitum, in accordance with standard notation.
So, for example, so4 = [6,1-5], and for k > 1,
-k-i
=[6,1,
5 -3,
F4k+1 2
F4k+3 =
[6,
F4k_1
\
F4k+3 /2
'F4k+)
1,5 -1, 2]
-k
k-1
/j
T-5-1 ]]
,31
=[6, 1,5
F4k+±
[
(F)
2
]I
k-i
It follows that
F3
< _-1
(F
3 )
2
(Fs>
F5
Fi
Fi
2
F7
F3
3
F5
4
F5
F5
For convenience, let's write out the first few terms:
k\iJ
2
1
(22
5
1
(5 )2
2
60
13
S-
13 )2
5
34
5
4
(*)
Comparing the continued-fraction expansions with Lemma 4.13, we see that 7-
is a fair lower approximation to (
for all k E N, and
approximation to
is a fair upper
for
all k E N\{1, 2}.
Lemma 4.21. Fix k E N, a = F2k-3, b = F2k-1, c = F2k+1, and d = F2k+3. Let n > m > 0 be
relatively prime integers such that n/m is a fair approximation to (c/b)2 in the interval (c/a, d/b].
(lo) If
j < L
< (q)2, then either there is an integer i - k (mod 2) with -1 < i < k such that
n)
F2i-1(a, c) + F2 i(b,d)
3
or else there is an integer f >, 0 such that
2 2
2
(M, n) = f(b2 , c ) + (a , b + 2).
(The two expressions are equal when i = k - 2 and t = 0.)
, then either there are integers i = k (mod 2) and e with 2 < i ( k and
1 < e < min{5, 3i - 4} such that
(hi) If
(f)2
<
-L <
,
n)
(F 2 i-
1
- eF 2 i- 5 ) (a, c)
+ (F 2i+l - EF2 i- 3 )(b, d)
3
or else there is an integer f >, 1 such that
(*,
n) = f(b2 , c 2 )
-
(a 2 , b2 + 2).
(The two expressions are equal when i = k and e = t = 1.)
Sketch of proof. Use the continued-fraction expansion of (c/b)2 and Lemma 4.13. The pairs (m, n)
of the form f(b2 , c 2 ) t (a 2 , b2 + 2) come from changing the "oo" in the continued-fraction expansions
of (c/b)2 Lemma 4.13 to a positive integer. The other pairs (m, n) come from truncations of these
expansions (along with, optionally, decreasing the last number of the expansion that remains). E
Proposition 4.22. Fix k E N, a = F2k-3, b = F2k-1, c = F2k+1, and d = F2k+3. Let (6; a = a)
be weakly triangulatingwith m > 1 and ! < a < i, and let D = m6. Then the following statements
hold:
}. The same inequality holds if a = { and (6; a)
i. If a # {{, }, then we have D > max{,
satisfies the bth triangulating condition, or if a = { and (6; a) satisfies the cth triangulating
condition, even though these triangulating conditions are not required by the definition of
"weakly triangulating."
ii. If a
{,},
then for all j E N, we have Sj ((c/b), (c/b)2 ) C Sj(6; a), i.e.,
b2 x + c2 y < bcj
->
mx + ny < Dj,
iii. The fraction n/m is a fair approximation to (c/b)2 . If a V
{,
}, then for (x, y) E N
b2 x + c2 y < bcj +=># mx + ny < Dj.
whenever bj < cm and cj < bn.
61
(4.4.1)
(4.4.2)
iv. Either
m <b and n <d,
(small)
rc2m - b2n1 < 1.
(large)
or
v. In case (small) above, (m, n) is one of
(a,c),
(a'b
(a ± b
cd)
and (b,d),
c
unless (m, n) = (3, 20). In case (large) above, (m, n) is either (b2 , c 2 ) or of the form
f(b2 , c2 ) ± (a2 , b2 +2),
f E N.
Proof. It is useful to distinguish between two cases:
c
n
a
m
cC2
\bI
and
(
<
<d
(hi)
Always keep in mind that (c/b; (c/b) 2) is strongly triangulating, by Proposition 4.6.
i. In case (lo), ' > Pn, so we just have to prove cm < bD. The simplest fraction in (a, (c/b)2 )
has a numerator exceeding c and a denominator not less than a. (This holds even the cases k E
{1, 2}.) Note that
(b - 1)(b - 2) = (a - 1)(c - 1) < (m - 1)(n - 1),
and recall that ac - b2 = 1, so in particular ac > b2 , and (6; a) must satisfy the bth triangulating
condition. By Lemma 4.15 for j = b, we have for all (x, y) E N2 ,
b2 x + c 2 y < b2c
==>
x + ay < b/a.
In particular, set (x, y) = (c, 0) to find that cm < bD.
In case (hi), ' < h, so we just have to prove bn < cD. The simplest fraction in ((c/b) 2 , a)
has a numerator exceeding d and a denominator exceeding b. Note that
(c - 1)(c - 2) = (b - 1)(d - 1) < (m - 1)(n - 1),
and recall that bd - c2 = 1, so in particular bd > c2 , and and (6; a) must satisfy the cth triangulating
2
condition. By Lemma 4.15 for j = c, we have for all (x, y) E N
b2 x + c 2y < bc 2 => x + ay < cV/Y.
In particular, set (x, y) = (0, b) to find that bn < cD.
ii. Suppose b2 x + c2 y < bcj. In case (lo), b2 n < c2 m, so
2
mc <Dj,
(b M2 x + c2y)
<m
bj<
mx + ny <
and in case (hi), c2 m < b2 n, so
mx +ny <
T(b 2 x
C
+c 2 y)
62
n - bj
C
.
that
and 4b are already fair approximations to
iii. Note
ii.Noethta
from (ii) by Proposition 4.19.
iv. Assume that either m > b or n > d. Also assume that
C.
b
If e (
{g,
a {},
b then (iii) follows
A := lb2n - c2 mI > 1,
so that (i, n) is not of the form
e(b2 ,c2 ) ± (a2 , b2 + 2)
for any f E Z.
In case (lo), Lemma 4.21 implies that there is an integer i = k (mod 2) with 2
that
i < k - 4 such
(mn) F2 i1(a, c) + F2 i+1(b, d)
3
since any smaller value of i would entail m < b and n < d. Let
j = cy'+ b - 3.
y' = F 2i-1 - 1,
We claim that
j
is small enough that (4.4.2) must hold, and yet we will exhibit a particular pair
(x, y) for which (4.4.2) fails.
First we claim that
C(F2i-3 - 3) < b(F
2 ji-
-
3).
Indeed, if 2 < i < 3 then this is clear. If i > 4, then both sides are positive, and since c/b <
suffices to show that
F2j-j -
3 >
F2i-3 - 3
or, in other words,
To that end, note that F2 1
(F2j-j - 3)2 + (F 2 i-
Since F 2i-
3
+ F 2i+
=
3
F 2 i-1 - 3
F 2 i-3 -
F2i-3 - 3
F2ji
-
3
2
+ Fi-
3
1
-2
1, so
=
- 3)2 - 3(F 2j- 1 - 3)(F 2 i-
3F 2j3y'
3F 2 - 1F 2i-
3
- 3
3
- 3) = 3(F 2 j-
1
, it follows that
b
c
F2i-3 - 3 + F 2j+i < -(F 2j-1 - 3) + F 2j+i-
Thus,
cj
=
c(cy'+ b - 3)
<c-F2ji - 3 +C
+ 2 .F2i+1 l +bc-3c
c-3
±c
" bc
3
3
so j <
<
+ F 2 i- 3 )
+ (c2 + 1).
<
bc F2i-1
=
bc- F2i-1 + bd - F2i+1
=
bn,
3
3
3
3
, and (4.4.2) holds.
63
-
10 > 0.
92 it
And yet, let
(x,y) = (c - 1, by' - 1).
Then
I(da
dm - bn = F 2
3
= F2n-1
-
+ 1
=
so by our assumption that A > 1,
(c2 m - b2 rn)y' = Ay' > 2y' > y' + 1 = dm - bn
=
(3c - b)m - bn.
This means that
b(mx + ny) = bm(c - 1) + bn(by' - 1) < cm(cy' + b - 3) = cmj,
so that, in light of (i),
CM
mx + ny( <
-b.j(Dj.
On the other hand,
b2x + c2y = b2 (c - 1) + c2 (by' - 1) = bcj + 3bc - b2 - C2 = bcj + 1 > bcj,
so (4.4.2) fails. This concludes the proof of (iv) in the (lo) case.
Case (hi) is similar, only slightly more tedious. Lemma 4.21 implies that there is an integer
i = k (mod 2) with 2 <, i k and a positive integer E ( min{5, 3i - 4} such that
(m, n) =3
(F
2
i-1 - CF2 i- 5 )(a, c)
+ (F 2 i+ - EF2i-3)(b, d)
If i = e=2, or if i = 3 and E = 5, then (m,rn) = (b,d), and if i = k and c = 1, then A = 1; by
assumption, we are not dealing with these cases, so in particular, k > 3.
Let
X' = F2i+1 - EF2i-3 - 3,
j = bx' + c - 3.
We claim that
j
is small enough that (4.4.2) must hold, and yet we exhibit a particular pair (x, y)
for which (4.4.2) fails.
First we claim that
F2i+1
-
3
CF2--
F2i-1 - EF2i-5
F2k+1
F2k-1
C
_
b
We fix k, and prove this by induction on i.
.
If i = 2, then the only case to consider is E = 1: since k > 2,
F2i+1
-
F5 - 1 - Fi - 3
F3 - 1 -F-1
EF2i-3-3
F2i-1 - EF 2 i_ 5
5-2
1-3
- 1
21<
F2k+1
C
F2k-1
b
* If i = 3, then we only need to consider the cases where 1 < E < 4: since k > 3,
F2i+1 -
F2i-1-
EF
2
- 3 -
cF 2 i_ 5
3
13 - 2c - 3
=
5 - le
64
2
F2k+1
C
<F=F2k-1
b*
e If i > 4, then note that F 2 ±i+ - 6F 2i2 i- 3
F2i+l - EF
F2i-1
-
= F2i-3 - F 2i- 7 .
3
(F2i+1 - 6F2i-3 - 3) + (6 -
3
(F2j-1
CF2i-5
-
Since
-
6F 2
-6
5 )
+ (6
)F2i-3
)F 2i- 5
-
(F2-3 - F2i-7 - 3) + (6 - E)F2i-3
(F 2i- 5 - F 2 i- 9 ) + (6 - E)F2i-5
the quantity in question is a weighted average of
F2i-3 - F2i-7 - 3
F2-3
and
F2i-5 - F2i-9
F2i-5
The first of these is less than b/a, by the induction hypothesis (with i - 2 replacing i and 1
replacing e). The second of these is an increasing function of i > 4 and so, since i < k, we
have
F2i-_
F2k-3
F2k+1
F 2i- 5
F2k-5
F2k-1
_
C
b
as well.
Thus,
bj
=
b(bx' + c - 3)
<
b (c(x'+ 3) + (3b
<
bc
c)x/
-
-+
+ ba
3
F 2i- 3
c~
3
-
bc bc - F2i+1 - EF2i-3 + ca 3
=
3
-
-
<
-
3
=CM,
so j < ' < '-, and (4.4.2) holds.
And yet, let
(x, y) = (cx' - 1, b - 1).
Then
-
an - cm =
F2
3
)(ad
-
cb)
F2i+
3
- EF2 i-
3
=
x' + 3.
Also note that
A
=
=
b2 n - c2 m
(F 2j- 1 - EF2j- 5 )(b2c
=
We claim that A . x' > x'
(F 2 i-
1
-
c2 a)
+ (F 2 i+1 - CF 2i-3)(b2 d - c 2 b)
- CF 2i- 5 )c + (F 2 i+1 - EF 2 i-
+ 3. If i
=
3
)b.
2 and e = 1, then x' = 1 and
A = 4b - c > 7
since k > 3, so indeed A -x' > x'+3. If i = 3 and e = 4, then x' = 2 and A = 5b - c > 12, so again
A - x' > x' + 3. Otherwise, x' > 3, so by our assumption that A > 1,
(b2rn - c2 m)x'
=
A . X' > 2x' > x' + 3 = an - cm = (3b - c)n - cm.
65
This means that
c(mx + ny) = cm(cx' - 1) + cn(b - 1) < bn(bx' + c - 3) = bnj,
so that, in light of (i),
mx
bn
+ ny < - - j < Dj.
C
On the other hand,
b2x + c2 y = b2 (cx' - 1) + c2 (b- 1) = bcj + 3bc - b2
-
C2 =
bcj + 1 > bcj,
so (4.4.2) fails. This concludes the proof of (iv).
v. This follows from (iv) if we can eliminate the case where k > 4 is odd and
(m, n)
=2(a,
d)
c) + (b,
3
In this case, (4.4.2) holds, so since b2 c = b - bc, we must have
cm < bD.
Choose
x:
a (mod 3). Since k is odd, one can check the numbers
E {4, 5} such that y
7c -
b -,q
4c+d-77
=
y=
9
8b-2c- 77
9
-
.2b±C -77
c
andj:=
2a+2b-r;
9
2
2
2
are all positive integers. Since ac = b2 + 1, bd = c + 1, and b + c + 1 = 3bc, we have
b2 x+c
b 2(4c + d - 77) + c2 (2a + 2b - 77)
9
2
-
3bc(2b+c- q) +b+2c+77
9
>
and
(m +1)(n.+
2
1)
_
bcj,
(2a+b+3)(2c+d+3)
18
(7b-2c+3)(5c18
b+3)
so that one can check
(j+ 1)(j + 2) _ (4b+ 2c + 9 -
(m + 1)(n + 1)
2
)77+ 2
18
2
and similarly
(j - 1)(j - 2) < (m - 1)(n - 1).
Therefore, by (4.4.2), we must have
mx
cm
+ ny > Dj >, j.
b
Since
mx +y
=
mj=
c(16b - 3c) - (2b + c)- + 5
)
9
2
-2c)7A+
-(7b
b(16b -3c)
9
66
> 0'
This means
0> crnj
But if k
=
5, then
)
b(mx
-
F2k4 =
2c - 5b - 2
2
b-2a-27
9
F2k4 - 21
9
8 and q = 4, so cmj - b(mx + ny)
=
9
0. If k > 7, then
F2k-4
> 55 > 277,
so cm]- b(mx + ny) > 0. This is a contradiction.
E
In summary, we showed that, in many cases, if a fraction n/m is sufficiently close to a strongly
triangulating ratio, then in fact n/m must be a fair approximation thereto. For fractions less than
P4, we were able to use strongly triangulating ratios of the form
4.4.2
2
(F2k+1/F2k-1) .
Weakly triangulating sequences (6; a) with W4 < a '; 9
Proposition 4.23. Let (6; a
-L) be weakly triangulating with m > 2 and p4 < a < 9. Then
=
i. If 04 < a < 7, then a is a fair upper approximation to 94
k > 2, we have
_
n
[6 1
m
5
k-1 1]
=
[6, 1,5], i.e., for some integer
-F4k+4/3
F4k/3
i. If 7 < a < 8, then a is a fair upper approximation to 4= [7, 8, 1], i.e., either m
some 2 ; m <; 8, or
= [7,8, 1, k] - 64k+57 for some integer k > 1.
iii. If 8 < a < 9, then either a
=
=
7+
1 for
9 or else a is a fair lower approximation to 9, i.e., g
m = 9 -
M,
Proof. Let D = m6. The cases where a E N have been dealt with before, so let us assume m> 1.
It is not hard to see that
(m+1)(n + 1)
(2+1)(15 +1)
(3+1)(3+2)
2
2
2
so the third triangulating condition applies. It follows that
n
- < 7
-
n
- > 7
=>
m
i. Suppose
4 <
m + n <n3D < 8m,
(4.4.3)
8
(4.4.4)
< 7 but a is not a fair upper approximation to p4. Let -p be the simplest
mm
fraction in (,o4, a). Then
such that
must be a fair upper approximation to W4 , so let k be the positive integer
F 4k+ 4
(pq) =(f
Let a :=
F4k-2,
b
;3D < m + n.
F4k
=
3p, c
F4k+2,
and d := F4k+4 = 3q. Note that ac + 1
and
3a - b = F4k_4.
We have
4
b/3
n
d/3
<b/3<rn<(3a - b)/3'
67
-
b2, a + c
3b,
ad b/3
ad/d3
and3a-b)/3 and 3 are consecutive fair upper approximations to
4 By Corollary 4.14, there are
4.
two cases; either
b)+
(m,n) = (3a3b
,j )
= (a, c)
or m > a and n > c.
First suppose (m, n) = (a, c). By (4.4.3), we have
D > m+n
3
_
a+c - b,
3
as one can easily check. Since
(a- 1)(a - 2)< (a- 1)(c - 1) - I
and
(a + 1)(c + 1) >(a + 1)(a + 2)
z
2z
the ath triangulating condition must hold. Yet we claim that Sa((p2
ISa(6; a)I > ISa(p2; 4 )I = (a+2). Indeed, note that
and
ab<aD
so (b, 0) Cj Sa (6;
e)\Sa (p
2
->
a
4
)
Sa(6; a), so that
C
21
a), then the above considerations show
4
2
; p41). If (x, y) E Sa (so ; 9P )\Sa0;
that x < b and y > 0, so that
cy > a(D - x) >, a(b - x)
and
4 <S
2
<c
b-x
a-x
a
y
Y
n
m
4
which contradicts the fact that Ip is the simplest fraction in (p , a), since b - x < q.
Now suppose
n > c.
m > a,
Since
(b- 1)(b - 2)= (a- 1)(c - 1)+ 2 < (m - 1)(n - 1)
and
(m+ 1)(n + 1)
2
(a+ 2)(c + 2) _ ac + 2(a + c) +4
2
2
the bth triangulating condition must hold.
ISb(6; e) I > ISb( 02. ,43
=
(b+2).
_
(b2 - 1) + 6b+ 4
2
Yet we claim that Sb(cp 2 ; p 4 )
Indeed,
m+n
b+d
m
b
3c
b
and
b
>
2,
4
2
so (c, 0) E Sb(J; a)\Sb(W2 ; W4 ). If (x,y) E Sb (W ; W )\Sb(J; a), then
m(x - p) + n(y - p) > bD - (m + n)p ;)0
while
2
+
2
(y-p)
68
b- 3p = 0.
(b+ 1)(b+ 2)
2
C
Sb( 6 ; a), so that
I
Max
Range of a
7< a< 8
V
15
22
Lower bound on D Upper bound on D
Max jj
8
Range of a
D
22m < 8D
Upshot
contradiction
11m > 4D
22
22<
5
13
7m + 2n
29
29< a < 22
18
7m + 2n ( 8D
25m + n > 12D
36
5
36<
<
9
24
43m < 16D
33m + n > 15D
ae > 29
43
43<
<
6
29
7m + 5n
16D
41m + n > 18D
a < 265 <
50
50< a < 43
34
57m + n
24D
48m + 3n > 26D
a > 65 > 43
40
64m + 6n
40D
56m + 3n > 29D
a < 64
(64k + 64)m + 6n
< (24k + 40)D
(64k + 56)m + 3n
> (24k + 29)D
4w
57
64k+57
9k+8
<
57 <
64k+57 < a<
9k+8
50
-
64k-7
9k -1
48k + 18
(k >1)
8D
17m + n > 9D
a <
<
< 2
< 6
Table 4.1: The 6 < a < 8 case.
Since n/m > p4, we must have x < p < y and
4 <
X <
y -p
-,
rn
which contradicts the fact that Ip is the simplest fraction in (W4, a), since p - x < q.
ii. Suppose 7 < a < 8 but a is a not fair upper approximation to
First we claim that a > 6. If not, then
6.
9.
(8- 1)(8 - 2) < (9- 1)(64 - 1) < (m - 1)(n - 1)
and
(8+1)(8+2)
(9 + 1)(64 + 1)
(m + 1)(n + 1)
.>
= 325 >
2
2
2
so the 8th triangulating condition must hold. But in fact S8(6; a) = S8(
;
6
+E)
Li {(0, 3)} because,
by (4.4.4),
8
21m < 14m + n < 7m + 2n < 3n < - 8m < 8D
3
but
8
3 - 8 5 8,
3
Thus, we must indeed have a > 64
) Table 4.1 lists the possible values of
<
< 8. Let I be the simplest fraction in (9,
So
and the corresponding ranges for -. The third column, "Max j,"lists the maximum j for which
the jth triangulating condition is required to hold; it is calculated with the aid of Corollary 4.14.
For example, in the last row, if
64k + 57
9k + 8
n
m
64k - 7
9k- I'
then Corollary 4.14 tells us that
m
(9k + 8) + (9k - 1) = 18k + 7
n
(64k + 57) + (64k - 7)
69
=
128k + 50,
so that
(m + 1)(n + 1)
2
2
2
2
(j + 1)(j + 2)
(18k + 8) (128k + 51) = (48k + 19) (48k + 20) + 70k + 28
as long as j < 48k + 18. Note that the coefficients of D in the fourth and fifth columns never exceed
the "Max j" value in the second. We'll show that, if in the fourth column a lower bound on jD
is violated, then the triangulating condition fails because Sj(6; a) C Sj (; 6 + E). Similarly, if in
the fifth column an upper bound on jD is violated, then the triangulating condition fails because
2+e) C Sj(6; a). Then the last column follows immediately from the fourth and fifth columns
Sj (;
and contradicts the second one. So, without further ado, let us prove the bounds on D in the fourth
and fifth columns.
Lower bound for p = 2: Suppose 22m > 8D. Then 7m+2n > 22m > 8D, so Ss(; a)Lf (7, 2) c
S8(; 6 ).
Lower bound for 3 <p < 4: Suppose 7m + 2n > 8D. Then (7,2) E 8(8; 6 + E)\Ss(S; a). If
(X, y) E Ss(; a)\Ss( ; 6 + E), then it is not hard to see that x > 7, y < 2, and
64
x-7
n
9
2- y
m'
since 2 - y ( 2 < p.
contradicting the fact that .is the simplest fraction in (4),
in Table 4.1 does not hold. We
given
bound
lower
the
6:
Suppose
p
<
5
<
Lower bound for
claim that (7, 5) E S16 (8; 94 + 6)\S16(6; a): In the p = 5 case, we have 7m + 5n > 43m > 16D,
whereas in the p = 6 case, this is by assumption. If (x, y) E S 16 (0;a)\Si 6 (); 9 + e), then it is not
hard to see that x > 7, y < 5, and
Since I is the simplest fraction in (4,
inequality 43m ( 16D actually holds.
64
x - 7
n
9
5- y
m
n), we must have p = 5 and (x, y) = (0, 43), but then the
Lower bound for p = 7: Suppose 57m + n > 24D. Then 7m + 8n > 57m + n > 24D, so
(7,8) E S 24 ( ;6 + E)\S 24 (6; a). If (x, y) E S 24 (J; a)\S24 ( ; 6 +E),
y < 8, and
Besides i
and
64
x- 7
n
9
8-y
m
then it is not hard to see x > 7,
, all fractions in the interval (L, n) have denominator at least 13. Therefore,
either (x, y) = (57, 1) or (64, 0). But by assumption (57, 1)
S 24 (J; a), and (64, 0) E S 24 ( ; 9
so neither of these pairs lies in S 24 (6; a)\S24 ( ; 6 + F).
Lower bound for p = 9k + 8, where k E N (note p = 8 when k = 0): Let
j
:= 24k + 40, and
suppose (64k + 64)m + 6n > jD. Then
7m + (9k + 14)n > (64k + 64)m + 6n > (24k + 40)D,
so (7, 9k + 14) E Sj(8; 6 + E)\Sj(6; a). If (x, y) E Sj(6; a)\Sj(8; 6 + e), then it is not hard to see
x> 7, y < 9k + 14, and
n
x- 7
64
m'
9< (9k + 14) - y
Besides
,all fractions in (L, -) have denominator at least p + 7 > 9k + 14, so
(X, y) = (q + 7, 9k + 14 - p) = (64k + 64, 6) E Sj(J; a),
70
contrary to our assumption that (64k + 64)m + 6n > jD.
Upper bound for p = 2: Suppose 11m < 4D. Then 3m + n < 11m ( 4D, so S 4 (); 64 + E)
{(11, 0)} C S 4 (6; a).
Upper bound for 3 ( p 6: Suppose (8p-7)m+n < 3pD. Then (8p-7, 1) E S 3p(6; a)\S3p( ;
e). If (x, y) E S 3p( ; 64 + E)\S3p(J; a), then it is not hard to see that x < 8p - 7, y > 1, and
64
8p-7-x
n,
9
y-1
m
contradicting
9Mf)
cotrdctn the fact that I is the simplest fraction in (64,
48 - x
n
9
y- 3
m
+
since 8p - 7 - x < 8p - 7 < q.
Upper bound for p = 7: Suppose 48m + 3n < 26D. Then (48,3) E S 26 (6; a)\S 26 (8; L +
(x, y) E S 2 6 (0;L + E)\S26(; a), then it is not hard to see that x < 48, y > 3, and
64
u
contradicting the fact that 2 is the simplest fraction in (L, a), since 48 - x < 48 < 50.
Upper bound for p = 9k+8, where k E N: Let j := 24k+29, and suppose (64k+56)m+3n <
)- If
jD.
Then (64k + 56, 3) E Sj(6; a)\Sj( ; L4 + E). If (x, y) E Sj( ; 6 + E)\Sj(J; a), then it is not hard to
see that x < 64k + 56, y > 3, and
64
64k + 56 - x
n
9
y - 3
m
contradicting the fact that p- is the simplest fraction in
(64,
9
a),
M since 64k
+ 56 - x < 64k + 56 < q.
iii. Suppose 8 < -2- < 9, but - is not a fair lower approximation to 9. Let Pq be the simplest
m
'
m
fraction in (-, 9); then p > 2 and q = 9p - 1 > 17. Then by Corollary 4.14, we have m > 2p + 1
and n > 2q + 9, so
(m - 1)(n - 1) > 2p(18p + 6) = 6p(6p + 2) > (6p + 1)(6p)
and
(m + 1)(n + 1)
2
(2p + 2)(18p + 8)
2
( 6 p + 3 )( 6 p + 4 ) + 10p + 4
2
Thus, the triangulating conditions must hold for all j < 6p + 2.
We claim that, if 6m + pn > (3p + 2)D, then S3 p+ 2 (J; a) C S 3 +2(3; 9 - E). Indeed, under this
assumption, we have
(q+6)m > 6m + pn > ( 3 p + 2),
so (q + 6, 0) E S3 +2(3; 9 - E)\S3p+2 (;
to see that x<q+6, y>0,and
a). If (x, y) E S3 p+ 2 (6; a)\S3p+2 (3; 9 - e), then it is not hard
n
q+6-x
M
y
-<
<9.
Besides , all fractions in (L, 9) have numerator at least 2q - 9 > q + 6, so we must have (x, y) =
3
( 6 , p) E S 3 + 2 (6; a), contradicting our assumption that 6m + pn > ( p + 2)D.
We claim that, if (9p + 2)m < (3p + 1)D, then S 3 +1(3; 9 - e) C S 3p+ 1 . Indeed, under this
assumption, we have (3, p) E S 3 +1(; a)\S3p+1(3; 9 - E). If (x, y) E S3p+1(3; 9 - E)\S3p+2(J; a),
then it is not hard to see that x > 3, y < p, and
n <x-3
m
p-y
71
which contradicts the fact that 9 is the simplest fraction in ( , 9).
Therefore, by the jth triangulating conditions for j = 3p + 2 and j = 3p + 1, we have
(9p + 2)m
3 p+ 1
6m + pn
3p + 2
and
9p+5=q+6<
6m + pn
m
(9 p +
<-
2
)(3 p +
2
)
99 p+ 5
3p+1
1
-,
3p+1)
which is absurd.
Classification of weakly triangulating sequences of length 1
4.4.3
If ( = 2; a = n) is weakly triangulating and n > m > 2, then (m, n) must appear somewhere on
the following list, according to the results of previous sections. We also give some bounds on D in
each case. In cases involving only finitely many pairs (m, n), these bounds on D can be checked
easily and straightforwardly, so we omit the verification. But in cases involving an infinite sequence
of pairs (m, n), we sketch the proof.
Let z, a, b, c, d denote consecutive odd-indexed Fibonacci numbers; more precisely, z = F2k-5,
a = F2k-3, b = F2k-1, c = F2k+1, and d = F2k+3 for some k E N.
We claim that, if m > 2 and a > 3, then (6; a) must satisfy the jth triangulating condition as
long as j < D - 1. Suppose that were not the case, let j be the integer such that
j+ 2
2 )<
(m + 1)(n + 1) <j+
2
3)
2
by assumption, we have j < D - 2. The fact that (6; a) satisfies both the jth and (j triangulating conditions means that there are j + 1 < D - 1 non-negative linear combinations
and n in the interval ((j - 1)D, jD], an interval which contains at least [DJ > D - 1 integers.
every integer that is at least (m - 1)(n - 1) is a non-negative linear combination of n and n,
must be the case that
j
2
+ j - 2
=
(j
1)st
of m
Now
so it
1)(j + 2) < (j - 1)D < (n - 1)(n - 1) - 1 = mn - (m + n)
-
And yet
j2 + 5j + 6 = (j+ 2)(j + 3) > (m + 1)(n + 1)= mn + (m + n) + 1
so we must have
4D > 4j + 8 > 2(m + n) + 1.
But then m + n < 2D, and, under the assumption that 3m < n, we see that (x, 0) E S 2 (6; a) for
0 < x < 4, and (1, 1), (2, 1) c S 2 (6; a), so we know that (6; a) violates the second triangulating
condition, which is a contradiction because (m + 1)(n + 1) > 3 - 4 > 6.
" (m, n) = (3, 20), and
39
D < 4.
D < 1.
"
(m, n) = (4,27) and L
"
2). The lower bound follows immediately
(m, n) = (a, c), and b < D < a-b (assuming a
from the third triangulating condition, and can be boosted up to 0' if (D; n) satisfies the
bth triangulating condition. To establish the upper bound, let j = b - 1. Note that j < D,
72
m > 2, and a > 3, so (J; a) is required to satisfy the jth triangulating condition, which in
light of Proposition 4.22, part (ii), means that Sj (j; a) =S (, ()2). Let
a 1)if
(2c
Recall that b2 + 1
1
(mod 3).
ac and b2 + c 2 + 1 = 3bc, so c2a > b2c, and
-
b2 (2c - 1) + c2 (a - 1)
3
b2 X+C 2Yy
2c
a
2
2b 2 c- b2 + c(b2 + 1) - c
3
3b 2 c - (b2 + c 2 ) + c
3b2 c - 3bc + 1 + c
>bc(b-1),
3
3
_
so
= Sj (; a),
(x, y) V Si
which implies that
3ac-3b
ac-b
3
b-1
b-1
3
b-1
b-i
1
2ac-c
ac-a
1
ax+tcy
D
y
where E E {1, 2} and Tj E {4,5}, with
; D < M+n- + 6+,
c+fd) and
-- 0 (mod 3), assuming a > 2. The lower bound was proved in Proposition
a+ eb
m+ n4.22, part (i). To establish the upper bound, let j = m+n 9. Note that a > i > 3, and
(mrn)
n = (a+,
3a
b(m + n 3b
)
bm+ dm - b
3b
3cm - bq < D -1
3b
_
so (J; a) is required to satisfy the jth triangulating condition, which in light of Proposition
Note that m + n = b + cc, and let
4.22, part (ii), means that Sj (3 ; a) = Sj (,()2).
(jj)
(c2d-
+ (d - c, a - b)
3
{(2c+2d-
2a+
7
a+4b-7)
1
if
if c = 2.
Since the Fibonacci numbers are periodic modulo 9, it is a simple matter to verify that
x, y E N. Then
2
3(b
x + c 2y) = (b2 + c2 )j + b2 (d - c) + c 2 (a - b)
= (3bc - 1)j + b(c2 + 1)- b2c + (b2 + 1)c - c2b
= 3bcj - b+cc-T+b+c
=3bcj+ 2b+ (3 - cc) +,q
3
> 3bcj,
so (x, y) V Sj
(g, (C)
mx+ ny
j
=
m+ n
3
Sj (J; a), which implies that
+
m(d-c)+n(a-b)
3j
m+n
(a + eb)(d - c)+
+
3(1+ c)(ad - bc)
9j
3
n
m+
3
73
(c+cd)(a - b)
9j
m+n +
3
+1
m+n -
* (m, n) = (b2 , c2 ), and bc < D < bc-+
'1, assuming b > 2. The lower bound was proved in
Proposition 4.22. To establish the upper bound, let j = bc - 1 < D - 1. Since a > ()2 > 3,
we know (6; a) is required to satisfy the jth triangulating condition, so S (J; a) =Sj (q,(c) 2).
Let
(xIy) = (c(b - a) - 1, 2ab - 1),
and note that
b2 x + c 2y
=
b2 c(b - a) + 2abc2 - (b2 + c 2 )
b2 c(b - a)+ 2(b2 + 1)bc - (3bc - 1)
=
b2 c(b-a+2b)-bc-+1
=
bc(bc - 1)+1,
=
so indeed
1
bc - I
bc(bc - 1)+1
bc- 1
< D < (d±+2)cn-b , provided b >, 2, and (in, n) #
the (4, 27) case will be treated separately. The lower bound was proved in Propo(4,27)
> 3
sition 4.22. To establish the upper bound, let j = (cf + a + 2)b - c. Note that a > b
and
= j+ (c - 2b) + a > j +1,
= b(cf + a) +
D >b
* (M,
n)
=
f(b2 , c2 ) + (a2 , b2 + 2), and c
so (6; a) must satisfy the jth triangulating condition, so Sj (6; a) = Sj(q, (,)2). Since we have
specifically excluded the case (m, n)
may let
(x, y)
=
(4, 27) from consideration, we have m > b, and so we
=
2
(2c - 1, m - b) E N .
2
Then c2 m - b2 = c2 (b2 -+ a2 ) - b2 = b2c2 -+ ac(b2 + 1) - (ac - 1) = b c(ce + a) + 1, and
b2X + C2y = 2b2C - b2 + C2
(f)2= Sj (6; a), which implies that
y Q
so y(, (
D<m
D
- bC2 = bc(2b - c) + b2C(C + a) + 1 = bcj + 1,
+ ny
.
1+ 2c)m - bn
(c +a+2)b- c
(2c - 1)m + (m - b)n
=
(n
i
-
2 2
* (m, n) = f(b2 ,c2 ) - (a ,b + 2) with b = 1, and LC < D <
cm-b
(mI, c 2 m -+1). The lower bound was proved in Proposition 4.22.
bound, let j = cm - b. Then
(cf + a + 2)m - bn
(c + a + 2)b - c
.(n-c)
Note that (M, n) =
T b establish the upper
(j - 1)(j - 2) = (cm - 2)(cm - 3) = c2 iM2 - 5cm + 6 < c2 m(m - 1) = (m - 1)(n - 1)
and
j +2
2
_
(cm-+
mm12
c2 m 2
(m-+-1)(c 2m+1)
-+cM
<
2
(m+1)(n +1)
2
2
so (6; a) must satisfy the jth triangulating condition, which in light of Proposition 4.22, part
(ii), means that Si (6 ; a) = Sj (1, 1). Now let (x, y) = (n - c, 0). We have
b2 (n - c)
= c2 m +I1-
74
c> c(cm - b)
=
bcj,
so we must have (x, y) V Si (b,
()
=
Sj(J; a), which implies that
D < mx
+ ny
m(n - c)
_
cm-b
j
(m, n) = e(4, 25) - (1,6), and 2n"- D < 2n-5m. The lower bound was proved in Proposition
4.22, with (b, c) = (2,5). To establish the upper bound, let j = 10f - 4. Note that a > 3 and
D2n- = 10f
-
12
>j + 1
so (3; a) must satisfy the jth triangulating condition, so Sj(J; a)
(25f - 16, 1). Then
=
Sj(q, (q)2). Let (x, y) =
b2 x + c2 y = 100f - 64 + 25 = 10(10f - 4) + 1 > bcj,
so (x, y) V Sj
(,
()) = Sj (J; a), which implies that
mx + ny
_
DK
(25f - 16)(4f - 1) + (25f - 6)
10e - 4
2en - 5m
5f -2
The lower bound
n) = f(b2 , c2 ) - (a2 , b2 + 2) with b > 5, and n < D < (b-z+1)bn-3cm.
d - z + 1 c - 3bT
pb
was proved in Proposition 4.22. To establish the a upper bound, let
* (m,
j = (b - z + 1)c - 3b = bc - ab - 3 + c - 3b = b(ce - a) - (a + 3).
Note that a > 3 and
hn
D >,
C
b
= bc - ab - - > j + 1,
c
so (3; a) must satisfy the jth triangulating condition, so Sj(6; a) = Sj
()2).
Let (x, y) =
(n - 3c, b - 1). Then
b2x + c 2 y = b2 (c(ce - a) - 1 - 3c) + c 2 (b - 1)
2
= b2c(cl - a - 3) - b
-
c2 + bc 2
-3bc+1+bc 2
= b2 c(cL-a-3)
= bc(b(cf - a - 3) - 3 + c) + 1 = bcj + 1 > bcj,
so (x, y)
S
mx +
Dj
.
(,
()2)
=
Sj (6; a), which implies that
(m
m(n - 3c)-+-n(b - 1)
ny
.
=
-
1 + b)n - 3cm
j
(bf - z+1)bn - 3cm
(be- z + 1)c - 3b
m+n
, where T E {4, 5} and
1
+ m+n-,qwerr
eIf (m, n) = (Fk , F4k+4
3
333 ), then F4k+2 < D < F4k+2
condition,
third
triangulating
to
use
the
(mod 3). To establish the lower bound, it suffices
< D - 1, so that
noting that m + n = F4k+2. To establish the upper bound, let j = n37
2
(3; a) must satisfy the jth triangulating condition, which means Sj (J; a) = S (0 ; p4), since
One may check that
M is a fair approximation to the strongly triangulating ratio 9o.
(j,j) + (m + n, n - 8m) _
(X, Y)
=.-
3
75
4m + 4n-p 4n - 23m - 7
9
9
E N2
Bounds on D
M
n
2
15
3
22
3
4
29
43
5
36
107
6
43
7
50
9k + 8
64k + 57
16
11
<_yn
D < 19"m =7
8 =
22f
< D < i'm
8"
7m+2n < D < 17_+n
9
D < 2 5 +n
16 = 8" < D < 33m+n
6
7m+2n <
241
318
299 = 7m+5n < D < 241+3n
17
17
16
16
= 8m < D < 32(2k+)m+3n = 24k + 4 + 12(6k+5)
24k +
Table 4.2: Bounds on D-values for fair upper approximations to
6
Now 1 + p4 = 3(p 2 and
3p6 = 3p2(3
so
4 (1+p
y = (I +
X+
4
2 _ 1) =
4
)j+(m~)+yo
+(m
)+
so (x, y) V S
4
+ n(n
-m
+n + m(m + n) + n(n - 8m)
m+ n- 7
3
j
*
- 32 = 8p4 -
1,
) = Sj(6; a). Therefore,
D< mX + ny
since m(m + n)
4
2
(n-8m)
24
= Q (j + n - p 4 m) > p 2,
3
(p 2 ;
9
- 8m) = m
2
- 7mn
+
m + n1
3
m
+ n - r1
722 = 1
2 is a fair upper approximation to 6, and D is bounded as shown in Table 4.2. The table
also explains which triangulating conditions to apply, in order to establish these bounds. For
example, let us establish the upper bound in the generic case (m, n) = (9k + 8,64k + 57). Let
j = 24k + 20 < D - 1; we know that (6; a) must satisfy the jth triangulating condition, and,
since -" is a fair approximation upper to a strongly pseudo-triangulating pair, we know that
S4 (J; a) = Sj(8; j96 + e). Now let (x, y) = (32(2k + 1), 3), and note that
32(2k + 1) .32 + 3 82 = 3 .8 -j,
so that (x, y) V Sj(); i + ) = Sj ( 6 ; a). Thus,
mx + ny
_
32(2k + 1)m + 3n
24k + 20
as claimed in the Table.
n) = (m, 9m - 1), and 1 < D < "4m-6n2+1.
To establish the lower bound, use the third
- 1 =
triangulating condition. To establish the upper bound, let j = 3m - 2 < (3m is
a
fair
lower
D - 1, so that (J; a) must satisfy the jth triangulating condition. Since -L
m
approximation to a strongly pseudo-triangulating pair, we must have Sj (6; a) = Sj (3, 9 - E).
Let (x, y) = (3, m - 1). Note that
* (m,
3 + 9(m - 1) = 3(3m - 2) = 3j,
76
so (x, y) V Sj (3, 9 - E)= Sj (6; a). So
m + ny
3m
j
+ (m - 1)n
j
mn - n
+ 3m
3m-2
as desired.
2
2
It may be helpful to restate the bounds on D in a more uniform way, in the cases where Ic m-b n1
1 and m, n > 2. Instead of requiring that b ( c and m < n, let us reverse the roles of m and n, and
of b and c, and stipulate that c 2 m - b2 n = 1. The key fact that makes this possible is the equation
f (b 2, C2) - (a 2, b2 + 2) = (f - 7)(b 2, C2) + (C2 +2,dd2).
Rewriting things in this way, we have the following lemma:
Lemma 4.24. Suppose a, b, c are consecutive odd-indexed Fibonacci numbers, with possibly negative
indices. Let (m, n) = f(b2,c 2 ) + (a 2, b2 + 2), where f is any integer, as long as m, n > 2. Suppose
(D; -) is weakly triangulating. Then
bn
cm
c
b
(cf + a
+ a)cm -
bn
b
Oc
(c+a±)b-
cm
1
b
(cf+aa+a)b- 3c
where
(a,/) =
(1,3)
ifa>b>c.>5,
(1,2)
(1, 1)
if (a, b,c) = (13, 5, 2),
if a > b > c=1,
(5,2)
(2,1)
if (m, n) = (4, 27),
if a < b and (m, n)
$
(4, 27).
If D is rational, then its denominator is at least b.
Proof. The lower and upper bounds on D itself are proved directly above. To prove the statement
about the denominator of D, assume b > 2, since this statement is vacuous if b = 1. It suffices to
note that there is no fraction of denominator less than b in the interval
cm cm
1
b ' b
b['(c+a+a)b- Oc)
This is obvious if
1
b
_
_
_
(cf + a + a)b
_
-
_1
Oc
b- (b-1)'
which (one can somewhat laboriously check) is the case in every instance except when (z, a, b, c) =
(13, 5, 2) and f = -6. But in that case, our bounds are 71 < D < 7 , and there are no fractions of
l
denominator less than 5 in that interval anyway.
Proposition 4.25. If (6; -) is weakly triangulatingfor an integer m6 E N, and m, n > 2, then
(6; -) is in the following list, with one caveat: recall that if (6; -) is weakly triangulating,then so
is (m; 1); the list below may list just one of these two. Let z, a, b, c be consecutive odd-indexed
Fibonacci numbers, possibly with negative indices.
1. (6; n)
=
( ; 2), provided that b, c > 2,
77
2. (8; ') = ( ; f), provided that a, c > 2,
3(J; Q)
=
(1; "M
4. (6; ' ) =(2;
)
l),
4
5. (J; ' ) =(j; '),
6. (J;
a)
=(); ').
Proof. This follows directly from the results outlined in this section. Lemma 4.3 may be useful in
D
the verification process.
4.5
Weakly triangulating sequences of length 2
In this section we will prove some necessary conditions for a sequence (8; n-, '2 ) to be weakly
triangulating. The main results are Proposition 4.33 and Lemmas 4.34, 4.35, and 4.39. In fact,
computer tests seem to suggest that we do not have too many false positives, so our enumeration
can be considered as a reasonable step towards a full classification, if one is interested in classifying
these sequences. We will write ' = a- and g3 = mn + 2 throughout. Evidently, one necessary
condition is that (A; Q) be weakly triangulating, so the results of the previous section will be our
starting point. If one is interested in strongly triangulating or pseudo-triangulating sequences, then
the main results in this section are Lemmas 4.27, 4.28, and 4.37.
Here is one consequence of the results we are about to prove:
Proposition 4.26. If (8; "', 2) is weakly triangulatingfor an integer m 2 j E N, and m, n, m2 > 2,
then (6; n,
2)
is in the following list. Let z, a, b, c be consecutive odd-indexed Fibonacci numbers,
possibly with negative indices, and write (M, n) = (m 1 , n 1 ) for convenience.
,
1. (6;
26
2. (8;
'
2
2
b
= () ln +4
4n+
1 . n
' 4n+1
,
) = (C;.b'
MM2
3. (6;'rnm2
4. (8;
c2 + 2 ±2,),
(C;
bj271 2
1
bf
)
fl,
fl
+b+2
2T_
b~e+_a
2
)
g2) = (m
5. (6; n6. a( "a.)
b z),
for some f E Z, as long as m, n
(2n+1)
2
) as lon as
4n2+1.
mm -1),
1
2
(m +1. n-1
)
for some f E Z, as long as m, n > 2.
as long as m ) 3 and m
(m+1)asongasm 3
31)
10. 2
4mm22-1
7. (8; a,
)
8. (6- -n "-.
1 m2
n,
9. (6- m'
n2
m2
n2
10. (6- m'
- m2
-
(2m; 4m-1
17.
4T1
7
-2
17)
(F4k+2. F4k+4/3
3 ' F4k/3
(2F4k+2.
3
1), as long as m, M2
mm
1)
for some k > 2.
1)
1F4k+4/3
F4k/3
6 for some k >
78
2.
) 2.
2
> 2.
2.
Proof. We go through the lists in Proposition 4.33 and Lemmas 4.34, 4.35, and 4.39, and look for
examples where m25 is an integer. This procedure is finite and trivial in the case of the three
lemmas, among which only Lemma 4.39 yields any solutions, namely (9)-(10) in our list above. It
remains to consider the list in Proposition 4.33, so let (i, n) = £(b2 , c2 ) + (a 2 , b2 + 2). By Lemma
, and 6 has denominator at least b, so m 2 > b.
4.24, we know that J >
Consider item (a) of the list in Proposition 4.33. Recall from Lemma 4.27 that Si(i; , m) is
, then we must be in case (1).
=
strongly triangulating, so, by Lemma 4.3, if
, n) whenever
ST(
Sj,()
If - is a fair approximation to ', then it must be that Sj(J;
(J;
-E,
n2 = bfz, then we must be in case (2) above.
satisfies the jth triangulating condition. If M2
=b
0
T. Let us consider two cases:
n
*3
c~n~
Pmnm C
theS
for all j, so whenever
Sj (J; -, -)
,
)2
i;
, t
I f B2b <
M2nMiM2
n2)
MM2
Otherwise, m2 > b + 1 and
bb1M
MbmP
(6;
n,2) satisfies the jth triangulating condition, so does (C;
we must in fact have 6 = C, so b I M2.
,
2).
By Lemma 4.3,
If b = 1, then we are in case (4) or (7).
with
If b > 1, then we claim there are in fact no fair lower approximations to m = e +
has no fair lower approximation of the
denominator divisible by b. It suffices to prove that
in
<
<
, this follows from Lemma 4.21; all the fair approximations to
form 1. For
that interval are well understood. If c = 1, then nothing remains to be done. Otherwise, to
eliminate the possibility that - < 2, let us calculate (from the continued fraction expansions)
below '.
some consecutive fair lower approximations to
{1, 2} such that 3 1 z + ca, b +,Ec.
Let 6 be the unique number in
If a > b > c, then
a
c
(z + ca)/3
(b + ec)/3
are two consecutive fair lower approximations to
a
c
(a - z)/3
(c - b)/3
are two consecutive fair lower approximations to
If a < b < c and 6=2, then
a.
If a < b < c and e = 1, then
a.
(2a - z)/3
(2c - b)/3
(a - 2z)/3
(c- 2b)/3
a
c
are three consecutive fair lower approximations to a. In the three inequalities above, the
smallest denominators (that is, bhc cb, and c 32b) are all less than b, and none of the
denominators is a multiple of b. Therefore, b admits no fair lower approximations with
denominator divisible by b, as claimed.
If
then since M2 > b, (J; -, n) must satisfy the triangulating condition cm, so bg3 I<
b23 E~ n2m2;
cmD, where g3 = mn + 2 as usual. Then S(
m
(( g
,
)
(
; , n)2
(
93
so
does
for all j, so whenever (6; -, n2) satisfies the jth triangulating condition,
CMM
M
> 7
,
M2
mn
By Lemma 4.3, we must in fact have
bg 3
b(mrnm
cm
2
+ n2)
cmm2
79
so cm I b(mnm 2 + n2), and in particular m I n2, so n2 > m. But if n is a fair upper
approximation to ' with n2 > m, then it must be the case that b2n 2 - mm2 = 1, and so
m { n2 after all, a contradiction.
Now let us consider item (b) of the list in Proposition 4.33. If
Q
= (beMz) 2
then by Lemma 4.3 we
must have m26 = bmn+U+z and c = 1, and we are in case (3) or (5) above. If a , (bf+Z)2, then by
inspection, only two of the sporadic examples has integral m26, namely cases (6) and (8) above.
E
Finally, item (c) of the list in Proposition 4.33 is easily eliminated by inspection.
4.5.1
Cases of the form n' =
EF2
+F2+ 2
Fk+1Fk±
) is strongly
Lemma 4.27. If a, b, c are consecutive odd-indexed Fibonacci numbers, then (c,;n, n
triangulating when (m, n) = f(b2 ,c 2 ) + (a2 , b2 + 2) for some f E Z (provided that m, n > 2).
-,22) is weakly triangulatingwith m2 > 2, then 2 > F 1 -1.
Moreover, if (D;
M~
M2
M2
T7-L
Proof. Proposition 4.6 says (2; 2) is strongly triangulating. It follows from Proposition 4.16 that
( ; , g) is also strongly triangulating, because c 2m - b2 n = 1.
Now suppose for the sake of contradiction that (D; ", 2) is weakly triangulating, with m2 > 2,
but there is an integer x such that
n2 <
a2
n
Let us note that
a2]
x-<
-1 =
6
if a > b > c,
3
0
if (a, b, c) = (2, 1, 1),
if a < b < c.
We may check that, in each of the three cases of (a, b, c) appearing in the above equation, we have
a(b2 + 2)- b2
2cn - b2(c+ a +1)
b2 c
b2c
and
b2 - c + 3
n - c +1
Let
cx + cf + a + 1 = cx +
and:
n + c- 1
c
bi = bcx + b(cf+ a + 1) <
Note, by Lemma 4.27, that D > ',
b '
so
j < 2D < M 2 D,
and
293 - jD
>
2(mn + x) - cm cX+n+c
=
2mn - m(n + c - 1) + (2 - c2 m)x
=
m(n - c + 1 - c2 x) + 2x
2x > 0.
>
80
c
) is required to satisfy the jth triangulating condition.
Thus, (D; ',
On the other hand, we claim that
n n2\S
(b2 xn + (c - 1)m, 1) E Si (D;
cm
m
Indeed,
cmi = cm
C + n2+ c
c 2 mx+m(n+c-1)
)
=b 2 nx + (c - 1)m+ (mn + x),
so
b2 nx + (c
-
1)m
93 < cmi < jD
+
and
b2 nx + (c
1)m + (mnn +
-
> cmi =
J
cm
b
cannot be triangulating, a contradiction.
So we conclude that (D; -,
f)
' M' M2/
The case (m, n) = (2, 3) is special:
Lemma 4.28. The sequence (10; 2, 46 + E) is strongly pseudo-triangulating.
Proof. Note that
10.3 46
Si
-;
={(w,z) ENx N| w# 1 and (9w+100z<30j or 3w = 10j)};
, -2 +E)
we have to show that
j+ 2
2 )
3 46
;'2' 9
#S10
#
for all j E N. Let us induct by 10's. The base cases 0 <
for j - 10, then since
z > 3 and (w, z) E Sj
3 46
(0.
j
< 10 can be checked by hand. If it holds
+ E)*, 46 + E
1;
(w,z - 3) E Sj-o
10(j - 10)o)
and (w, z)#
3
there are
(j -10)
+ 2)
J
10
elements (w,z) E Sj
('; , L
z=O
30j - 100z
L
9
_
-I
0j-1 +
i
3
3
if j
0
else
1
(mod 1),
+ E) with z ;; 3. Meanwhile, there are
+ (1
li
2
1
- 12)
-
10'-13
3
23) = 10j - 35
10-13
-22 )
10j~p-2
3
-22)
if j
0
(mod 3),
10j - 34
if j
1
(mod 3),
=10j - 35
if j
2
(mod 3).
elements (w, z) E Sj (L2;
3 '2' , 4-9 + e) with z < 3. So, in all cases, we see that the total cardinality of
Sj (01;
.,
+ 6) is
(j - 10) + 2 + 10j - 35 =
10
((j
-
10) +2)
10
and induction is complete.
81
-1 + 10j - 34 =
j+ 2
10 ))
Corollary 4.29. If (D; 2
then
2
or, equivalently, (D;
'2)
,
2),
is weakly triangulating and m 2 > 2,
.
is either a fair approximation to 3 or a fair upper approximation to
Proof. We know from Lemma 4.27 that (3; 2, 3) is strongly triangulating, but that is the same thing
as saying (3; , 3) is strongly triangulating.
If m2 < 20, then a computer test verifies that the corollary is true. If m 2 > 20, then
30D < 30- 4 < 20 -6 <2093
m2g3,
), so (D; i, 2) satisfies the jth triangulating conditions for all
S 3 o(D; 3,
j < 30. Then a computer test shows that either 29 < Mn22 < 10 or 49 < M2
n
< 1.
67~ In the first case,
must be a fair approximation to 3, by Proposition 4.19 and
n2 < 4, and "2
we have a fortiori 5
the fact that 3; , 3) is strongly triangulating. In the second case, we have a fortiori 5 < m2 < 3
we have (0, m 2 )
and -
must be a fair upper approximation to 46, by the Lemma and Proposition 4.20.
E
Lemma 4.30. Let z, a, b, c be consecutive odd-indexed Fibonacci numbers, and let f E Z. Let
2
2
(i, n) = (mi, ni) = f(b2, c 2 ) + (a 2 , b + 2) and (m2, n2) = (m, (bf + z) ) = (m, (f + 7)m - n) and
bmn+b + z
cm
93+n
c(bf + z)
Suppose that m, n > 2. Then:
i. (D; n,
n
) satisfies the jth triangulating condition for all
j(
ii. In fact, (D;
,
&)
mg
cm 21
b
C
(b+
z)(m -1)
1)
Dg3
satisfies the jth triangulatingcondition for all
j < c(bL+ z)m,
except for those j of the form
j = c(b + z)m - bi,
for integers i such that 1 ( i <
C =
+ a+
1
C,
Sj(
)
D; In2 )=
Sy;n
m m2
in which case
( j + 2)
2
-21 .
Proof. Note that be+z = m+1 > 0,
b and ct +a = n-1
C > 0. Note also that -7b
so f > -6 in all cases, and
rn> -6b 2 + a 2 = b(z - 6b) - 1.
2
+a 2
-
(c 2 +2)
< 0,
i. First we claim that (i) holds when m = 1. In that case, we can easily see that a = 1 and
f = 0. (The cases where c = 1 and f = -6 fail because they force n to be 0, which we are assuming
2
is not the case.) If (z, a, b, c) = (2,1,1,2), then n = 3, 93 = 7, D = 5/2, and cm /b = 2. If
2
(z, a, b, c) = (1, 1, 2, 5), then n = 6, g3 = 7, D = 13/5, and cm /b = 5/2. In both cases, we can
directly verify that the jth triangulating condition holds for all j < 2, which is enough.
So we may assume that m > 1 below.
82
Step 1: under the assumption that m > 1, we claim that the jth triangulating condition for
holds for any given j < c(bf + z). Recall that (--; ", ') is strongly triangulating and,
2)
(D; -,
therefore, satisfies the jth triangulating condition. It suffices to show, therefore, that #S, =
where
Sj :={(X,Y,Z) E N3 Y <m and mX+nY+g 3 Z <jD},
Tj :=
(XYZ)E N3 Y
.
Z<
m and mX + nY + (mn+
The equality of cardinalities obviously holds if the sets themselves are equal, so let us investigate
the set differences. Suppose that (X, Y, Z) E Sj\Tj or (X, Y, Z) E Tj\Sj.
, it must be the case that (X, Y, 0) E Sj\T, and
Case 1: Z = 0. Then, since D >
< mX + nY < (
so
+
j=jD,
cm
mX + nY
cm
1
b
j
b
bcm
But c + 1 = bmn+b+z is a fair upper approximation to ',
we're assuming that j < c(bf + z), this implies
so it must be that
j
> cm. Since
b2 f + a2 = m < be+ z.
2
We claim this implies b = 1. Indeed, if b > a > 1, then a > z and, since m > 0, we must have
£ > 0, so b2 f + a 2 > bf + z. And if a > b > 1, then one can easily check z > 6b; then, since f > -6,
we have
-6b 2 + a 2 = b(z - 6b) - 1 > z - 6b,
so -6b 2
+ a2 > -6b + z, and a fortiori we have m > bf + z. So indeed this proves that we must
have b = 1, which is to say (z,a,b,c) E {(5,2, 1,1), (2,1,1,2)}.
If (z, a, b, c) = (5,2, 1,1), then we have j < c(bf + z) = m + 1 and n =
M <
1 and
1
mX + nY
-
in-
.
M + -,
so there is just one possibility:
(X, Y,Z; j) = (2, m - 1, 0; m).
If (z, a, b, c) = (2, 1, 1, 2), then we have j < c(b- + z) = 2m + 2 and
cm
2m=b <
mX+ nY
cm
b
+
1
bcm
2m+
1
2m
so there are two possibilities:
(X, Y, Z; j)
{(5, i - 1, 0; 2m), (7, m - 1, 0; 2m + 1)}.
Case 2: Z > 1. Then the mere fact that (X, Y, 1) E Sj U T implies that (0, 0, 1) E SJ U T, so
at the very least
mn < jD.
83
Since c(bf + z) - 3 = b(ce + a) > 0, a little calculation shows
mn - (c(bf + z) - 3)D = (b 2 m
So it must be that
j
mX + nY< n =
cm
= c(be + z) - E, with c E {1,2}. Note that
mX + nY + g 3Z < jD
so Z
)(c+a)+am>0.
mX + nY < (g 3 + n) - ED - Zg 3 = n - cD - (Z - 19 3 ,
mX + nY + (mn + )Z <
1mm
M)m
b+
- (Z - 1) mn +
< n - (Z - 1) (mn + M ,
1 and Y = 0, and we can only have (X, Y, Z) E T\Sj and
Ecm
sob
b
1
b2PmX
FC
Ec. If not, then as gcd(b, cm)
and
n
=
+ bcm'
v,
__[EM+
bcmb
lb
T2
E
b
e
11
1c
b
We claim that b
Ecm
1, we also have b
c
{ Ecm,
and b > 1, so
1
bcm
b
which is to say 1 < cm < E. But then E = 2 and c = m = 1, so the properties of Fibonacci numbers
are such that b E {2, 1}, and in both cases we indeed have b I E 2, as claimed.
If b = 2, then c = 2, and we have two cases for (z, a, b, c). If (z, a, b, c) = (13, 5, 2, 1), then
m
-
1
-
4
-rnX
<m +
1
-.
m
If, on the other hand, (z, a, b, c) = (1, 1, 2, 5), then
1
5m - - < n - mX < 5m
4
+-.
1
5m
Both cases are impossible under our assumption that m > 1, since ged(m, n) = 1.
If b = 1, then again we have two cases for (z, a, b, c). If (z, a, b, c)
(5, 2, 1, 1), then n
and
em - 1 < n - mX = m(1 - X) - 1,
which implies c = 1 and X = 0:
(X, Y, Z; j) = (0, 0, 1; m).
If, on the other hand, (z, a, b, c) = (2,1, 1, 2), then
2cm - 1
4m - 1 - mX < 2Em+
,
or in other words
0 < (4 - 2E - X)m < 1+ 2m
84
2.
=
m - 1
Since m > 1, we must have X
=
4 - 2c, which implies
(X, Y, Z; j) E {(0, 0, 1; 2m), (2, 0, 1; 2m + 1)}.
At any rate, we have #(Sj\T)
= #(T\Sj)
in all cases, so #Sj = #Tj, as desired.
Step 2. Having established that the jth triangulating condition holds for all 0 < j < c(be + z),
we induct in increments of c(be + z). Suppose that c(be + z) < j < E,
and the jth triangulating
condition holds for j - c(be + z). Note that for Y, Z > 1, we have
mX + nY + g 3 Z < jD -=>
mX + n(Y - 1) + g 3 (Z - 1) < (j - c(b + z))D.
To ward off duplications, let us stipulate that 0 < X < n. To show that the jth triangulating
condition holds, it remains to show that the number of integer solutions (X, Y, Z) to
0 < X < n,
min{Y, Z}= 0,
mX + nY + g 3 Z < jD
(4.5.1)
equals
)+
-j
n.
(j - c(bf + z) + 2) + --- + (j + 1) =1 + jD - (m-1(
2
cm
To that end, note that there are n solutions to (4.5.1) with Y = Z = 0; namely, X could be any
integer from 0 to n - 1. So it suffices to prove two things: First, the number of solutions to (4.5.1)
with Z = 0 is 1+ LjD] _ (m-1n-),
and second, the number of solutions to (4.5.1) with Y = 0 is
2
]. These are steps 3 and 4 below.
Step 3: we claim the number of solutions to (4.5.1) with Z = 0 is 1+ LiD_ (m 1n-)
. Indeed,
these solutions are in bijection with the elements of the semigroup generated by {m, n} within the
interval [0, jD]; since
jD > c(be+ z)D =-g3 + n > (m - 1)(n - 1),
exactly (M
n-
integers in that interval are missing from the semigroup, as required.
Step 4: we claim the number of solutions to (4.5.1) with Y = 0 is
subtle. It suffices to prove, for 0 < X < n and Z E N, that
X +nZ< -- j <->4=
cm
Let r = bnj - cm(X
+ nZ), and substitute j =
cm(X+nZ)+r
to obtain (after some computation)
m
CM=b
mD
+ z
This is a bit more
mX + g 3 Z < Dj.
bn(mX + g 3 Z - Dj) = n(bg3 - cmD)Z - m(cD - bn)X - Dr =(U +
and note that
Fgj1.
z)nZ - (bf + z)X - Dr,
bn
c(bC+ z)'
so our task is to prove
r > 0 <=
Note that r = -cmX
nZ < mX +
cm -
bn
c(bf + z)
r.
(mod n), so
X+cr= (b2 n +1)X+cr=c2 mX+cr =0
85
(mod n),
(4.5.2)
and in particular, the key point is that
X + cr = 0
|X + crI > n.
or
Let us consider two cases deDending on r.
Case 1: r > 0. In this case evidently X + cr > n, and (since we are assuming
j (
)
cm(X + nZ) < bnj < cm - mn,
so Z < m. Since
bn< bn +3c-b=c 2 (bf +z),
we have
nZ ( n(m - 1) <
bn
(m -
± ) n
c(bf + z))
2(
bnz
c2(bf + z))
so (4.5.2) holds.
Case 2: r ( 0. Thanks to our lower bound on
j
cmn
-
r,
and upper bound on X, we have
bn - c(b+ z) - cm(m - 1) -0
cmn
Z = bnj - cmX -r
bn
+
(cm
(X + cr) ( mX +
2cn >0
cmn
so Z > 1. We distinguish between two sub-cases:
Case 2a: - < r < 0. Then -nm<X-n<X+cr(< X <n, and X+cr = 0 (mod n), so we
must have X = -cr. Since Z > 1, we have
nZ-mX-
(cm
bn
c(bf
>
+ z)
m+cmr-cmr+
> n-
bn
r
c(b + z)
>0,
bn + (3c - b)
so (4.5.2) holds, as required.
Case 2b: r < -. Then the mere fact that X < n implies that
nZ -
- (mX
cm-
bn
+r >
c(be+z) I
>
n-mn+m+mn-
n
bn
n-m(n-1)+ (CM
c(be+z) ) c
bin 2
2
(b
)
-n+mbn2
>0
bn + (3c - b)
so (4.5.2) holds, as required.
ii.
In light of (i), we may assume
j
= c(be+ z)m - i, for some 0 < i < c(be+ z). We know that
j - c(be + z) < c(be+z)(m - 1) K
cm2
2
so by (i) we have
Sj-(U+ z) (D;
j - c(be+ z) + 2).
1nIn2
M M2
S2
86
In the proof of (i) above, the only place we used the upper bound on j was step 4, case 1, where we
needed to show that Z < m. So, in the case at hand, it suffices to find all pairs (X, Z) E N 2 with
0 ( X < n and Z > m, such that (4.5.2) fails, which is to say, more precisely,
r := bnj - cm(X + nZ) > 0
and
nZ>mX +
cm-
bn
c(bf + z)
(4.5.3)
Well evidently, for this to be true, we must have
Z<
so we must have Z
=
bjb2b
j= b(b + z) -b2-=M
cm
+1-
m
m exactly. Plugging in for Z and
j,
m
we have
r = cm(n - X) - bni
and
nZ-mX-
cm-
bn
c(bf + z) )
r=
((c 2 m(be+ z) - bn)i - cm 2 (n - X)),
bn
c(bt + z)
so we are just looking for cases where
c
b
Now note that
to 2. Therefore,
-
1
bn
cm
bcm
n-X
i
K
c2 m(b
+z)-bn
cm 2
c
1
-+I-M
b
bcm 2
1 is a fair lower approximation to 2, and 2 + is the only fraction in the interval
c
1
C
1
bcm'b
b
is a fair upper approximation
bcm2)
with numerator (when put in lowest terms) at most bn. Since evidently
n - X < n < bn,
we must have
n-X
which is to say, X = n required.
.
So indeed (X, Z)
=
_c
(n - 7,m) is the unique pair satisfying (4.5.3), as
D
Lemma 4.31. Let z, a, b, c be consecutive odd-indexed Fibonacci numbers, and let f E Z. Let
(i, n) = (mi, ni) - f(b2, c 2 ) + (a 2 , b2 + 2), and suppose that m, n > 2. Suppose that (D; L, n2) s
a weakly triangulating sequence with m2 > 2 and
max
bmn+b-+z
g3+n
cm
c(be+ z)
}
b
-
cm
_ mn = bz,b
and the
Then of course -M2 > bmn+bt+z
r true.
re
h following
olwn are
b
results.)
the
for
(x)-(xiv)
(Parts (i)-(ix) are needed for the proof only. See
where g = mn
+
n2-.
M2
i. For every integer j with 0 < j <
D; _,
m I
M2
Cmb
, we have #Sj (D; ',
<(%
).
That is to say, if
fails the jth triangulatingcondition, it can only be because Sj (D;
small, not too large.
87
, P-2
MI M2)
is too
ii. The number m2
=
b2 +bz+1
b22
and (bt+z) 2
2
is a fair lower approximation to (be+z)
= f+7- C
2
(be7)
haM
Moreover, (b+z has a fair lower approximation with
is a fair upper approximation to '-+-.
denominator b, namely bez if b > 1, or f + z + 1 if b = 1.
i.
There is no fraction I
strictly between
p
iv. If m 2 < m and
'2
M2
<
(bU+z)
m
2
and (bU+z)
with p < min{m, m2}.
m
2
M2
2
m+= b2t+z+
then m 2 < b2 , and -2Ml2 is either m+
b b±
2
approximation thereto. If m2 < m and
2
>
2
(bt+z)
then
or a fair lower
is a fair upper approximation
n2
to M.
v. If m >, b, then for each integer i with max{{, p} < i < min
have
{
c(
+b)c(b£+ z) - 1}, we
icm - bn > (ib - c)D.
In particular
n2
b£+z
c(b£+z)
M2
b
b(ib - c)
}, we have
+
vi. For each integer i with 1 < i < min{bf + z - 1, "
i(bmn + bi + z) - cm > (icm - b)D.
In particular,
n2
2
b£+z
_(b£+z)
m(icm - b)
m
M2
vi. Suppose i is an integer with 0 < i < min{m
i <m 2 -b-1. Then
-
2,m
-
b,m
2
1, in fact
- b} and, if a
i(g3 + n)+ bmn+bf + z -cm > [ic(b + z)+ cm - b]D.
In particular,
(be + Z)2
nn
1
m
m2
zcm
viii. Assume that m2 > m and i E N with 1 < i
Then there exists exactly one pair (W, Z) E
g.
n
Sc(bU+z)m-bi (D; I-, r ) with
0
m~ + - (bf+
0 <Z)2(b
ci) -W
m2n+(b+z)2+m(n
-
(mn+
+ Z)2
Z <
.
If i < c, then moreover
(W47, Z)
ix. If m2 > m and bf
E
{(m(n - ci), m), (m(n - ci) + m 2 n + (bf + z) 2, 0)
+ z >, 2 and m > max{b, 2}, then
(m(n - ci), m), (m(n - ci) + m 2 n + (bf + z) 2 , 0) V S(bU+z)cr-ib (D;
for each i E N such that
.
M
m - max{b,2}
88
(bf
+ z)cm
b
n2
M, I2)
n
+
m2 < b, then 2 =
with 3M2 = a (mod b).
X. If
xi. If b
m2 < m and
, which implies m2 is the unique integer in the interval (0, b)
n
2
< (bt+z)
in
n2
Mn2
then these are the only possibilities:
-
(z, a, b, c) = (34,13, 5, 2), with
-
(z, a, b, c) = (13,5,2, 1) and
f E {-6, -5},
4
2
21 >
xii. If b < M2 < m and M2
xiii.
then c
(bt+z)2 ,
in
6 41
2
4q27
-
n
and
1 and
-
2
= (be+z) 2 -(b f+bz+1)
n2
rn-b
M2
If m 2 > m and c > 1, then these are the only possibilities:
-
(z, a, b, c) = (34, 13,5,2),
-
(z,a,b,c) = (2,1,1,2),
= -6,
f = 1,
-
in
-L = 19-1
'm
and
= {2
and
n2
Mn2
- (Z, a, b, c) = (1, 2, 5, 13), f = 0, .2 = 2-, and
27 .
-
Mn22
E
{ 133T
17
n2
E
5$1}
2
}.
T
xiv. If M2 > m and c = 1, then these are the only possibilities:
-
n
m
n
-
2 and 1-
3
= 3
.
is a fair upper approximation to
Mn29
and
2=
31
Proof. i. Suppose for the sake of contradiction that, for such
#Sj
Then, since (D; -,
n)
D;
,
(j
2)>
m MM2
2
j,
we had
2)
is weakly triangulating, and
D >,
bmn+ b£+z
cm
1
> j ,
bcm
m2
cm
b
=--+
it must be the case that
{(WZ) E W2 x N I W+Zg3 <m2g3} < (
2).
In particular,
D;
(0, m2) E Si
.n
,
Now the assumption that cmD < bg 3 tells us that
cmD
-m
b
which contradicts our assumption that j <
ii. Note that
2
<rm2 g 3 (jD,
'
(b + z)2
m
.
(bz + 1) + z 2
b2 £+a 2
+
so
(bf + z)2
m
(e+bz+1
2
b
_
(bz + 1) + z 2
2
b e+ a
bz +1
2
2
b
89
(bz)2 - a2(bz + 1)
b2 m
1
b2 m'
(bz) 2 are fair approximations to each other. If b = 1, then this is all there was to
+and
so 2
prove for part (ii). Otherwise, note that in all the cases we have to consider m > b - 1, so if b > 1,
then
1
m+1
bC+ z
b+ z
(b + z)2
2
bm
b(b - 1)'
bm
b
m
which easily implies that -+z is a fair lower approximation to (be+z) 2
Indeed, if there existed some
2
p < b, then
and
fraction I
anpbte
m
b < p < (bE+z)
p with bt±z
1
bp
q
bC+ z
(bf + z)2
p
b
bm
_
1
b(b - 1)'
b + z
b
which is a contradiction.
iii. Suppose for the sake of contradiction that such a fraction Ip exists; without loss of generality
we may take p minimal, and I to be a fair approximation to (be+z)2 Let us consider two cases,
2
depending on whether - is less than or greater than (be+z)
m
M2
Case 1: We have
n2
m2
q
(bC+z)2
P
m
This inequality, along with our assumption that bmn+bU+z < D, implies that
cm
sj bmn+b+zi n
for all
say,
i =
j
2
)
(b+z)
Si(D; n , n2
E N. Let i E N be smallest integer such that W := i(bmn
mnp+q
bmn + bf + z
Then let
j
< [(mn+ (bf+z) 2
<M
-
bmn + bf + z
=
[
- +
b
b]
+ bf + z) > mnp + q; that is to
+1
+=z
P
bm(bmn + b + z)
<-
b
b
b + z.
= icm < (bf + z)cm. Note that cm = b2 n+1 > n, so
(bf + z)cm - j
_
(bf + z - i)cm
b
b
It follows from Lemma 4.30, part (ii), that the
cannot be a positive integer less than or equal to 4.
C
jth triangulating condition holds for (bmn+b+z. n (bf+z) 2 ) Now we have
W=j- bmr-be+z
cm
jD,
so
in (b+z))2 csjD
(Wo) E sJbmn+b+z
that
and, due to our assumption
2
<
M2
(W - (mnp + q), p) E Si (D;
< (e+z)
n
n2
2
m
, n2 ) \Sj (bmn+be+z;z
CM
m MM2
90
(b
mn
m
2
.+Z)
ThusI # (D
M, )
, so by part (i) we must have
(2
>
m2 <
Since
bf +z b< M2
approximation to
<
p
bi.
, it follows from part (ii) that
(bf+z) 2
n
M2p
lies between - and some other fair lower
having denominator at least b, so by Corollary 4.14,
m2
Thus, b
=
-
cm
> b+ p.
+ p < m2 < bi, or in other words
p
b +1
p1-i]
b
'
i
which is patently absurd.
Case 2: We have
(be+z)2
m
q<2
p
m2
Let
so
bmnc+b+z
tha t
2
cm
j = pc(be + z) < (m - 1)c(b + z) <
b
satisfies the jth triangulating condition, by Lemma 4.30, part (i). Note
(bT1z) 2
2
=p(be+ z) - bm=+ be
m
np + p (mn h (be+z)
+
bmn ± bf + z
z
cm
and np + pg3 < jD by assumption, so
(np, p) E Si (D;
n S4
bmn + bf + z n (be+z)2
(
m
cm
-,
m
)
.
Now p < m, so
(be+z)2
bmn+be+z
np-mn+h
-=
m
in
2
(p+1) (mn + (be+hz)
+
so
(0,P +1)
V sy
n (be+ z) 2
m
bmn + b +z
cm
M
A little thought shows that, due to the assumption that
Sc
bmn+be+z n (be+z) 2
m
i -n,
2
But since (bU+z)
< Ip <
m
cm
n2-,
M2
M
m
m)
(bf+z)
m
C
S
2
<
n2
M2
D;
we have
.
,
M
. bn+be+z
m
M2
we have
n
((m + 1)np + q, 0) = (mnp + q + np, 0) E Sj (D;
n2)
M IM2)
91
2
bmn + be + z. n (be+ z)
cm
-
m
I
m
.
Thus,
# (D; n,
> (j+2), so by part (i) we have
n)
p(m +1)
pb(bf+z)
m
bj
cm
m2<
P+-<P+1,
m
m
which contradicts our assumptions.
iv. Assume m2 < m. The case where
so suppose n <
m 2 < m, we must have
>
2
(bt+z)2
follows directly from parts (ii) and (iii),
+
Then by (ii), (b-z) is a fair upper approximation to
so, as
b 2 e+bz+1
n2
By (iii), then, we must have m 2 = min{m, m2} < b2 , and
mation thereto.
is either
-
2
-+bz+1 or a fair approxi-
v. Let j := ib - c. Note that
c(mm2 + b)
.
.
_
_
cmm2
cm 2
b
b
b
1
bmn~bf+z. n
so Lemma 4.30 tells us that {Cm
(bf±Z)
(n
.
2
)
satisfies the jth triangulating condition. More-
over,
(c 2 m 2 + 1)b
cm
bmn+be+z
cm
and
c(b+
ic-+i
z)
- c(b + z)b
bmn+bf+z
cmm 2
cm
b
so that
(b +z)
bmn+b+z n
(icm - bn, 0), (0, M2)
< icm - bn,
cm
cm
Si
CM
i m
2
m
even though
icm - bn = (ic - n)m + (m - b)n E W 2 .
Let us suppose for the sake of contradiction that icm - bn < (ib - c)D, so that
(icm - bn,0) E Sj (D;
n (bf + z)2
Sj(bmn+be+z
n n2
M, M
m m2/
,
;
\SCM
.
Then part (iii) implies that
bmn+b +z.
Si
C
n (bf +z)
;- M
M
2
)
C Si
D;
,
,
M M2)
which contradicts part (i).
So indeed icm - bn < (ib - c)D, so
r2
-2-
(Ja\icm
<c(b+z)D---n-n<c(b+z)
M2
.
-bn
ib - c
as desired.
92
-b(be+z)
- i(be+z)
ib -c
c(bt+z)
b +z
= +
,
b(ib -c)'
-b
vi. Let
j
:= icm - b, and note that
j
;
=
min{(b+z-1)cm - b, cmM2
cm 2 - [cm(b - 1)
min
+ b2]
cmm2
b
so Lemma 4.30 tells us that
cm - min{m, m2}
b
b
satisfies the jth triangulating condition. Note,
(+z)
emnc+bt+zn
<
too, that
b(bmn + bl + z) = (b2 rn)m + b(bf + z) = (c2 m
so
1)m + (m + 1) = c2 m 2 + 1,
-
bm - +be+z =i(bmn+b + z) - b(bmr+be+Z)< i(bmn+b+
z) -cm,
cm
cm
and
bmn+b+z. n (bf + z) 2 )
(Zi(bmn + bf + z) - cm, 0) V Sj
; -- ,
.
Likewise,
bmn+b+z
'
cm
,; min{m, m 2} . (mn
+ b)
< min{m, m2} -
so
(0, M), (0, M 2 )
(b
(mn
+
+z)2 )
,
(bf + z)2
SM n ,
( ~(bmn+C b+z
-
Let us suppose for the sake of contradiction that
i(bmn + bf + z) - cm ; (icm - b)D.
Now bn + bl + z = dm, so
i(bmn + be + z) - cm = [(b - 1)(n - 1) + 2c - 1 + (i - 1)(bn + d)]m + (m - bi)n E W 2 ,
and
(i(bmn + bf+ z) - cm, 0) E S (D;
-n- -n2
M M2
)
Then part (iii) implies that
bmn
Sj
+ bf + z n (bf + z)2)
cm
m
m
; Sj
D; - , -2
M
M2)
which contradicts part (i).
So indeed
i(bmn + bf + z) - cm > (icm - b)D > (icm - b)
+
b3
c(bso+-z)'
so
(b + z)2 )
(icm - b)
(M2
m
2
<b (m+ 1)n+
m
_be + z mM
93
-(be+Z)
c2 (b + z)m
'
vii. Note the claim is vacuous if m2 < b, so we may assume that m 2 > b.
Let j := ic(bf + z) + cm - b, and note that if i < m - b then
j < c(bf + z)(m - b - 1) + cm - b
= c(be+ z)(m - 1) - c - b < c((b
cm2
+ z)(m - 1)
whereas if i = m - b, then (by assumption) b > 1 and
j = ic(bl + z) + cm - b = c(bC + z)m - c - b,
which is not of the form c(be + z)m - i'b for any i' E N. In either case, Lemma 4.30 says that
bmn+b~z.
(bez)2 ) satisfies the jth triangulating condition.
Note that
S-
bmn+be+z
cm
.
=Z
(b(be+
z)n +
(be+z)2'_
Z2+bmn+bC+z
m
)cm
=i (mn + (be+z)
2
m
= (i + b + 1) (mn +
<(i+b+1)
(mn+
+n
-
(be+ z)2
cm
be + z
1
-cm-cm
m
(be+ z)2
+ z)2
- (m - )n .- (b +
m
<(m2+1)
m
so
Sj
-
,
(mn+
n (be+z) 2
bmnr+b +z
(in + bmn + bf + z - cm, i), (0, M2 +1)
n
I
+bmn+b+z-cm-
/
(b + z)2 )
cm
and a fortiori (0, m 2 + i) V Sj ( (bmn+bt±z.
b(bmi+bl?+z)
' m'
CM
m
/
2
(be+Z) \
(m)2m
Suppose for the sake of contradiction that the inequality
i(g 3 + n) + bmn + be+ z - cm > [ic(bf + z) + cm - b]D
fails for some i < min{m - 2, m - b, m2 - b}, and let i be the minimal number for which it fails;
1. As
thanks to part (vi), we know that i
in + bmn + b? + z - cm = ((b - 1)(n - 2) + 2(c - 1))m + (m - b + i)n E W 2 ,
we have
\Sj (bmn+b+ z n )
(in+bmn+b +z -cm,i) E Sj (D;
\SCM
Then part (iii), along with the fact that (0, M 2 + i)
5
bmn+bU+ z n (b+ z)2
cm
CM
Then, since (D;
f)
n,
M' M2
7m
cmM
n
mI
-m
Si (bmn+bt+z.
(
cm
C
z) 2 )
(b+
mn
(
_+z)2
) ,, implies that
Sj D; n-, n2
)
m MM2)
is weakly triangulating, by Lemma 4.2, it must be the case that
{(W,Z) EW
2
x Nj W+Z93 <m 293} < (
94
+ 2)
2
j
and in particular
(D ; M
(Om 2) E Sj
, -2
M2)/
Thus,
m293
jD = [ic(bl + z) + cm - b]D.
<
By minimality of i, we may assume
(i - 1)93 + bmn + bf + z - cm > [(i - 1)c(bf+ z) + cm - b]D,
and recall that
bg 3 = bmn+b - -2 > bmn+ b + z.
M2
So it must be that
(m2 - b - i + 1)g3 + cm = m2g3 + (bmn + bf + z - bg3) - [(i - 1)93 + bmn + bf + z - cm]
< jD - [(i - 1)c(bf + z) + cm - b)]D + cm
=
which, given that i
<
c(bf + z)D,
m2 - b, implies
D; n,
(cm, 1) E Sc(bf+z)
and if a
=
1 then in fact i < m2
n2)
b - 1, so
-
D n' n2
(cm, 2) E Sc(bf+z)
D;
m2 - b, we have m2 > max{m2, b}, so
Recall that, under the assumption that 0 < i
cmm2
,2.
> cm - max
> c(bi + z)D,
1,
and it is easy to see from our original assumptions on D that
Sc(bt+z)
(bmn+bf+z
cm
n (b+
z)2)
m
C Sc(bU+z) (Dn
,
so part (i) says we must in fact have
Sc(bf +z)
bmn+bf+z n
m
cm
(b + Z)2
M
Sc(bf+z) (Dn
,
If a > 2, then it is not hard to see that
cm =-
b2 n+1
_
-
(ac - 1)n +1
C
C
-
1)
so we must have
cm~mi-i-(be+z) 2
+2>
cm+ mn+
M
n+mn+
(bf+ Z)2
M
95
= c(bL+ z) -
bmn+bf+ z
cm
so
n (b_+Z)2
(bmn+b+z
(cm, 1)
Sc(be+z)
b
=
(b
mi
Sc(be+z) (D; n)2
which means
cm + 93 > c(be+ z)D,
a contradiction. And even if a = 1, it is clear that
cm + 2 (mn +
(bf + z)2)
m
so
2
> n+mn+ (be+ z)
m
)
bmnr+be+z n (be+z) 2
(cm, 2)
SC(bt+z)
cm
m
=
M
Sc(bf+z) (D; n
2
again a contradiction.
So indeed we know that
i(g3 + n) + bmn + b + z - cm > [ic(b + z)+ cm - b]D,
and recall that
z
D ;bmn+ bf +
cm
so we have
i(g3 + n) > [ic(bf + z)+ cm - b] .
Z)2
= i (mn ++(bf +
m
bmn + bf + z
- (bmn + bf + z - cm)
mcm
+n
+
CM-
which implies
(b + Z)2
n2
1
M
m2
icm
viii. Let's write
Q := m2n + (bf +
z) 2 + m(n - ci) - W - (mn +
(b + z) 2
)
Z,
and let j := (bf + z)cm - bi. Since m2 > m, we have
cmm
2
b
and
jD <m2 so (D;
2
b n+
', E)
cmD
b
< m293,
must satisfy the jth triangulating condition. Recall that b(bf + z) = m + 1 and
C2M, so
+ z)cm.
(be
bm'n+ b + z2
= (m + 1)mn + (b +z)2
cm
bmn + be + z
c 2 m 2 +1
cm
CM
96
_
CM
+
1
-,
CM
and consequently
* bmn+ be
cm
Thus, for any (W, Z) E Sj (D; -!,
M'
+ z = M2n+ (b + z) 2+ m(n - ci) +
.
cm
'2
M2
(
(W, Z)
Q
), to say that
<
is the same as saying
n (b +z)
bmn+b+z
Si
C
;~ -,
2
M
and by Lemma 4.30, part (ii), there is precisely one pair with this property. Fix (W, Z) to be this
particular pair.
If W E {(m(n - ci), m), (m(n - ci) + m 2 n + (bf + z)2 ,0)}, then we have Q = 0 exactly, as
so
desired. Otherwise, by linearity we must have Q > 0. Now Q is always an integer multiple of
> Q > , which is to sayi> c.
if Q>0, then
a,
ix. Note that
b
m2
+ - >
b
cm
so part (vi) must hold with i = bI
m+ I
b
= b + z,
+ z - 1, which means
(be+ z - 1)(bmn + bf + z) - cm
bmn + bf + z
cm
b- <
<D <
(bf + z - 1)cm - b
b
CM
so
(Acm - pb)D < A(bmn + bl + z) - pcm
whenever
A < (bf + z - 1)[i.
In particular, if bf + z > 2, then (A, p)
conclude that, for i, bf + z > 2,
=
(bf + z, i) satisfies the above inequality when i > 2. We
+ z)(bmn + b + z) - icm
(be
(bD + z)cm - ib
(m + 1)mn + (be+ z) 2 - cm
(bf
+ z)cm - ib
provided the denominator is positive. In particular,
(m(n - ci) + m 2 n + (bf + z) 2 , 0)
D;
V S(be+z)cm-ib
I,
We also know that part (vii) must hold with i = m - max{b, 2}, so we have
cm
b
bmn + bf + z g3+ n
'c(bf + z)
cm
(m - max{b, 2})(93 + n) + bmn + bL + z - cm
(m - max{b, 2})c(bf + z) + cm - b
(m - max{b, 2})(g3 + n) - cm
(m - max{b, 2})c(bf + z) - b
and therefore
(Ac(b+
z) - tb)D < A(g3 + n) - /cm
whenever
A < (m - max{b, 2})p.
Therefore, as long as m > 2 and i > m-m
,
we have
97
- ci)
D < m(g3 + n
(bf + z)cm - ib'
provided the denominator is positive. In particular,
(m(n - ci), m)
D; n,
V S(bU+z)cm-ib
M M2
x. The case n = 4 is the only case where m < b, and must be dealt with separately. However,
the problem is finite, and a computer check readily shows that
=
is the only possibility here,
which agrees with the lemma as stated.
So assume m2 < b < m. We first claim that m > a:
(5a)2 + a2 =
" If a > b > c > 1, then m > -6b2 + a2 > -6
* 1a~~c>,u~en~-uLJl-~--~13+
* Ifc=1,thenm= b n + 1 > 2b2 +
2
1
a > 9a > a.
19a
169
-13
> a.
" If a ( b, then m > max{2, a 2 } > a.
Thus, as m2 > 2, we have
[2m - (m + 1)1m 2 = 2(mm
so c(mm
50 2-+b)
2
+ b) - b[(bf + z)m
2
+ 2] = m 2 (m
-
1) > 2
a
- ) =,
C
C
- c[(b1+z)M2+2]
2b
-
Now let
. c(mm2+ b) I
c(mm 2 + b)
b2
b2
:=
c(mb 2+ b)
b
=c(be+z).
Note that
c2 (b+ z + 1)= bc2 + c(ab+ 3)+ c2 = bc2 f + b(ac +1) + (3c - b)+ c2 = bn + d +c 2,
so we have
bn++d+c
c(b+z+1)
b
c[(b+z)m2 +2]
2b
and
bc
2
n
c
c
b
+2 z)m2
.c(be
ib - c >
so, by (v) we have
n2
M2
bC+z
b
c(be+z)
b(ib - c)
2
bm2
Recall that n > b z, so we must have
n2
bC+ z
M2 b
1
bin2 '
as claimed. Thus, bn 2 - (bf+ z)m2 = 1, and in particular zM2
sides by -a (which is relatively prime to b), then we have
3m
2
= -3(bz
-
-1 (mod b). If we multiply both
1)m2= -3a 2 m 2 - -a(3a - b)m2 =-azm2 = a
98
(mod b),
as claimed.
xi.
then
We first claim that
U
m2
> b. If b = 1, then we explicitly assumed
is a fair lower
loe approximation
aprxmto to
to ((ez +z),2
by (ii), so as
+_z
m2
> 2, and if b > 1,
2
(bt+z)
m
B
-:
M2
we must have
m2 > b.
Suppose for the moment that c > 2 and m2 b+ , and that either a > 1 or m 2 > b+1
Then m2 < C(M2 - b), and either a > 1 or m2 < c(m2 - b - 1). Since m 2 < m, we can find i E N
such that
c
i
2
b,
M2 -
and either a > 1 or i < M2 - b - 1. In either case, the hypotheses of part (vii) are satisfied, so it
must be that
1
(be+ z) 2
n2
1
ZCm
m2
m
mm2
a contradiction.
So one of three things must happen: c = 1, or b < m2 < b + Cb-I-, or both a =
b+ 1+ c-i, .b+
*If z > a > b > c > 5, then note that M2 < b + gb- < b+ max{,
,
2
<
so it is not hard
.},
,
and b < m
to see that, at any rate,
b+ 1 < m2
b+ 3.
Let y := 3z - a, and let e E {1, 2} be such that 31 a + Eb, and note that 3
claim f + (y+Eb)/3 is the simplest fraction in the open interval
so
y+Cz
z
1
a +Eb
b
b(a+ b)/3'
bf,
y
(b+Z)2
+ ez as well. We
Indeed,
is a fair upper approximation to (. Moreover,
8b 2
-
(Eb
-
c)(a + eb) > 8b 2
(2b
-
c)(a + 2b)
-
=b(c
+ d) + 1 > 0,
so
bz+1
b2
and (+E
f+
y-+Ez
1
[(a +b)/3]
eb-c
2
b (a+cb)
a+eb
is a fair lower approximation to
.
2
'
So, since a+Eb < b2 we know from (ii) that
is indeed the simplest fraction in the open interval
b+z, (bf+Z)2.
It follows that
in fact
m2b>
M2-b
a+Eb
3
-3b=
eb-c
33
>3,
a contradiction.
* If (z, a, b, c) = (34, 13, 5, 2), note that / > -6.
Part (iv) tells us that
£ + 6 + - or a fair lower approximation thereto. Now f + 6
lower approximations:
4
5
5
6
+
21
25
f+ 6 + - < f+ 6 + - < f+ 6 + -.
99
2
is either
b2 e+bz+1
- has these consecutive fair
' >
Recall that we are assuming that b= 5 < m 2 < b + 6 = 10, and
= E+6+ 5. But in this case, note that
the only possibility is
cm + c(bf + z)- b = 60e + 401 < m2
=+
+6 + A so
Y+-z
brn + bE + z
< M2D,
cm
so by Lemma 4.18, we must have
(bmn+ be + z)+ (g3 + n) - cm
cm+ c(be + z) - b
bmn + b + z
cm
Thus,
(g3 + n) - cm
bmn + bf+ z
cm
c(be+ z)- b
which can be simplified to
(be + z) 2
1
<g3
cm
m
e from
If we subtract
e<
+
-mn
m2
41
6
both sides, then multiply by m, we find that
171e+ 1156 which implies
n2
-
=1
- = (bz + 1)e + z2 _ 1 <
c
2
( 171
6
e+
-)
6
-
e ;,
-4 and by part (iv), m2 K b2
-
4, and
or a fair approximation thereto. Since we are assuming m 2 > b, we have
e + 20, then part (vii) must hold with i = 1, so
But if 2
3m
6
-4, as desired.
* If (z, a, b, c) = (13, 5, 2, 1), then
n +1_
7
1156 - -
m- 3 n_(
\
3m
+7-
_ (b + Z)2
M
20)
n )_(+
3
m/
which is a contradiction, as n > 2. So, in fact, we must have
2
M2
e+
1
cm
_
e+
27
,eE+L}.
{e+
n2E
_n2
=
is either
n2
1
m
?-, as claimed.
b2 = 1, which is absurd.
" If b = 1, then, by part (iv), m2
b2 -4 and 2 is either
eIf (z, a, b, c) = (1, 1, 2, 5), note that e > 1. Part (iv) tells us that m2
,2+,z+=e + 3 or a fair lower approximation thereto, which implies 2 = f + '21.
Recall
that we are assuming that b < m2 < b+ 1 + b+ = 3 +
that case, part (vii) must hold with i = 1, so
e+1
3m
=f
(i7-
(
n-
< 4, so we must have
2)
(b ± )
3
m
n2
1
1
m2
cm
5m'
which is impossible. So this case does not arise.
* If z < a < b < c, then we have b < m
2
< b + -_- < b + 1, which is absurd.
100
n2
=
.+2
In
xii. We consider three cases.
Case 1: Suppose c > 1, and that there exists an integer i with
(bf
+ z)M 2 +b
m2
b
cm
Since b < m2 < m, we have bl + z
= m+1
b
cm
> 1. Note that
cm(be + z - 1) - [(be + z)m + b] = m((c - 1)(b + z - 1) - 1) - b.
We claim the right-hand side is non-negative. The only way it could fail to be so, under our
assumption that c > 1, is if c = be+ z = 2, in which case m = b(be + z) - 1 = 2b - 1. The fact that
Z, which is impossible. So
= -32
m > 2 implies that b > 1, so (z, a, b, c) = (34, 13, 5, 2) and
indeed, we always have
(be+-z)m2 -+b
(be+z)m-+b <be+zcm
cm
so we may assume
(b+z)m
2 +b
i
cm
b
-,be
±
cm
M2
b
-
1
.
Then, by part (vi), we have
1
mm
n
2
(bf +Z)2
be+z
m
m(icm - b)
m2
b +z
m(b+ z)m2'
which is absurd.
Case 2: Suppose c > 1, but there is no integer i with
(be+ z)m 2
cm
+b
Let
i :=
b
M2
b
>I1
b
b+
b
cm
cm
so that
bi -
cm
<_
m2 <
b(i + 1) - b
We also must have
< b(i + 1) - 1
cm_]
_
(be+z)m
2
+ b
> ,
cm
so that
icm - b
beb < m2 < b(i + 1) - 1,
bl+ z
icm - b < (bf + Z)M2 < (m + 1)(i + 1) - (bf + z),
i(c - 1)m - i < m + 1 + b - (be+ z) = (1
m + b+ 1
-
Since i > 1, we have
(c-2
m<b+2101
1
1
b
which, since m > max{b, 2}, implies c < 4, and so in fact c = 2, and
i(m -1
If (z, a, b, c) = (2,1,1,2), then i =
1
(i
I - -)m + b + I - I.
<
+
[2
m2 > 2, so
>
2(m - 1) < 1,
which is impossible. If, on the other hand, (z, a, b, c) = (34,13,5,2), then m > 19 and
4m + 29
5(m - 1)
so i = 1 and m < 34, so in fact f = -6 and I = 1. In this case we have
8 <
b2
cm -b
38 -5
= 10< m 2 < b(i + 1) =
cm
4
bC+z
25
- < 10
38
is a fair upper approximation to both
'2
so m2 = 9. But we know from parts (iii) and (iv) that M2
(b+z)2 and f +
. Since M2 < M, this means in fact M2 < m and 2 is among
130
1f 9
171
2-5
48
T
7
89
13
7
1
But evidently this is not the case, if m2 = 9.
Case 3: Suppose that c = 1. By (iv), 2 is a fair upper approximation to
M2
so it is not hard to see that m2
=
1
m (mod b2 ) and
-
b2 + bz+1
n2
b
m2
1
2
b2M2
and consequently
n2
M2
(bf+z) 2 =
b2+bz+I
1
b2
b M2
m
b2f+bz+1 +1
Let
.m
and note that i E N and
1<i
+2M
b
2
b2M
b
M2
+1_1
b
m+1
min
1
-1,
bb
m2
+b
b
m
so by part (vi) we have
M-M
2
n2
(bL-+ z) 2
bf+z
be+z
b2 m
2m
m2
M
m(icm - b)
m(im - b)
102
1
m
m-M2
b
'
which is to say
2
irn-b < b (be + z)M
m
-
2
m2
Thus, if n2 = m - b2 x, then
0< x (b2 (b + z)m
m-m2
m
(bf + z)(m - b2x)
2
bx
M2
m
m
=be±Zb(m+l)x-bx
=bfz
bl+
z~
-- x _ rn+1
m
x
b
m
=b+ z - bx - (be+z)x +bx2 + x
=
(b+ z - bx)(1 - x)+
=
1 - (bf + z - 1 - bx)(x - 1),
Y
+ 1
b
+ bx 22 +
x
x
so
(bf+ z - 1 - bx)(x - 1) < 1.
Now we are assuming m2 > b, so
b + z - 1 - bx = be+ z - 1+
Therefore, it must be that x = 1. Then m2
n2
M2
=
m2 - m
b
>
1
M
+bez - -
> 0.
m - b2 and
b2 + bz +I1
1
b2
b2(M - b2)
(be+ z) 2 - (b2 f + bz + 1)
m - b2
as claimed.
xiii. If bf
+ z > 2, m > 3, and there is an integer i with
m
m-
m
m - max~b, 2}
< i < min{c, ce + a},
then we immediately have a contradiction between parts (viii) and (ix).
So one of three things must hold: bf + z = 1, or m = 2, or there is no integer i with
m Max
m - maxf b, 2}
< i < min{c, cf + a}.
If the last of these is true, then, under our assumption that c > 1, we must have either ce + a
=
1
or m < max{2b, 4}.
0 If z > a > b > c > 5, then (by plugging in -6 for f) we easily see that be + z > ce+ a > 4
and m > 2b > 4, which we have shown to be impossible.
* If (z, a, b, c) = (34, 13, 5, 2), then (by plugging in -6 for f) we easily see that bf + z > 4 and
m > 2b > 4. Thus, we must have ce + a < 1, which implies f
case - = 3 and be + z = 4. A computer check shows that '2
caem
32
M2
19
* If (z,a,b,c) =
-6 and ce + a = 1. In that
27 is the only possibility.
=
thonypsilt.
(2,1,1,2), then be+ z =m+1 > 3 and c +a= 2m-1 > 3. Let usconsider
four sub-cases:
- If m = 2, then
=
. A computer check shows that
103
2
E
f
}.
- If m = 3, then
. A computer check shows this is impossible.
rn=
15=2. A computer check shows this is impossible.
- If m = 4, then
- If m > 5, then in fact m > 4 > 2b, which we have shown to be impossible.
e If (z,a,b,c) = (1,1,2,5), then £ = "4- > 0,soinfactfe> 1. Then bf+z > 3,c +z > 6,
and m > 4 > 2b, which we have shown to be impossible.
" If (z, a, b, c) = (1, 2, 5, 13), then cf + a > 2, so either b + z = 1 or m < 2b = 10. Both of these
=- 27.A computer check shows that
imply f = 0, in which case
"
If 2 < z < a < b < c, then f)
E
-
,
f
.
0, so bf + z > 2, m > 25, and min{cf + a, c} > a > 5, so we
must have
m
->
m - max{b, 2}
5
'
which is to say
5b > 4m > 4a 2
which is clearly false.
xiv. The case a =
then we have
2
< n, and by part (viii) there's a unique pair (W, Z) E
with
0 < m 2n+(bf + z)
2
(mn+
W-
-
(b + ) 2
) 2 -W-
0 < m 2nn+(b+
(mn+
(bV +
=
b2n +1
m
denotes the fractional part of nz, so 0 <
Z-1
b2
nZ = Z-1
b2
M
{
1
bi2
Thus, in order for n to have fractional part less than
we may write Z = b2 Y + 1, with 1 < Y < n - 1. Then
W+
+ z)
(mn + (be
+
2
}
)
<
b2
m1
< n. In particular, 1 < Z < m, so
m-Z
mn )
Z
<b2
< b, we must have Z =1 (mod b2 ), so
Z=M 2 n+(b+z) 2
-
n__
m
Let
(W1 , Z1) = ((bf + z - 3)(bmn + bf + z) - m + (b2 + 1)n, b2 + 1);
104
L,
so
W+ (mn+ (bf + Z) 2 ) Z=m 2rn+(bf+ z) 2
{n}
(D;
f{(m(n - ci), m), (m(n - ci) +
Z)2Z< -n
Z < m
and 0 < Z < m. Recall that mn + (bf+z)2 =mmn+f+7
Sc(bt+z)mbn
Z<n
so that in particular 0 < Z < m. By part (ix) we know (W, Z)
m 2 n + (bf + z) 2 , 0)}, so that
where
,;
was dealt with in Corollary 4.29. Let us, therefore, assume that
note that this satisfies
2
z1 =
mn + (b+Z)
m 2n + (bf + )2 - W-
m
We claim that in fact (W, Z) = (W 1 , Z1 ). If not, then we have Z > Zi, Y > 1, and, by (vii),
(W + 93Z) - (W1 + g3Zi)=-
n -Y
m
Y-1
n -i1
m
+
(b + z) 2 (Z
(n2
Zi)
(Z -
m2
Y- 1
b2 (Y - 1)
(m - 2)m
m
But then (W 1 ,Z1 ) E Sc(be+z)m-bn D; ",
+
-~
=
b2
-
i
>0.
also, contradictory to the uniqueness of (W, Z). Thus,
we conclude that (W, Z) = (W 1, Z 1 ), so
D>
(b2 +1)(g 3 + n)+ (bf + z - 3)(bmn + bf + z) - cm
(b2 + 1)c(bf + z) + (bf + z - 3)cm - b
W 1 + g3Zi
c(b + z)m - bn
according to a tedious calculation.
Meanwhile, note that
((bf +z)(bmn+bf+z)-c 2m 2 -2, b2 )
0
S(be+z)cm
bmn -- be + z
n
m
(b+Z2
mj
= S(bf+z)cm
n n2
(D;
,
so after another tedious calculation
b2(93 + n) + (bl+ z - b)(bmn + b + z) - cm
2
(be+ z)(bmn + be + z) - c2M2 - 2 + b 9 3
(bf + z)cm
blc(bf + z) + (bL + z - b)cm - b
Recall from part (vii) that
D< (m-2)(g3 +n)+bmn+be+z-cm
(m - 2)c(be+ z) + cm - b
The time has come for us to consider the set R C R2 of pairs (a, 3) such that
[ac(b + z) + /cm
- b]D < a(g 3 + n) + 0(bmn+ be + z) - cm.
According to the three inequalities, R supposedly includes (b2 , bf + z - b) and (m - 2, 1), but
not (b2 + 1, bf + z - 3). Now R also includes (0, 0) and is by nature convex, so we can obtain a
contradiction by showing that (b 2 + 1, bi + z - 3) lies in the convex hull of (0, 0), (b 2 , bf + z - b) and
(m - 2, 1). By Cramer's rule,
(b2 + 1 b + z - 3)=
* If (z,a,b,c)
=
+z -3) -(b
(m -2)(be
2
+1)bbeb)
b + z - b)+
(m-2)(b+z -b(b2
(m - 2)(bf + z - b) - b2
(b2 + 1)(bf + z - b)- b2(b + z - 3)
2
(m-2)(b+z-b)-b
(5,2,1,1) and n > 4, then we have m
(a~en-2-=2n-1
(a, b + z - 3) = n2 - 2
=
2
)
n+1 and be-+-z = n-+2, so
n+3
(b2 ,b +z-b)+ n2 2 (m - 2, 1).
Since we're assuming n > 4, we have n 2 - 2n - 1 = (n - 1)2 - 2 > 0 and
(n 2 - 2n - 1) + (n + 3) = n 2 _ n + 2 <; n 2 - 2,
so (a, be+ z - 3) lies in the convex hull of (b2 , be+ z - b), (m - 2, 1), and (0, 0), a contradiction.
105
2
"
then a computer test shows that
=,
If (z, a, b, c) = (5, 2, 1, 1) and
is the only exception.
, but this
can be
-
" If (z,a,b,c) = (13,5,2, 1), then we have m= 4n+ land b +z = 2n+ 1, so
5
(ab+ z- 3) = (4n - 1)(2n - 2)-
(b2
bfz-
b +5(2n - 1) - 4(2n - 2) (
11
(4n - 1)(2n - 1) - 4
(4n - 1)(2n - 1)- 4
2
8n -l1ri-3 (b 1+
2
- 2, 1).
b + z - b) + 22n +3
-n-3(b
8n
8n - 6n -3
8n2 - 6n - 3
Now (4n
-
1)(2n
2) - 5> 7. 2-
-
5 > 0 and
(8n2 - 10n - 3) + (2n + 3) = 8n 2 - 8n < 8n 2 - 6n - 3,
so (a, bf + z - 3) lies in the convex hull of (b 2 , bf + z - b), (m - 2, 1), and (0, 0), a contradiction.
D
This completes the proof.
Lemma 4.32. Let z, a, b, c, d be consecutive odd-indexed Fibonacci numbers, and let f E Z. Let
(m, n) = (mi, ni) = f(b2 c2 ) + (a2,b2 + 2), and suppose that m, n > 2. Suppose that (D; ", 22) is
a weakly triangulatingsequence with m2 > 2 and
D<
where g3 = mn +
"
2.
n
c(bi + z)'
Then here are the possibilities:
(z, a,b, c)= (34, 13, 5, 2), f = -6,
" (z,a,b,c) = (13, 5,2, 1) and
n
M2
n2 =
additionalpossibilities: M2
3
-
"
93 +
,and
&
} with L'
n2
D < 11j.
= 2n3.
,
2rm In the special case f =
with
n2 =
with
2 < D < 8,18' or M24
'33-
(z, a, b, c) = (1, 1, 2,5), e = 1, a =M
5
, and M2
n2 =
3',
2 there are two
9'
2
5
<
D
-
< 313
68
with 37 5 < m 2 D
2 < 377.
8
Proof. Let us remark from the outset that a few cases pose certain technical difficulties, but the
lemma can be checked by computer in these cases:
" If (z, a, b, c) = (89, 34, 13, 5) and -6
f < -2, then there are no solutions.
" If (z, a, b, c) = (34, 13, 5, 2) and -6 <f
the lemma statement.
-3,
then the only solution is the one described in
" If (z, a, b, c) = (13, 5, 2, 1) and f = -4, then the only solutions are the ones described in the
lemma statement.
" If (z, a, b, c) = (5, 2, 1, 1) and -1
" If (z, a, b, c) = (1, 1, 2, 5) and f
statement.
( f < 0, then there are no solutions.
=
1, then the only solution is the one described in the lemma
106
Henceforth assume we are not in any of these exceptional cases. That is, if (z, a, b, c), then f > -1,
and so on.
1. Note that
Let a,3 be as in Lemma 4.24, and let y := [8 - "
by- z > 7b For j
2c(b + z - })
2 (mn+
=
m
m
2 (b(ce+ a) + 2), we have
-
(+)
m
dm -bn
brn
- - z=
> 0.
(bmr+b
)
Abmnr+b-+z
-Z)
(2c(bf
=
cm
CM
2c
b
so
(0, 2) V Sj (bm;
(b + Z)2
n
bz
-,M
m
m
CM
bmnr+ib +z
cm
and Sj (binn~be~z.
C
(bII z) 2
contains only pairs (x, y) with y E {0, 1}.
The proof proceeds in eleven steps:
i. We show
c(+-y) - cm - (by- z) -bn
c(e+ -) - b - (by- z) -c
D >mn-+n+y
c(bC+z)
and
mn + (+
For every j E N, if Sj(D;
)
,
m
z)
Si (bmn
< mn +
(X, 0) E S
By the way, (D;
-; n)
(bnz)2 ) we show there must exist x E N
bez.
such that
n n2 ) \S
D; m ,)
M2
+ -y < g3.
bmn+b+z n (bf + z) 2
cm
m
m
must satisfy the jth triangulating condition for all 0
j <
C 72
ii. If A > 0 and y > 0 are such that
P <
A
1
c(be+z)
c&+
a+ a
A + c'
then we show that
SbA-cp
D; -, I
SbA-cP
z n (b+ z) 2 )
(bmn+be+
c
cm
M
I'
m
iii. We eliminate all but four cases: (z, a, b, c) E {(89, 34,13,5), (34,13,5,2), (13,5,2,1), (1, 1,2, 5)}.
After this point we assume (z, a, b, c) E {(89, 34, 13, 5), (34,13, 5, 2), (13,5,2,1), (1, 1,2, 5)}.
iv. We show that
D
bn
(93 + n 3 ) + (by - z)
c(bf + z) + (b-y - z) -c'
v. We eliminate the case (z, a, b, c) = (89, 34, 13, 5).
107
After this point we assume (z, a, b, c) E {(34,13,5, 2), (13,5,2,1),
= b-y - z = 1.
(1,1, 2,5)}.
Note that, in
these cases,
vi. Let
2f+ 13
i := min
cf + a + a -C
c((b-1)f +z-7)-
=
~ b2
f+ 5
if (z, a, b, c) = (34, 13, 5, 2),
if (z, a, b, c) = (13, 5, 2, 1),
U - 1
if (z, a, b, c) = (1, 1, 2, 5).
We show that
D < 3 + n + Acm + pbn
c(bf+ z)+ Ab+ pc
as long as A, p > 0 and A + c(e + -y)[t < i. Moreover,
+
93 > mn + £ +
b
c(b + z)
A+1j if (z, a, b, c) = (34, 13, 5, 2),
fo+68
=
mn +
+-y+
t
5f10t+5
if (z, a, b, c) = (13,5,2, 1),
if (z, a, b, c) = (1 1, 2,5),
vii. We show
+(1+b)n_
c(bf +z+1)
93+(l+b)n(D
bc( + )
c(f + Y)
(cf + a + a) - b
93
We note for future reference that when (z, a, b, c)
=
'
-
c
(13, 5, 2, 1), this bound says 93 < mn
+
viii. In the case (z, a, b, c) = (34, 13, 5, 2), we show that
- 2bn
D < (2(ce+ a) + 1) cm
(2(cf +a)+1)-b-2c
49 + 28
93 < m
i + y+ 20f + 131'
and M2 > 5.
After this point, we know M 2 > b. (In the cases where b = 2, this is given.)
ix. In the case (z, a, b, c) = (1, 1, 2,5), we show that
-cm-bn
D < (c+a+3)
(ce+a+3) b-c
x. By considering the jth triangulating condition for j = cm, we eliminate the case (z, a, b, c)
=
(34, 13, 5, 2) and show that m 2 > 3 if (z, a, b, c) = (1, 1, 2, 5).
xi. We finish the proof in the cases (z, a, b, c) E {(13, 5,2,1), (1, 1,2, 5)}, by considering the jth
triangulating condition for
j
=
c(m
+ b).
108
Without further ado, let us prove each of these in turn.
i.
Our strategy is to compare Sj(D; -,
that
c(b+
'2)
MM2cm
to Sj (bmn+bt+z.I'nn
z)(bmn + b + z)
. For example, note
=mn-+n+f+y,
+1
cm
_
2
(bf+z)
m
)
S
+ n +f+ y is the smallest integer x for which x E W but (x,O)
Whenj = c(be+ z), we have
so mn
2cm
cmm2
b
b
bmn+bf+z. n
2
(bt+z)
and
jD <
so (D; n,
n2
12)
c(be+ z)
<93 + n < m293,
) must satisfy the jth triangulating condition. We already know
(n, 1) E Sj
n
(bmn+be+z .n;M-, I (b+ M Z)2
2
E S3(D; -,
so we must have (inn + n + f+ -y,0)
\Sj (D;
)\S, (bmnbe+z. n
n2
,
),M
in return, which
(bf+z)2
implies
bn
c
mn+n++-y
c(bf +-z)
bn +
bc
c
f+Y>
c(bC+z)
cm
b
since, by - z > 0, as we noted above. Incidentally, this implies that
93 > c(bf + z)D - n > mn +
For 0 < j <
,
2)
12
(be+z) 2
m
m
y > mn + + 7 - - = mn +
.
"2 < m 2 D, we have
jD <
so (D;
e+
3 + n) <
mm2(93 + n)
c(bf+ z)
M + 1
= m29 3
+
mn - 93 < m293,
m+1
must satisfy the jth triangulating condition. Now if there is any
\Si (bmn+
\SM
(x,y) E S (D;
2
b + z n (bf + z)
; M
then the above inequality ensures that
(x + y(mn + f+y), 0) E
(D;
,
2)
( miMM2)
\S (bmn + bf + z. n (b + Z)2
cm
M Im
as well.
> Cn, so two weighted averages of £ b
and n can be compared on the basis of weights alone. (Later on in this proof, we will also use this
principle with other ratios that are just as well understood.) For example, it helps to write
ii. The main time-saving calculation principle is that
m
++y
+1mn+
c(b + z)
bm
c(+-y) - cm - (by - z) - bn
c(f+7)-b-(by-z)-c
109
This is not quite a "weighted average" in the ordinary sense of the term, for the weight -(by
is negative:
bn
-bib
b +z
by - z > (7b - z) > 0.
- z)
_dm
m
m
m
Nonetheless, the principle remains valid.
As a first application of this principle, we have, by the assumptions on A, P, that
A - cm -p
bn < c(b
A - b - -c
+ z) - cm - bn
_
bmn+ bf + z
c(be+ z) - b - c
cm
whereas
A - cm - p - bn+1
_
(A + c) -cm - (p + b) - bn
A-b-[t-c
A-b--
(cf + a + a) - cm -
c
(c+a+a)-bn' n)\S
so there cannot possibly exist any integer x with (x, 0) E S (D; ;,
It follows from part (i)
M2
\S
b
> D,
c
'
(bmn~be~z. n
C
______
-Mm)mz m
that
SbA-,(D ;n
Cbmn+b+z
n2
Cm
M m2
n
(b+Z)2
M
m
iii. As another application of the same principle, perhaps even more direct than the previous
one, recall from part (i) and Lemma 4.24 that
c(f +7) - cm - (by - z) bn
c(f + y) - b - (by - z) -c
D < (ce + a + a) - cm bn
(c+a+a)-b-#-c
Then the principle says it must be the case that
c(e+y)
cf + a + a
b-y - z
0
This allows us to rule out most cases immediately:
If z> a> b> c >-13, then (a, ,y) = (1,3,7), so
c(e+ 7)
d
c(e + 7)
7b - z
c( + y)
cf + a + a
by - z '
cf + a + 1
3
and 3c(f + 7) > d(cC + a + 1). As d > 5 > 3, this inequality becomes more and more stringent, the
larger f is. Since f > -6, we must have at the very least 3c > d(-6c + a + 1) > 5(-6c + a + 1), or
33c > 5a + 5. But this is absurd because a > 8- > 3.
If (z, a, b, c) = (5, 2, 1, 1), then (a,3, y) = (1, 1, 7), so
f+ 7
c(e+ y)
ce + a + a
f + 3
2
by-z
#
1'
and f < 1. That leaves just f E {-1, 0} as possibilities, which were discussed at the beginning of
this proof.
If (z, a, b, c) = (2,1, 1, 2), then (a, , y) = (2, 1, 4), so
2(e + 4)
2
_
c(e + y)
cf + a + a
by-z
3
110
so
e<
1 and m = 1, which we are assuming not to be the case.
If (m, n) = (4, 27), then (a, /, y) = (5, 2, 1) and f = 0, so
13
4
c( + -)
by-z
13(i+ 1)
5.1 - I
which is simply false.
If z < a < b < c but (n, n)
$
c + a+ a
13 + 2 +-5
2
3
7
2'
(4, 27), then (a, 3, -y) = (2, 1, 1), so
c(f + 1)
_
c ++a
a
c( + -)
b-z
by-z
_
c +a+2
#3
1
Since b - z >, 5 - 1 > 1 and f > 0, this implies at the very least
c > (b - z)(a + 2).
If (z, a, b, c) = (1, 2, 5, 13), then in fact 13 < 4 - 4, and otherwise, b - z > 13 - 2
So this is absurd.
=
11, but c < 7a.
iv. The remaining cases, where (z,a, b, c) E {(89, 34,13, 5), (34,13, 5, 2), (13, 5,2, 1), (1, 1,2, 5)},
are a bit more involved. Note that, barring the exceptions we listed above, we have
0
c(£+y)
O+b
+
1
c(bf + z)
c+
a+ a
c(f +
) + c*
Only the past part requires any explanation:
b(cf + a + a) - 0[c(f + - ) + c]
=(b - O)c(f + -y - 1) + b(a + ce - c(-y - 1)) - 20c
if (z, a, b, c) = (89, 34,13, 5),
50(i+ 6) + 35
6(f + 6) + 2
( + 6) - 2
5(f + 0) - 4
if (z, a, b, c) = (34, 13,5,2),
if (z, a, b, c) = (13,5,2, 1),
if (z, a, b, c) = (1, 1, 2, 5).
So, in order for this quantity to be positive, we need to have
-6
if (z, a, b, c) = (89, 34, 13, 5),
-6
if (z, a, b, c) = (34, 13, 5, 2),
-3
I
if (z, a, b, c) = (13, 5, 2, 1),
if (z, a, b, c) = (1, 1, 2, 5).
The first, second, and fourth of these are implied by the fact that m > 2. The third fails only in
the exception we listed at the beginning of this proof.
Note that b-y - z < 2 in all four cases, so
bc( + -)<c(bf + z + 2) (
given the bounds on f above, so, by (i),
for
j
(D;
c(m + 1 + 2b)
Cini2
<
b
b
) must
,
satisfy the jth triangulating condition
= bc(f + -y). It follows from part (ii) and the triangulating condition that
Sb oc(+y)y)D;n
D
n2 )=
,2
Sbc(e)
()
(bmn + b + z. n (b + z)
cm
' i'
m
111
2
1
to see that
Now write bc(e + -y) as (by - z) - c + c(e +-y),
2
(bmn + bf+ z n (b + z)
bn + n, 1) E S(by-z)-c+c(e+y) (
Cm
mJ'
'm
((by - z)
so we must likewise have
((b(by - z) + 1)n, 1) E S(by-z)-c+c(e+y)
D;
n
n2
which implies
(93 + n) + (by - z) - cn < D
c(bf + z) + (by - z) -b
as claimed.
v. Let us now consider the case where (z, a, b, c) = (89, 34, 13, 5). Part (iv) tells us
d - bn + (n + 93)
d.c+c(be+z)'
so, as d < 12, we have a fortiori
(15e + 65)m + 157n + g3
104e + 674
(3e + 13)cm + 12bn + (n + g3)
(3f + 13)b + 12c + c(be + z)
and
((15e + 65)m + 157n, 1) E S104e+674 (D;
On the other hand, since we are assuming f > -1,
we have 8 ±53
3
c(f+7) -cm-d-bn
c(f+7)b-d-c
(8f + 53) - cm - 3 - bn
(8e+53) -b-3-c
_ c(bo+z)
< 5(13e+89)
But we also have 8e53
11
3
_
n2
,
2
.
so
>5e +35
d Ista
2 _ c(e+7)
that
mn + n + + 7
< D.
c(bf + z)
s
bmn + bf + z
cm
c(bf+z)-cm-bn
c(b+z)-b-c
(8f+53)-cm-3-bn
(8f+53)-b-3-c
n
Since (8f + 53)b - 3c = 104f + 674, this means that
((8f + 53) cm - 3bn, 0) E S104f+674
We conclude that So4e+674(D; ",
1)
D;
;
n
n2
,I
bmn+bf + z rt
\S104f+674 (
cm
.
(b+Z)
Sio4e+674
cmn+bt+z
note that (D; -, -LI) must satisfy the jth triangulating condition for
fact that
104f + 674 < 10(169f + 1158)
13
_
2cm
b
j
,
m
a contradiction, if we
= 104f + 674, by (i) and the
cmm2
b
vi. Now three cases remain, namely
(z, a, b, c) E {(34, 13, 5, 2), (13, 5, 2, 1), (1, 1, 2, 5)}
112
m
(bf + z)2
and in each of them we have, as if by coincidence, 3 = by - z = 1. By assumption i < c(bf + z) c(f + -y), so we have
c(+ Y)
c(bC+z)
c(Iy)+i
1
by - z
by -z
so
bmn+ib +z
cm
(c(t+ -y) + i) -cm
c(b+z)-cm-bn
c(bf + z)-b-c
-
(c(+-y)+i) -b-
(by
-
z)
bn
(by-z) c
y) cm- (by-z) bn <D.
(by-z) c
c(+
Thus, if we let
j = c(bf + z) + bi = (c(e + y) + i) - b - (by - z) - c,
then
((c( +
(bmn + bf+ z n (b + z)
n n2
)
+ i) - cm - (by - z) - bn, 0) E Sm(D; my),
M
cm
M2
M
2
M
By the way, we have
j < c(be+ z) + b (cf + a+
so (0, 2) V Si
S(mn+bf+z.
(mn+bt+z.
c(b+
b-=
.(b+Z)2
By Lemma
n
(b+z)2 ) \S(D; -,
2
2
b>- b-+1)
n).
a)+3+
z) +b(c+
a - 3b
-
c =2c
Note that the bounds we are assuming on f imply that
347132 = -142
21
52-5+1
13-7-52 = -19
3
22-2+1
1-1-12 __-1
2T 2+ 1 - 3
if (z, a, b, c) = (34, 13, 5, 2),
if (z, a, b, c) = (13, 5, 2, 1),
if (z, a, b, c) = (1, 1, 2, 5),
i < c((b - 1)f + z - y) < cm,
and
while
icm + I
ib
(c
1
4.2, then, there must be some element (x,1) E
so (b - 1)f + z - y < b2f+ a 2 = m and
so
bf+z-
c+i
c(bLf+ z)
b
1
+ i) - cm - b - bn
(c+i)-b-b-c
K'
bmn + b( + z
cm
c(bL + z) - cm - bn
+z)-b-c
c(b
2
n+mn+(bC+z)
/m
_
bmn+bf-l+z
c(be+ z)
cm
so x < icm + n. Thus, we must have (icm + n, 1) V Sj (D; -,
D < icm+ n+g3
ib + c(bL + z)'
which is a special case of our desired result, when p = 0.
113
2).
The upshot is
Now if we recall from part (i) that
D > c(f + y) - cm - (by - z)-b.
c(+y) b - (by - z)- c
_
c(e+ y) - cm - bn
c(f+-) - b - C
we find that
cm - bn)
D < n + g3 + i- cm - P(c(f +)
c(b+ z) + i - b - p(c(C+ y) - b - c)
for any p > 0, as long as the denominator is positive. Set A < i - c(E + y)p to recover the result we
claimed.
Note that
(b2 nr+ 1)(b +z)
b(m+ 1)n+bf +byc(bf-+z)
mn+n+f+y
c(bC+z)
by
-
z
c(be+ z)
1
c(b +z)
As a consequence of the above inequalities and part (i), we have
93 > (ib+ c(be + z))
>m(ib + c(bn + z)) =
mn +
e+
D - icm - n
icm - n
c(bf + z)
y + c(b + z) .
vii. The bound on D follows directly from part (iv) and the fact that by - z = 1 now. The
upper bound on 93 then follows from Lemma 4.24, after some calculation:
g3 < bc(e + y) -
(ce+ a + a) -cm -
(cf + a + a) -b-
bn
-c
- (1 + b)n
c(f + Y)
(cf +a+a)*b-c
viii.
In the case (z, a, b, c) = (34,13, 5, 2), we in fact claim that
- 2bn
D < (cf + a + i)cm
(cf + a + i)b - 2c
for 2 < i < cf + a + 1. Obviously, i = cL + a + 1 provides the tightest bound, so that is the
only case of any importance. At any rate, let j := (cf + a + i)b - 2c. Note that j < (al+ a +
i)b - 2c < 2 (c(bf + z) - f), so (0, 2)
#s, (D;
,)n2
Si (bmn+b+z.n
(bf+z) 2 )
and in particular, by Lemma 4.2,
(j 2).
We claim that every (x, y) E SI bmn+b+z. n (bt+Z)2 also belongs to Sj (D;
this is obvious, so it remains to consider the case y = 1. Now note that
(cf + a + i)b - 2c = c(bt + z + 1) + (i - 3)b + 3c,
and
(b+1)n+ g 3 + (i - 3)cm + 3bn
< D
'
c(b + z+ 1)+ (i - 3)b+ 3c
114
, 2). If y = 0,
because
that
(i
(2-3b3c1>Note
-bn< cmc3b<D and (i-3)b+3c>
c
b
+(b++1
< D by part (vii),
3)cm + 4bn + 1
-
(i - 3)b + 4c
because
(i - 3 + c) -cm - (b - 4)bnr
c(b1+z)-cm-bn
bmn+bf+z
(i - 3 + c) - b - (b - 4) - c
c(be + z) - b - c
cm
i- 3+ c
-1
+ 68- c(be+ z)
+2<
I
0- 4
So every (x, 1) E Si
bmn+b+z. n
)
(x, 1) E S4 (D;
2,
Nt
0
satisfies x < n + (i - 3)cm
(f+z)2
+ 4bn + 1, and consequently
also.
So #Sj (D; -, -)
> #S3
(bt+z)2
(bmn+b+z. n
2), and we noted the reverse inequality
(
earlier, so equality must hold, and we must have Sj(D;
)
-
S (bmn+b+z.n (bf+z) 2 )
=
Now
note that
2e + 13+ i
c + a+ i
2
2
so that
< cL+ a +
2
= 5+
1 < 10f +68 = c(be+ z)
2+
1
2
(ce + a + i) - cm - 2bn> c(b + z) - cm - bn
(c + a + i)-b - 2c
c(bf+z).b-c
bmn + bf + z
cm
which means
((ce + a + i)cm - 2bn, 0) V S(cd+a+i)b-2c
+ be+ z n (bf + z)2
bmn
CM
m
- S (D;
n
n2
M I M2)
and consequently
- 2bn
D < (ce+ a + i)cm
(cf+ a+i)-2c
as claimed.
Consequently,
93
<
c(b
z1)
We may as well let i
=
cl
-
(1 + b)n
2c(e + -)
(ce + a + i) - b - 2c
-y
m-
(cf + a + i) - cm - 2bn
(c + a + i)-b - 2c
+ a + 1 for the best bound, so in light of (i)
+ 7)
n2 _2c(
(2(cf + a) + 1) - b - 2c
M2
4 + 28
20f + 131
1
5
since we are assuming f > -4 at least. This implies m 2 ) 5.
ix.
In the case at hand, namely (z, a, b, c) = (1, 1, 2, 5), let j := (cl + a + 3)b - c = 10f + 7.
Since f > 1, we have j < 20f + 5 = 2 (c(bli+ z) - (),
particular, by Lemma 4.2, #Sj (D;
c +a
bf - 1
)
5f+1
2f- 1
_,
(
)
so (0, 2)
Moreover,
cf +a+a
5f+3
1
115
Sj (bmnjMez.
-
(
),
and in
so
3n-m
_
7
(c + a + a) -cm (c+ a + a) -b -
(cf +a) -cm- (bf - 1) -bn
(n-1)m-(m-3)n
(cf +a) -b- (bf - 1) -c
7
-bn
-c
> D'
and therefore, by part (vii),
93+
a) - cm+ (bf - 1) - bn
(b+ 1)n - (c+
_
g3
j
c(b + z + 1) - (c + a) -b + (bf - 1) -c
We claim that every (x, y) E Sj
(
bmnc+bf+z.
"z)
2
+ m
) also belongs to Si (D; y, %). If y = 0,
this is obvious, so it remains to consider the case y = 1. Now n < 7m, so
cm
5m
n-2m
2
bmnr+ b+
b
<2
z
cm
and consequently
(mn +
m
(b+z)2
(bM z) 2
so (2m, 1) V Si (bmn be+z
(mri+ (be+z)2 + n) - (n - 2m)
c(bf + z) - 2
bmn + b + z
cm
Mz
. Thus, for every (x,1) E Sj (bmn+
m(z)
2
),we
know
m, and (x, 1) E Sj(D; n ,
xis in the semigroup generated by m and n, and x < 2m, so in fact x
n)
also.
So #S3 (D;
n,
__
M) M2)
>
#S
77(Si
2
nMIn (bf+z)
(bmn+bf+z.
m
cm
CM
and we noted the reverse inequality
(j+2),
2
earlier, so equality must hold, and we must have Sj(D; ",2)
M M2
=
Sj
bmn+b+z. n
cmM
J
2
(b+z)
m'
Now
note that
c+
so that
(ce
a + 3 = 5+
4 < 10f + 5 = c(bC+ z),
c(b + z) - cm - bn
c(bf + z) - b - c
+ a + 3) - cm - bn
(cf + a + 3) - b - c
_
bmn
+ b + z
cm
which means
((ce
+ a + 3)cm - bn, 0) V S(ce+a+i)b-2c
2
bmn+bf+z n (b +z) )
=
m
M
cm
(jD; n'
n2 ),
and consequently
bn
D< (cf + a + 3)cm (cf + a + 3) - c
as claimed.
x. We first claim that, given the lower bounds on
SCM
(D;
C
Sc
c ~
ewe
are assuming, we have
bmn+b+z n (b+
ScCm
'm'
m
z)2
If (z, a, b, c) E {(13, 5, 2,1), (1, 1, 2, 5)}, this follows from part (iv) if we verify that bc(bf+z)-c = cm
and
1 z)
c(b + z)
i+b
1
c(be+z)
c + a+ a
116
c(b + z) +c '
as usual, only the last part requires any explanation:
(b+ 1)(cL+ a + a) -
[c(be+ z)+ c] = c + (b+ 1)(a + a) - c(z + 1)
{
+4
e- I
if (z, a,b,c)=(13,5,2,1),
if (z, a,b,c)= (1, 1, 2,5).
> 0.
In the case (z, a, b, c) = (34, 13,5,2), we have
2
as long as
2
6
1+ b
4 + 27
10e+ 70
c(be + z)+ c
_
2(ce + a) + 1
e > -5 (whi -h is part of our assumption). Then by using the upper bound on D we
described in part (viii)
in lieu of the weaker bound of ct+a+,)cm-f8bn
+a+a)b-,3c
mn+bt+z.
cm
=
,
D;
Sc,
have
must
shows that we
the same reasoning as in
part (ii)
n(bf+Z)a
Provided that m
2
> b, the triangulating condition cm must hold, by (i), so
SCM D,
n
n2)
,7
M
M2)
bmn+be+z n (be+z)2
; -,
SCM
CM
m
-
Now
bmn + bl + z - (b - 1)(mn +
e +-y)
e+
mn +
=
-y - (by - z)
mn +
=
e + -y -
1,
and
mn + (be + z)2 < mn +
+
m
so we have
(mn+
e+
- - 1, b - 1) E
SCM
bmn+be+z n (be+ z) 2
, -,)
cm
M
=
~
-~~
M
(D; ;
n 7 n2 )I
M2
which implies
(b - 1)g3 +mn + +-ycm
1 (;D.
If (z, a, b, c) = (34, 13, 5, 2), this means, in light of part (viii), and after some tedious computation,
(b- 1)93 <
=
(2(ce+ a) + 1) - cm - 2bn
(2(ce+ a) + 1) - b - 2c
(b - 1)(mn +
+-y)
(mn + e+ y - 1)
125'
20e +
+ 109
+ 16e
which implies
16e+ 109
93 < mn + f + -+80e + 625
and by part (vi),
1
we have
3
10e +68
2e +13
loe +68 < 93 - (mn+ f +y)
117
16e +109
< 80e+ 625
1
5
80
80e +68'
but this is blatantly false.
If (z, a, b, c) = (1, 1, 2, 5), this implies by part (ix), after much calculation,
(cf+ a + 3) -cm - bn
93 = (b-i)93 < cm -(ci+
a+ 3).b - c
-(n+f+7-
5e + 1
=(b - 1)(mn + e + y) + 10+ 3
Together with part (vi),
U
we have
-I
5f+ 1
n2
<93- (mri+e+)
ioe+55IM
So, as long as f > 1, this implies m2 > 3.
( +-Y)< 10f+ 3
-
=
xi. Assume now that (z, a, b, c) E {(13,5, 2, 1), (1, 1, 2,5)}, with t > -3 if (z, a, b, c) = (13,5, 2, 1),
and t > 2, if (z, a, b, c) = (1, 1, 2, 5). With this bounds, it is straightforward to verify that
1
13
1+Ib
ce+ a + a
c(bf + z + 1) + c
1
c(bf + z + 1)
c(b+
Z)
so we know from part (ii) that
Sc(m+b) (D;
since bc(be+ z +1)
bmn+bf+z n (bfC+z)2
CSc(m+b)
m
-
-
cm
m
I
m'
-c = c(m + b).
b <i, when f has the bounds we mentioned. Then
With i as in part (vi), note that c(f +y)
part (vi) tells us
bg 3 +b 2 n
b(g 3 +n)+ (b - 1)- bn
>D')
c(m+b)
b-c(bf+z)- -0b- 1) c
so
(b2 n, b) Sc(m+b) (D; -n- n2.
M IM2 )
/1
On the other hand, we know (bn, b) E Sbc(be+z)
so, by addition,
(b n, b) E Sc(m+b)
i\
(bf+z)2
(bmn~be+z. n
mb
)
bmn + bf + z n (bf +z)
cm
m
mj
(in' )2j and (bn, 0) E Sc
bmn+be+z. n
'in,
I
kci
(bf+z) 2
2
\Sc(m+ 2 )
(D; n,
Hence,
Scrm(D; n,
in, n)
bmn + bf + z m (bf+Z)2
M
cm
fails the triangulating condition for c(m + b).
m
and (D;
) C SCm
m2
It follows from (i) that we must have
C
m
b2 < c(m+ b), which is to say m2 < b(m+b) = b+
2+ i, which implies m2 = 2. If (z, a, b, c) = (1, 1, 2, 5) and f > 1, this contradicts part (x).
(z, a, b, c) = (13,5,2, 1), we have from parts (i) and (vii) that
0 < 93
-
+)
(mn +
in
If
< 2f+ 11'
so the only possibility is
= 93 -
mn = f + 7 +
2n
2
2
M2 samd2
3
2
E
as claimed.
118
in
)
Putting it all together, we have the following.
Proposition 4.33. Let z, a, b, c be consecutive odd-indexed Fibonacci numbers, f E Z, (m, n) =
2,-) a weakly triangulating sequence with m, n, M 2 > 2. Then
(b 2f + a2 , c 2f + b2 + 2), and (D;
- 1. Moreover, at least one of three statements must hold:
2 >
equals m or a fair approximation to it, not exceeding
(a)
(b)
~z
b
equals (bf+z) 2 , with bmn+be+z < M 2 D, or we have one of the cases listed in Lemma
parts (x)-(xiv), which we repeat for convenience:
2
xi. If b <
m2
+
= f
X. If M2 < b, then E
(0, b) with 3m2 =
which implies
m2
is the unique integer in the interval
a (mod b).
then these are the only possibilities:
< m and -Mn2 < (bf+z)2,
in
* (z, a, b, c) = (34,13, 5,2), with f E {-6, -5},
11
,2 6
<m 2 D
=M2, and86
=T,
have
*(z,
4.30,
and
4
67 < m 2 D
445,
<
a, b,c) = (13, 5, 2, 1) and
xii. If b < m2 < m and
2
and
81
<
6e+41
2=
-5,
. If
then we
If f = -6,
=
then we have
10667
101
2
> (bf+z)
,
4
n2
then c
+
with 2m±
1 and
=
n2
2D
< 2mn+2r-
2
(be+z) -(rn+2)
_
If (z, a, b, c) =
(13, 5, 2, 1), then
16n3 - 6n - 2 =8n 2 -4n2n + 1
1
1-
2n +
nM
2D
1
< 8n
24n
- 4n - 1 -
-4
- 4
8n -4n-3
24n
.
If (z, a, b, c) = (5,2, 1, 1), then
m3
= M + 1M
2
m 2 (M 3 - _ 2 -1)
1
-M2D 1m+
2
<
+ 1m2
xiii. If M2 > m and c > 1, then these are the only possibilities:
*
(z,a,b,c) = (34, 13,5,2), f= -6,
- =M 193 and
S(z,a,b, c) = (2, 1,1,2), f = 1, and hi =
the intervals
64 141)
5 '11
'
E17,
with
3'
and nt E
174
194 7
< 576.
T37
< M2D
8
6';
5
,} with
21,25
362 213)
434 332)
'
17 ' 10'
17 ' 13'
=27
and
273)
16
2,
32'
n2
M2
M 2 D in
respectively.
(z,a,b, c) = (1, 2, 5, 13), f = 0,
intervals
I
2570 37364)
'
13 ' 189
[23931
1 O'O
n2
with m 2 D in the
E {
54802)
229 /'
respectively.
xiv. If M2 > m and c = 1, then these are the only possibilities:
is a fair upper approximation to L6. If
2
-250k+17
< M2D < 900k 30k-5
'0M_
thenthn
3f12~i
=
=
and
n
2
119
n
-
46_-5
for some k>1,
*
=M 3Z and
n
M2
= ',5
with 362
17 < m 2 D < 213
10
(c) One of the cases listed in Lemma 4.32, which we repeat for convenience:
- (z, a, b, c) = (34, 13, 5, 2), f = -6, 2m = 19'
3, and
2n+ 3 ,
- (z, a, b, c) = (135, 2, 1) and 2
In the special case f = -4,
-
152 < M2D <
m
, or
=
n2
M2
= 4', with i2 < M2D < 885
29mD
with 4n +1+
m
2D
< 4n + 1+
, there are two additional possibilities:
9'
= L3 with 2- < M2D < 3 3
n2
-(Z, a, b, c) = (1, 1, 2, 5), f = 1, "
L , and
M 5
n2
M2
.
'2
10
M2
3
with
= 7,
with - 5 <M2D
3'
~-28 < '0.
Note that we are not claiming the converse. This is a necessary, but not sufficient, condition on the
pair (m2, n 2 ).
Proof. We omit the proof of the bounds on D. Let us establish, however, that these are the only
2 that arise.
M2
Note that
Fm
1
b2
mz be~z
b2<
(bt+z)2
b
M
Obviously, on the low side, [g] - 1 is a fair lower approximation to m. In the middle, we claim
bf-z = f + ! is a fair approximation to
= f + 2, because
- 2
bz-a 2 = 1 and no fraction
of denominator less than b can be within b of Z.
Similarly, we claim that no fraction between b
and (bf+z)2 has denominator less than b.
Indeed, this claim is vacuous of b = 1, and if b > 2, then it suffices to note that
(bC+z)2
m
which holds because m = b(bf+z) -1
(bf +z)
b
_
be+z
1
<bm
b(b-1)'
> (b- 1)(b+z). In the b > 2 case, we can conclude that bftz
is in fact a fair lower approximation to (beMz)2. (If b = 1, there might be another integer strictly
between them.)
In summary, no fraction in the interval (', (be+Z)2) has denominator strictly less than b, and if
b > 2, then "+z
is the only fraction in that interval with denominator b. Since we are assuming
b
that m 2 > 2, it follows that, if
n2
(be+z)2
m
M
b2 M2
'
then either 2- 9
or m > b. In the former case, 2 is indeed a fair approximation to 7m
', which
b or
m2
falls under case
(a) of our classification. In the latter case, 2 must satisfy the jth triangulating
condition for j = cm. Comparing with Scm(; -, '), we see that
min{bg3 , bmn + be+ z} ( cmD < max{bg 3, bmn + be + z}.
The two quantities bg 3 and bmn + b + z are exactly equal when 93 = mn + bf-.
Lemma 4.27 that (Q;g, g) is strongly triangulating and
trichotomy:
Case (a): We have
1
-1<
M2
<b
120
[7
[
(4.5.4)
Recall, too, from
, so we have the following
We have already mentioned the case
-IL
M2
b+z, so assume now that
b 'M2
<
'2
e+z.
b
Recall from
Lemma 4.22, part (i), that D > M, so we have D > 9 regardless of how 2 compares to
and if n > m, then we use (4.5.4).) Thus, D > cm 9,s
>
b7, then D >
(
,
C) m
b
M2
Sj(';
,)
S(D;
M2
b 'cm'0
bT
for all j E N.
Then it is easy to see that Lemma 4.19 applies: for the lower bound we have
m+ n +
- 1 = (b2 + C2 + 1)e + a 2 + ac + 1+
=(a + c)(cf + a) +
> (1+
so
1- mn +
[b2
-1
a]
c) -+ 1 = c + 2,
]-1 > mn - (m + n) + c + 2 = max(Z\W ) + c + 2.
2
Case (b): We have be+z <fn2 < (bf+z)2 and g 3 +n < c(b + z)D. (The condition on D is meant
to be an additional assumption; we are not claiming it as an immediate consequence of the bounds
on n.)
In this case, according to (4.5.4), we know
bmn + be + z ( cmD < bg3.
Then Lemma 4.30 applies.
Case (c): We have 93 + n > c(be + z)D. Then Lemma 4.32 applies.
This completes the proof.
EF2 J+F2
4.5.2
Cases where n
k+
E]
+
-1+
In stark contrast to the case of (m, n) = f(b 2 , c 2 ) + (a 2 , b2 + 2) discussed in the previous subsection,
most of the remaining cases do not admit many extensions to triangulating sequences of length 2.
Lemma 4.34. Let 2 < a < b < c < d denote consecutive odd-indexed Fibonacci numbers. There
are no weakly triangulating sequences (D; ,n ) with m2 > 2 for the following I:
1. (m, n) = (a, c) with a >, 5.
with a > 34.
2. (m, n) = (a{ , i)
3. (m, n) = ( a+2b,
+2d)
with a > 89.
4. n is a fair upper approximation to 6.
5. (m, n) = (m, 9m - 1) with m > 2.
For the following -,
(D; n, n):
there are only finitely many
n
completing a weakly triangulating sequence
1'. For (m, n) = (a, c) = (2, 13), there are only two possibilities(M2, n2) making making (D; Q, g)
a weakly triangulatingsequence, namely (n2, n 2 ) = (2, 3) with 1 < m 2 D < 9, and (m2, n 2)
(3,4) and 2 < m 2 D < 1.
121
2'. For (a, b, c, d) = (5, 13, 34,89), (m, n) = (ab, +d)
(6,41), there is only one possibility
) a weakly triangulatingsequence, namely (M 2 , n2) = (2, 1), with
(M 2 , n 2 ) making (D; ,
6
2
D< 72.
3'. For (a, b, c, d) = (13, 34, 89, 23) and (m, n) =(a%,
c)
= (27, 185), there is only one possibility (m2, n2) making (D; ', 2) a weakly triangulatingsequence, namely (M 2 , n2) = (5, 1),
with 12514
m2D < 332 3. (Please note that the case where (a, b, c, d) = (2, 5,13,34) and
(M, n) = (a%2b, c2d) = (4,27) is quite different; it happens to be covered under Proposition
4.33, with t = 0.)
For (m, n) = (3, 20), there is only (D; 2, 1) with
m2D
M
2
< 17.
Proof. The finitely many cases of (m, n) listed above that do admit weakly triangulating sequences
(D; ',
a) can all be checked by a finite sequence of operations on a computer, so we will not discuss
M M2
them any further. Let us concern ourselves with (1)-(5). Suppose for the sake of contradiction that
we had some counterexample (D; Q, ,) in any of those cases. Write
ni2/in2 and 93 = Mn+(.
We will apply the jth triangulating conditions for several j, omitting the routine verifications that
j is small enough that these triangulating conditions are required of (D; _, n).
Case 1: (m, n) = (a, c) with a > 5, and (since (D; ', n) must satisfy the bth triangulating
condition) by Proposition 4.22, part (i), and the results in subsection 4.4.3, we have
ac
ac - b
b
b--1
Since b > 13, note that
ac+ b
b+i
<
ac
b
<D<
ac-b
b-1
<.
ac+5b+1
b+5
Forj = b+ 1, we have
max{g3 + 2a, ac + b} > (b + 1)D.
Indeed, if this were not the case, then Sj(D; a,
3+ac+b±1- (a-1)(c-1)
2
_
) would contain at least
6+2ac+2b+2-ac+a+c-1
2
ac+5b+7
2
_
b2 +5b+8
2
(j+2)
2
elements, contradicting the jth triangulating condition. (Similar reasons can be given for the
remaining such inequalities in this proof, and we will not belabor them by writing out the calculations
in each case.)
Putting together the inequalities so far, we see that it must be that
93 + 2a > (b + 1)D.
Forj =b+5, we have
min{g3 + 6a + c, ac + 5b + 1} < (b + 5)D,
and by the above it must be that
93 +6a + c < (b+ 5)D.
122
Subtraction yields
ac - b
b -
4a + c = (g3 + 6a + c) - (g3 + 2a) < (b+ 5)D - (b+ 1)D = 4D <
4
So
0 > (4a + c)(b - 1) - 4(ac - b) = z(b - 1) - 4 > 2 .12 - 4,
which is false.
Case 2: (m, n) = (aQ,
with a ) 34, and
L),
cm
m+rn
1
3
+b
where q is either 4 or 5, and
we have
b<D
<
3
j
+b
+ c-
7'
=- a = b+ c (mod 3). Notethatm+n=b+c. Forj=3m=a+b,
min{g 3 + m2, m2 + m
and for
2
m+n
3mD,
1}(
= 2b - 3, we have
max{g3 + (3c - 2n - 1)m + (m - a - 1)n, (2c - n - 1)m + (m - 1)n} > (2b - 3)D.
Let's rewrite these two inequalities in terms of
instead of g3:
min{(, 1} < 3mD - m(m + n) <
b+ c-,q
2(a+b) <1,
=
b+ c-
max{(, 1} > (2b - 3)D - (2c - 1)m + n + 1
where we have incorporated the bounds on D that we computed in subsection 4.4.3. It follows that
( < 1 and, consequently, max{, 1} = 1, and
D<(2c -
m
)m -n
2b - 3
3
3(2b - 3)'
We can plug this back into our first inequality involving
=mn(,}
2
+n +
:
2m
mDmm+n) < 2b - 3
1
2'
as one can easily check the last bit under the assumption a > 34. In particular, the denominator
is at least 3, so we may apply the triangulating conditions for slightly higher j.
m2 of
For j = c, for example, we have
min{29 3 + (b - 2m)n, bn + 1}
and for
j
cD,
= a + 3b - 3, we have
max{2g 3 + (2b - 1)m + (b - 2m - 1)n, g + (b + 3c - 2n - 1)m + (m - 1)n} > (a + 3b - 3)D.
Let's both of these inequalities in terms of , and use the upper bound for D that we derived earlier:
min{2 , 1} < cD - bn < c
m m 1
}>
a +3b-br3) D
max{2( - 1, (} > (a + 3b - 3)D
m+n
3
- 3 (2 b 3 )
(b + 3c - 1)m + n.
3
-
123
2c - 2b +3
2
+ 3(b-3
bn =
3(2
3
-
( 2 b)
<11,
So in fact
S=max{2 - 1,
} > (a + 3b - 3)D - (b+ 3c - 1)m+ n,
so we have
cD - bn
(a + 3b - 3)D - (b+ 3c - 1)m+n <
from which we can calculate
m+n
3
m+n
3
1
3a+3b-6
1
which contradicts the bound in subsection 4.4.3.
2d) with a ; 89, and
Case 3: (m, n) = (a+2b,
1
b+2c
+b
3
where 7 is either 4 or 5, and a
For j = c, we have
c - b+
cm
b+2c
b <D<
3
3
+ b+2c-77'
2c (mod 3). (Note that an - cm = 2 and dm - bn = 1)
min{g 3 + (b - m)n, dm} < cD,
and for
j
= 3(b - 1), we have
max{g 3
and for
j
= b+ c
+ 3cm - mn - m - n - 1, 3cm - m - n} > 3(b - 1)D,
3, we have
-
max{g3 + (c -
)m + (b - m - 1)n, (c - 1)m + (b - I)n} > (b + c - 3)D.
Let's rewrite all this in terms of , and use the upper bound for D we found in subsection 4.4.3:
=
2c-q)
min{,
11 <, cD - bn < c (b+2c +
max{,
1} > 3(b - 1)D - 3cm + m + n + 1,
0} > (b+ c - 3)D - (c - 1)m - (b - 1)n.
max{,
bn < C -
bn+±1
+1bn
c
= 1,
so
< cD - bn < 1,
and so
1 > 3(b - 1)D - 3cm + m + n + 1,
which means
D<3cm-m-n3(b - 1)
1
m+n
3
1
3(b - 1)'
and in particular
(m + n)
c
( <- (
3
-c
3(b-
) 1
3(b -1))
124
1)
2
3
b- a +2
3(b - 1)
1
as one can easily check the last bit under the assumption a ;, 89. At the same time, if we recall
that c < D, then
>
c - 3)
(b+
(c - i)m - (b - 1)n
-
(b - 1)(dm - bn) - m
b
1
5
b-a-1
3b
b-m-I
b
under our assumption that a > 89. In particular, the denominator m2 of ( is at least 9, so we may
apply the triangulating conditions for slightly higher j.
For j = 4c, we have
min{5g 3 + (4b - 5m)n, 4bn + 1} < 4cD,
which is to say
min{5 , 1} ( 4(cD - bn) < 1,
thanks to the upper bound on D that we derived above. But then
4(cD
(b + c - 3)D - (c - I)m - (b - 1)n<
from which we can calculate
cm
D< c+
b
-
5
b)
cm
rn+-5- b <+-b
C
b
b(5b+ c - 15)
which is a contradiction.
Case 4: a is a fair upper approximation to L. When 2 < m < 7, a computer program can verify
(by brute force) that the triangulating conditions for j < 5m are contradictory, so there cannot be
any D and 2 for which (D; -, n) is triangulating. So assume (m, n) = (9f +-8, 64f + 57), with
1
12(6ef+ 5)
8m
D< 3
8m
3
For j = 24f+ 40, note
mn + 7m + 6n=min{g3 + 7m + 6n, mn + 7m + 6n}
so
1
8m
3
-
+1
12(6f+ 5)
<
24f + 40
1
8m
mn +7m-+6n
=
(24f+ 40)D,
D<
-
3
+
12(6f + 5)1
which is a direct contradiction.
Case 5: (m, n) = (m, 9m - 1), with
m3m-1 <D<3
3m--<D<3m-.
3m - 2
3
For
j
=
3m
+ 2, we see
mn + 6m = min{g 3 + 6m, mn + 6m} < (3m + 2)D,
so3m
M -i+1
+ 1
(3m - 2)+4
mn +
6
m<D
3m + 2
125
3
m-I
3m - 2'
but this can happen only if
1
rn-i1
4
3m - 2'
D
or m < 2, which is not allowed.
Lemma 4.35. If (D;
is weakly triangulatingwith b, c,m2 > 2, then n2 = 1 and
P,)
bc +
Proof. First of all, (D;
2)
1
bcn
g3
1
<-D < bc +.
bcm 2 bc ";-
=
2
must be triangulating, so the results in subsection 4.4.3 tell us that
bc < D < bc +
In particular, (D;
2,
n)
I
1
bc - 1
(4.5.5)
.
is required to satisfy the jth triangulating condition for all
j < bCM 2.
Our strategy is to exploit the fact that (C;
is strongly triangulating. This is equivalent to saying,
"(bc; 7; 0) is strongly triangulating," although I suppose we have not defined triangulating sequences
involving zeros, so perhaps this should be taken as merely an intuitive guide to what follows.
C)
First let us prove that n2 = 1. Suppose for the same of contradiction that n2 > 2. Let p =
so p is the smallest positive integer such that i < g.
to the strongly triangulating sequence (;
Note that m > p. For j
=
,
pbc, a comparison
) shows that
pb2c 2 + 1 = min{pg3 , pb2c 2 + 1} < pbcD < max{pg3 , pb 2c 2 + 1} = pg.
Let us consider two cases, depending on p:
9 Case 1: p = 1. For
j = bc + 1, we see that
min{g 3 + 2b 2 , b2c2 + bc + 1} < (bc + 1)D < max{g 3 + 2b 2, b2 c2 + bc + 1},
but we already know from above that
1
(bc + 1)D > bcD + D > (b2 C2 + -) + bc,
p
so it must be that
b2 c2 + bc + 1 < (bc + 1)D < g 3 + 2b 2 ,
and
For
1
n2
M > b(c - 2b) + 1 + A > 3.
j = 2(bc - 1), we have
min{g3 +(bc-2)bc-1, (2bc-2)bc+2} < 2(bc-1)D < max{g 3 +(bc-2)bc-1, (2bc-2)bc+2},
but since we know n2 > 3, it must be the case that
(2bc - 2)bc + 2 < 2(bc - 1)D < 93 + (bc - 2)bc - 1,
and in particular
(2 bc - 2 )bc + 2
2(bc - 1)
which directly contradicts the upper bound in (4.5.5).
126
1
bc - 1'
9
Case 2: p >, 2. For
shows that
j
=
pbc + 1, a comparison to the strongly triangulating sequence (f;
2)
(pbc + 1)D < max{pg3 + 2b 2, (p - 1)93 + b2 c2 + bc, pb 2 c2 + bc + 1}.
By the minimality of p, we know that
we have
<
2
and g3 < b2 c 2 +
p
, so, as 2b 2
p1
+
I
< bc,
2 2
pg3 + 2b 2 < (p - 1)g3 + bc < pb c + bc + 1.
So it must be that
(pbc + 1)D < pb2 c2
+ bc + 1,
but this blatantly contradicts the fact that
pbcD > pb2 c 2
+ 1
and
D > bc.
Thus, we have shown that n2 = 1, SO
strongly triangulating sequence (g;
is a fair upper approximation to 0. A comparison to the
'2
shows that
2)
b2C2 +
=
93 ( bcD
m2
and
(bcm 2 - 1)D < (bcm 2 - 1)bc + 1,
as claimed.
Lemma 4.36. Let (m, n)
-
#Sj
F4k+4 ) and 0
(E,
m~inn
(j
<m ± n. Then
;-nn) - #Sj ( 0;2#4
; )
if 3 1j,
=mL
otherwise.
0
~n
Remark. Note that m, n are always relatively prime integers, but
= F4k+2
is never an integer.
Proof. Let us prove this with induction by threes. For j < 3, this can be checked directly. Suppose
that j > 3, and the claim holds for j - 3. Note that, if x, y > 0, then
(X, y) (E Si
3m ;
(x, y) E Sj ( 9 ;
<=4
(X - 1, Y - 1) (E Sj-3 (3m
1) E Sj-3 (2
)
)
4
so we are done by induction if we can show that, for x, y E N,
(0,
y) E Sj
m+n;)
( 3m
(X, 0) E Sj
m +n n
(3m
+=>
(0, y) E Sj-3
(p2;
4)
+->
(x, 0)
(2;
4)
m)
'm)
127
E
Sj-
3
U
((m + n)j, 0)}
For the first equivalence, we have to show there is no integer y > 0 such that
2
_F4k+4
3n
m+n F4k+2'
<
y
and indeed this is true, because
j
is a fair upper approximation to 92, and we are assuming
F41+4
<m + n < 3n.
For the second equivalence, we have to find all integers x _> 0 such that
m +n
3m
x
j
Note that
F4k±2, so
F4k+2
F4k
and
in fact
F4k+2
F4k
sp2 ,
are two consecutive fair upper approximations to
F4 k4
F4k +2
2
X
m+n
j
3m
and
j < m+ n
=
m+n
x
3m
as claimed.
El
Lemma 4.37. Let (m,n)
-
(F4,
F 4 k+
4
)
Then the sequence (M+n n ,+')
is a strongly pseudo-
triangulatingsequence.
= mn+ , which is of course a necessary
Remark. Note that (m+ n) 2 = 9mn+1, so indeed (
condition for (-+n; f, 1-?) to be a strongly pseudo-triangulating sequence.
1-s) satisfies the jth triangulating condition for every
Proof. First let us prove that (-+n;
j K m + n. For such j, in light of Lemma 4.36, it suffices to show that
with 0 <
#S
nJ
m
-#
-)
m+n
n l+E
(3
mn
9
)
j
thwif3
ej,
otherwise.
0
Let
T := {(X, Y, Z) E N3 1 X < n}.
Every member of Sj (9+; n) can be expressed as (X + nZ, Y) for a unique (X, Y, Z) C T, just as
every member of Sj (-" 1_-; ,
) can be expressed as (mX + nY, Z) for a unique (X, Y, Z) E T.
Now for (X, Y, Z) E T, it is clear that
(mX +nYZ) E
(m+n
3
n
'm
1+E
9
<--+
mX+nY+
mn S+
__
m(X +nZ)+nY <
#
(X + nZ, Y) E Sj
9
z
3(+n)j
3
(m~n)j
3
3
m+n
n
3m 'm)
Thus, if we let
U :=
{(X,Y,Z) E TI (X+nZ,Y) E Sj (-"i"; n) and (mX+ nY, Z)
then we must prove that
={L 33-i#Ui #u
128
if 3 1i,
otherwise.
S ("-
n ,f6)},
Indeed, for (X, Y, Z) E T, we have (X, Y, Z) E Uj if and only if
m(X + nZ) + nY <(
1 +E
(m + n)j
Z,
3
< MX + nY + (mn + 9
which, if we remember that (m+n)j < (mn)'
3(mm + ) is equivalent to the condition
otecodto
(mi +
euvlnt
3(mn + 1)
Z E {1, 2} and m(X + nZ) + nY - (m + n)j
3
=
Obviously, this can happen only when 3 1j. For any fixed Z E {1, 2}, there exists at most one pair
(X, Y) E N2 with X < n and
(m + n)j
mZ,
-3nZ
3
mX +mY=
and there is such a pair (X, Y) precisely when the right-hand side is a non-negative linear combination of m and n. The minimum j for which this occurs is j = 3mZ, since
(m
+ n) *3mZ
3
but
(m+)(3mZ-3)
mnZ = m(mZ - 1) - n =m(mZ - n - 1)
3
+ (m - 1)n,
and both -1 and mZ - n - 1 < 2m - n - 1 are strictly negative. Thus #U includes one element of
the form (X, Y, 1) when 3 1j > 3m, and one element of the form (X, Y, 2) when 3 1j > 6m, exactly
in accordance with our claims.
So indeed we have proven that ("n; n, '+) satisfies the jth triangulating condition for every
j with 0 ( j < m + n. To prove the same for higher values of j, let us induct in increments of
) satisfies the (j - m - n)th triangulating
"
m + n. So suppose that j > m + n, and (
condition. Let us count the number of elements (W, Z) E S j3 -; mL, 9):
e If Z = 0, then W < (m
"M.
There are
+ 1 non-negative integers W with that
"2j
fail to be non-negative linear combinations of m and n. So
property, of which (m-1)(n-1)
2
there are
(m-+n)j +
-
3
elements (W, Z) E Sj (m-n"-; ", ')
(m - 1)(n - 1)
+
2
with Z= 0.
* If Z E {1, 2}, then, W < (m+n)j - (mn + -)Z. There are
(m+n)j _
3
-
9n
elements (W, Z) E S (m+l;
J3
-(m
+n)
3
2lmet
(m-1)(n-1)
mn2
, +T)
with Z = 1, and
9'
22mm
]
(m - 1)(n - 1) = [(m + n)j
(m + n)j
3
min
(m-1)(n-1)
2
3
2
elements (W, Z) E Sj ("" "; ;1-,6) with Z = 2.
129
2mm
(m - 1)(n - 1)
2(w2
* If Z > 3, then (W, Z - 3) E Sj-3('n; n, ')
n))(m+ n) 3
((p
(h
(W, Z)
and
By the induction hypothesis, there are
j- m - n+2
2
elements (W, Z) E Sj(mn; g,
if j-
1
m+n
otherwise
10
with Z > 3.
1±+)
So it remains to prove that the four numbers above add up to
I
(m + n)j
2)
3
I
(m +n)j
+2
(mod 3)
which is to say
(i2),
+1-3mn - 3(m -1)(n
2
1
n+2)
- 1)
+
if j
M 2
m+n
(mod 3),
0 otherwise.
Note that
3mn +
3(m - 1)(n - 1)
2
and
(i +
2)
_ (Cl+ 2) - (m + n)
1+3mn+
2
2
2
2
j + 2
so
9mn-3(m+n)+3 _ (m + n)2 - 3(m + n) + 2
_
(
m- n+ 2
i
1)
(m
+72)
+ (m + n)(j + 1) -
3(m - 1)(n - 1)
2
-
2
= (m + n)j,
and it remains to prove that
(m+n)j =
(m+n)jJ
3
+2 [(m+n)j]
3
Well, recall that msn is never an integer. If
j
1
if j = m+ n
0
otherwise.
(mod 3),
= 0 (mod 3), this if obviously true. If j = m + n
(mod 3), then
(m + n)j
(m + n)j -1
_
3
3
so this is true again. Finally, if
j
- -(n
(m + n)j - 2
3
3
so yet again it is true.
triangulating.
(m + n)j + 2
3
+ n) (mod 3), then
(m + n)j
_
2(m + n)j] -2-
2(m + n)j~
-
2- (m + n)j + 1
3
Thus, we conclude by induction that (man; -L, '+)
Lemma 4.38. If (m, n) = (&, F4k+ 4 ), and (D; n,
fair upper approximation to
.
130
n
is strongly pseudoE
) is weakly triangulating, then
riust be a
Proof. By Proposition 4.19, we know this to be true if
n21-,
0 < n2<
m2
6
and - itself is a fair upper approximation to -. So let us suppose that n > 1, and n is not a fair
upper approximation to !. Let p E {1, 2,3, 4,5, 6} be the least positive integer such that 1 < n.
Note that if p > 1, then
1 <n2
p
1
m2
p-i
so that m2 > 2p - 1.
Our proof proceeds by applying the jth triangulating condition for certain values of
below. For each of these values of j, we will endeavor to show
S(m
S ;n n2 )
Si (D
MM2
j,
listed
+n. n I1
3
m 9
or something close enough to that for our purposes. Then we will derive several inequalities involving g3 and D, culminating in a contradiction one way or another. Let us begin by noting that
+n), and in
D; , 2) is required to satisfy the jth triangulating condition whenever j < Mn
particular in each of the following six cases:
1. When
j
n - 2m, as long as m2 > 2, since n - 2m < 2(m+n)
2. When j =2(m+n)+r-9 as long as m2 > 2, if for this proof TI denotes the unique number such
that 77 E {4, 5} and rI = m + n (mod 3), just as in subsection 4.4.3.
3. When
j
m + n, provided that
=
m2
) 4;
4. When j = 3(m + i) < (p+l) m+n), provided that p > 2;
5. When
j
= 2(m
6. When ] = 3i
+ n), provided that
m2 > 7;
+ m + n < (p+l)(m+n), provided that p > 3.
Let us consider each of these six cases in turn, and derive some consequences from each.
Step 1. Let
j
= n - 2m = F4k+1, and note that
j - (m + n) - 3m - (2n - m) = F4k+1 - F4k+2 - F4k - F4k+3 = 1,
so
m(2n - m) <m(n -m)+
1
< M(2n -m)+
9
mn+
1
3
j(m +n)
n)
=j
3
Therefore,
(m(2n - m), 0), (m(n - m), 1) C Sj
and a comparison of Sj
(Mn
,
+ e) with Sj (D;
( 3
;
, +E
M 9
-, ) shows that
min{m(2n - m) + 1, m(n - m) + 93} < jD < max{m(2n - m) + 1, m(n - m) + 93}.
131
Now, from the bounds in subsection 4.4.3, we know that
m+n
3
It is not hard to see that m+1n-7 <
D<
m+n
3
+
2_
2
D
m+n +
3
m+n-
so
j(m+n)+2
3m(2n-m)+3
3j
3j
3i
m(2n-m)+
3
so we conclude that
m(n - m)+ g3
jD < m(2n - m) +1,
or in other words
=2g3 - mn < (n - 2m)D - m(2n - m) =
+ (n
-
2m)
3
M2
D-
m3n
(4.5.6)
3.,1
In particular,
(4.5.7)
3
3
M2
and p > 2.
Step 2. Let j :=
2(m+n)+(7-9), and note that 5m+14n+i7-9
3
9
5m+5n+q-9
5n-31m+rq-9
+
9
j+m+rn
so that
j(m + rn) + 1
j(m + n) +1< (j(m + n) + 1
j(m+n)+1 _
3
3
miEW 2 ,
3
while
, and
Nan
j+n-11m
i+
3
7=
5n-31m+7-9 E
9
,
-mn)
+ (mn+
1)\
j(m + n)
9J<
3
Therefore,
(j(m + n) + 1
)
and a comparison of Sj (mn; -,
min{j(+
3 )
j(m + n) +1
- mn, 1)
3
+ E) with Sj (D;
jrn + 3 n)+1_mn +g
,
<)max
3
n 1
m'
+ E
,
) shows that
} jD
-1
3
m+ n
v Si
j(m + n) +
3
j(m + n) + 1
3
_i n +
-
g
- Jn
From (4.5.7) we know that 93 < mn + 1, so
j(mA-n)+1
3
1<D
j(mA-n)A-1
3
3mn+93-1jD<
Thus,
D
m+n
3
1
3j
m+n
3
132
1
2(m+n)+r7-9
(4.5.8)
Putting this together with (4.5.6), we have, after a straightforward calculation,
n2
1
- + (n-
-
3
m2
In particular,
2
2m) (D
m+n
1
3)
3
n- 2m
2(m + n) + rq - 9
2
-.
3
7 2, so M2 > 4.
Step 3. Let j = m+ n; since we showed in Step 2 that M2 > 4, we know that (D; I, D)
satisfy the jth triangulating condition. Note that
must
j(m + n)
<3
3mn <imn + 2 (mn
so
(3mn, 0), (mn, 2) E Sj
A comparison of Sj
(-"m-+'
,
t+h
;
'm 9
)
3
+ E) with Sj (D; -L,
'2)
E
shows that
min{3mn + 1, mn + 293} < jD < max{3mn + 1, mn + 293}.
But with a bit of arithmetic, we know from Step 2 that
D<
rnr
3
+
rn-fn
2
1
+
<
3(m+ n)
3
2(m+ n) +q - 9
_3rnn+1
m- + I
so it must be that
mn + 2g 3
In particular,
'2
(4.5.9)
(m + n)D < 3mn + 1.
< -, so p > 3 and M2 > 2p - 1 = 5.
Step 4. Let i be any positive integer such that 3(i + m) < (p+l) m+n); such i exists because
p > 3. Let j = 3(i + m). Note that
(m + n)i + m2 + mn+ 1
m+-n
3
3i + 3m
<
so
((m + n)i + M 2 , 1)
Sj (
+ 71;
3
'm
,
),
We claim that
((m + n)i +
i
2
Si
,1)
+
9
D; ",
m MM2)
as well. If this were not the case, then by the jth triangulating condition, the sets Sj (D; a
ntte
andand~
5 (m+n.
+)
~3 in
M
Sj
have the same size, so there would need to exist some
(Xy) E Sj
Since
n
m+n
n
3
',
> 1, we must have x < (m + n)i
mn+-<
1
9
1
9 +E)
\S
D;n n2
D;
,2
+ m 2 , y > 1, and in fact
(m + n)i + m 2
y-I
133
X
<mn+
n2
M2
.
But by assumption 'is the simplest fraction strictly between
y - 1 > p. In particular,
m+n n
(0, p + 1) E Si
which contradicts the assumption that j <
S (D; n,
}
and " , so it must be the case that
+E),
So indeed we must have ((m+n)i+m2 , 1) g
()m3n)
, which is to say
(m + n)i + m 2 + g 3 > 3(m + i)D.
If we combine this with (4.5.9), we find that
mn + 29 3 < D < (m+ n)i + m 2 +g
m+'n
3(i + m)
3
so, with a bit of arithmetic,
m+n
6n2/m2 - 1 < D
3(m + n)
n2/m2
3(i + m)'
3
and in particular
+ M
n2 <
6(i + m) - (m + n)
m2
We already showed in Step 3 that p > 3. If p = 3, then we may set i
m2
4(m + n) -,
i+ m
n2<
a contradiction. If p = 4, then we may set i + m
n2
m2
i+m
6(i + m) - (m + n)
=
+ n <n2/m2
3
_
3(i + m)
so that
< 1
1
3
p'
5(m + n) +,q - 9
21(m + n) + 6(q - 9)
1
4
15(m + n) - 6q
5(m + n) + n - 9
6[5(m + n) + 7 - 9] - 9(m + n)
m
4(m+n)-,
5(m+n)+-q-9 so that
a contradiction.
So we conclude that p > 5. Thus, we may set i + m =
D
=
_4(m+n)-
6[4(m + n) - 7] - 9(m + n)
6(i + m) - (m + n)
+ m
6(m+n)+3(q-6)
_
9
1n2
2(m + n) +,q
- 6
1
p'
so
2(m+n)+±--6
3
m2
For future reference, let us note in particular that
3
Step 5. Let
and (D; -,M M2
n)
j
4[2(m + n) +y7
6(m + n)
- 6]
= 2(m+n). We know from Step 4 that p > 5, so in particular m 2 > 2p-1 > 9 > 7,
satisfies the jth triangulating condition. Now note that (0, 6) just barely fails
to belong to S-(m
Sj (Mn ;
1
1
m+n
+ e), and for each integer y such that 0 < y < 5, we have (x, y) E
, 1 + E) if and only if x E W2 and
;
,
/+
I\
9+r+)y
2(
+ n)2
2(9mn + 1)
3
3
134
2
=6mn+,3
all cases
at hand,
with strict inequality if y > 0. But in Butinal
csesathad,9
equivalent to
x + mny < 6mn,
<
<
9
31,
so really the inequality is
with no strictness requirement even when y > 0. This means that a comparison between Sj(m+n; ni+
L,
E) and Sj(D; , 2) shows that
min{mn + 593, 6mn + 1} < jD < max{mn + 593, 6mn + 1}.
In Step 4, we proved that
D
so
2
<2(m + n)
.
mi+n
1
3
6(m +n)'
2(m + n) = 6mn + - 2 + -1 =6mn + 1.
6(m + n)
3 3
3
It follows that
mn + 59 3 ( jD < 6mn +1,
In particular,
n
M2
< 1,5' so p = 6. Let us rewrite the above result as
m+n
3
mn + 593
j
D - m+3
Step 6. Let i be any positive integer such that i <
15n 2 /m2 - 2
(4.5.10)
6(m + n)
such
(p-2)(n+n);
9
i exist because p
=
6. Let
j
so the
jth triangulating condition must
< M2(m+n)
= 3i + m + n, and note that j < (p+)(m+n)
f
3I
3
hold. We claim that (3i i- m i- n) D K (m -+n)i i- 393.- Indeed, if this were false, then we would have
n n2
D ; m ,I
M2)
((m + n), 3) E Sj
Since
>
m
for every (x, y) E S, ( -" ; - ,
+ E) \Sj D ;
r tee
and 3 < y K 7, but then
M2
,
(2i(m+n)-x,6
C Sj
-y)
#Sj
D;
n 1
mW9
+n
(rnF
(
3
,we must have x < i(m
,n
m+n n 1
,
3
M M2
'm' 9
+ E), so the jth triangulating condition
The upshot is that #Sj (D; n, a)l
>
cannot possibly hold for
contrary to our assumption. We conclude that in fact
D;
a, ),
(mn -, -
jD < (m + n)i + 393
which is to say
3
3
= (3z+ m i-n) (D
-
3
)
< 3(93 - mn) -
Let us combine this with (4.5.10):
15n
2
/m2 - 2
m + n
6(m + n)
3
135
+ n)
9n 2 /m2 -
3j
1
-
3
3n2
m2
1
3
from which it follows that
n2
2i
15i - (m + n)'
m2
provided that the denominator is positive. Since p = 6, we may in fact set i
2[4(m + n) - 77]
15[4(m + n) - 77] - 9(m + n)
n2
m2
8(m + n) - 2q
51(m + n) - 9
= 4(m+n)-7,
so that
1
6'
a contradiction.
I
, F 4 k+ 4 ), and (D; -,
Lemma 4.39. If (m, n) = (
is weakly triangulating, then
n)
D > m+n ,
3
and one of the following must hold:
*
-2 = 1 and m 2 D < (mn)m2 +
2
2
3
2(m+n)+nm-9d
m2
*
11-2M2
+
= -1 and m2D < (mon)"2
3
*
-2-
= 1 and m 2 D < (mn)m2 +
*
-2= 1 and m 2 D < (m+n)M2
+ 5(m+n)+77-9'
3
3
5
M2
M2
2
Ml2
*
*
3
3
4(m+n)-r '
'
+
= 16 and m 2 D < (m+n)m2
2(m+n)-1'
3
-2 =
M2
M2
3
4
I
m+n-1'
=
17 and m
D <
2
(mn)m2 +
3
2
7(m+n)-77
n2
+
3
9n2-1 and m 2 D < (m+n)M2
1
m2(m+n)+77-9
By
n2. Let j be the integer closest to m2(m+n)-4
Proof. Note (mi+n)2 = 9mn+ 1 and 93 = mn+ M23
Lemma 4.38, n must appear on the above list, and the lower bound on D comes from the lower
bound in subsection 4.4.3, so it just remains to derive the upper bounds on D.
* If
-=
, then j
=
2(m+n)+--9 E N, and the jth triangulating condition implies that
(q - 9)(m + n) + 5
9
jD< (m + n)j+
3
so
m
D<
+n
3
1
+ 3j
as desired.
" If
=,
-
then j = m + n - 1. If
jD <
whereas if 77
=
ri
= 4, then the jth triangulating condition implies that
(m+n)j312
+
=
3
M2
3
+ 2mn -
m+-1
3
E 93 + W2,
5, then the jth triangulating condition implies that
(m + n)j + 1
.
3
= 3mmjD <
136
m+ n- 2
cW 2 .
3
So in either case,
D <
3
+
-,
3j
as desired.
If
,
m2
E N, and the jth triangulating condition implies that
then j= 4(,+n)-:
jD< (m + n)j + 2=
3
M2
3 + 3mn -
E 93 + W2
9
so
D9<
+
3
4j
,
as desired.
0 If
n
M2
= 1,
5
then j = 5(m+n)+rj-9
E N and
3
(7
jD< (m+ n)j +
= 93 + 4mn +
3
m2
so
D <
mTn
+I 'n
-+
3
9)(m + n) + 5
-
E
93+ W 2 ,
1
5j
as desired.
If m
6', then
j
=
2(m + n) - 1. If a
r(m + n)j + 3 n2
(M
3
jD
-
=
1)
9
4, then the jth triangulating condition implies that
= 393+
3mn -
E 393 + W 2,
3
whereas if 7 = 5, then the jth triangulating condition implies that
jD< (m+ n)j +
=g3 + 5mn 3
M2
So in either case,
m+ n
1
D<
3
6.]
as desired.
* If , = , then j
3
E 93 + W 2.
and
7(m+n)-
1N
n) - 4
77(rn +I9n
E 393 + W 2 ,
= 393 + 4mn 9
9,,
jD<(m + n)j +3(n2
3
M2
so
D <
rn-I-n
+
3
2
21j
as desired.
*
If
n2
then
=
n2-,
M2
9 n 2 - 1 ,te
jD < (rn
3
jj3
= m2(m+n)+7-9 and
t~
t triangulating
h A
n the
i +i3( n2
M2
9
-
393 + (rn2 - 3)rnn + n2 +-(-
condition implies
mle that
9)(rn
9
n) -
E 393 + W 2 ,
so
D<
3
+
.
3M2j'
as desired.
l
So in all cases we have proved the Lemma.
137
Assorted results on triangulating sequences of length 3
4.6
In principle, we could keep going indefinitely, and try to classify triangulating sequences of all
lengths. Perhaps there is a general pattern waiting to be discovered. In the meantime, let us prove
some partial results on length 3 triangulating sequences.
Lemma 4.40. There are no weakly triangulating sequences (D; LL.,
,
&)
with M 1 , M 2 , M 3 > 2
in the following cases:
&M =
M1&
C,
where b and c are consecutive odd-indexed Fibonacci numbers, with 2 < b < c.
nj F4k4/3'
F k/3, where k
n* =
M1
2
and
n>
M2
> 2.
46
9
Proof. Suppose for the sake of contradiction that (D; n'-,
Part 1: n'
=
.
really is triangulating.
n)
n,
Then by Lemma 4.35, we have n2 = 1 and
bcm 2 +
1
b
-
bc
( D < bcM
2
+
2
2
bCM2 - I
) must satisfy the jth triangulating condition for j < bcm 2 M3, and, in partic,
,
Thus, (D;
ular, for j < bcM 2 + 5.
Note that the first 20 elements of the semigroup generated by b2 and c2 are, in ascending order,
0, b 2 ,
2b2 , 3b2, 4b2 ,
5b2 , 6b2, c2 , 7b2 ,
b2 + c2, 8b2, 2b2 + c2 , 9b2, 3b2 + c 2,
10b 2 , 4b2 + c2, 11b 2, 5b 2 + c2, 12b2 , 6b2 + c2
Using this list, and paying particular attention to the positions of the boxed expressions above, we
may apply the jth triangulating conditions for j E {bcM 2 + 1, bcM 2 + 5}. For j = bcM 2 + 1, we
find that
min{(g 3 + bc)M 2 , 94 + 2b 2 iM2 } - (bcM
2
+ 1)D < max{(g
1
(g3 + bc)M
3
+ bc)M 2 , 94 + 2b 2 iM2 }.
But as
D>bcM2 +->
bc
2
bcm 2 +1
the only possibility is
(93 + bc)M
2
< (bcM
2
+ 1)D < 94 + 2b 2 iM2 .
For j = bcM 2 + 5, we find that
min{(93+5bc)m 2 +1,
94+(6b 2 +c 2 )m2} < (bcM 2 +5)D < max{(9 3 +5bc)M
But as
(93 + 5bc)M 2 + 1
M2
bcm
2
- 1
138
bcm 2
+ 5
2+
1, 94+(6b2 +c 2 )M2}-
the only possibility is that
94 + (6b 2 + c2 )m2 < (bcm 2 + 5)D < (g3 + 5bc)m
2
+ 1.
Thus,
(g3 + bc - 2b2 )m 2 < 94 < (93 + 5bc - 6b2 - c 2 )m2 + 1,
and in particular
bc - 2b2 < 5bc - 6b2 _ c
2
or
(c - 2b)2
0,
which is patently absurd, since c > 2b.
Part 2: a
y
-
F
=
to 9, so either n 2 =
I
and m
is a fair upper approximation
< 7, or else m2 = 9n2 - 1. Note that g 3 = mn -m 2 + n2. Let us
Then by Lemma 4.38 we know that
/3
2
n
consider the various cases:
if D7- = }, then let j = m + n. The jth triangulating condition implies that, no matter how
small n is (as long as it is positive), we must have
mn - m 2
or else the set Sj
, n ',
mD;
m-
}
+ 293
,
(m + n)D,
will simply be too small. Thus,
(m + n)m 2
m2 + 2g3
m + n
3
+
4
3(m + )
.
Lemma 4.39 says that
2
(m+ n)m2
so
3
2(m +n)+rq-9
4
3(m +n)
2
2(m +n)+F-9'
which is false.
" The case of
If m
=
,
n
= ' will be treated together with the case of
then let
j
= 2n - m. The
=
; see below.
jth triangulating condition implies that
m(n - m) - m 2 + 493 , (2n - m)D,
which is to say
D
m(n - m) - m2 + 493
2n -m
_
4
(m + n)m2
+ 3(2n - m)
3
Lemma 4.39 says that
(m + n)m2
3
so
3
4(m + n)-
4
3
3(2n - m)
4(m + n) - 77'
which is false.
139
* If
=1,
n
then let
j
=
2(m + n). The jth triangulating condition implies that
mn -m2 + 5g3 < 2(m + n)D,
which is to say
mn - m2 + 59 3
(m + n)m2
2(m + n)
3
D>
+.
5
6(m + n)
Lemma 4.39 says that
D(m +<n)m2 +
3
so
3
5(m + n) + r7 - 9'
3
5(m + n) +,q
5
6(m + n)
- 9'
which is false.
-2=
M2
that
*If
-I
3k with k E {1, 2},
then for
j
= k(m + n) the jth triangulating condition implies
1 + k 2 (m + n)2 = 9k 2 mn + 3k - 1
=
mn - m2 + (3k - 1)93 < k(m + n)D.
Note that mn - M 2 + (3k - 1)93 is large enough that every integer exceeding it belongs to W 3 ,
and that the first 20 elements of the semigroup generated by m and n are, in ascending order,
0, M,
3m, 4m,
2m,
5m, 6m, n, 7m,
m
+ n, 8m, 2m + n, 9m, 3m + n,
10m, 4m + n, 11m, 5m + n, 12m, 6m+m.
Using this list, and paying particular attention to the positions of the boxed expressions above,
we may apply the jth triangulating conditions for j E {k(m + n) + 1, k(m + n) + 5}. For
j = k(m + n) + 1, the jth triangulating condition says that
min {g 4 + 2mm 2 , 1 + k(m + n)(k(m + n) + 1)} (k(m + n) + 1)D
<max{g 4 +2mm 2 , 1+k(m+n)(k(m+n)+1)}.
But recall and compute that
D >-
+_k2(m_+_n)
k(m + n)
= k(m + n) +
1
1
> k(m + n) +
k(m + n) +1
k(m + n)
so it must be that
1 + k(m + n)(k(m + n) + 1) < (k(n + n) + 1)D <94 + 2mm 2 .
For
j
= k(m + n) + 5, the jth triangulating condition says that
min {g 4 + (6m + n)M 2 , 2 + k(m + n)(k(m + n) + 5)}
< (k(m + n) + 5)D < max {g4 + (6m + n)M2, 2 + k(m + n)(k(m + n) + 5)}.
140
But recall and compute that
D < k(m + n) +
1
k(m+ n)-1
2
< k(m + n) +
5,
k
k(m +n) + 5
so it must be that
g4 + (6m + n)m2 < (k(m + n) + 5)D < 2 + k(m + n)(k(m + n) + 5).
Thus, putting this together with our results for
j
= k(m
+ n) + 1, we see that
1 + k(m + n)(k(m + n) + 1) - 2mm2 < 94 < 2 + k(m + n)(k(m + n) + 5) - (6m + n)m2.
In particular, this implies that
3k(4m + n) = (4m + n)m2 < k(m + n)(5 - 1) + (2 - 1) = 4k(m + n) + 1,
and
3(4m + n) < 4(m + n),
which is easily seen to be false.
if a?= 7,
then let
e
7)hn
M2
min{(8mn + 1)m
j = 3(m + n). The jth triangulating condition says that
2
+ g3, mn - M 2 + g3 + g4} < 3(m + n)D
< max{(8mn + 1)m2 + 93 , mn - m2 + g 3 + g4}
but
(7mn + 1)M2,
94 > 93M2 =
so in fact we must have
(8mn I+1)M2 + g3< 3(m + n)D,
which is to say
(8mnm+ 1)M2 +3
3(m + n)
_
(m+n)M2
3
+
1
3(mi+
n)
.
Lemma 4.39 says that
D < (m + n)M2
3
so we must have
2
7(m + n)
2
7(m + n)-rn'
1
3(m + n)
which is false.
* If
2=
,2
then let
j
=
(mn
(3n2 + 2)(m + n). The jth triangulating condition says that
+ n) 2 iM2n
2
+ 693 < (3n2 + 2)(m + n)D,
which is to say
D
(+n)2
I>
3
2
+ 3(3n2 + 2)(m
+ n)
141
Lemma 4.39 says that
(m
1
+ Dn)m2
++
m2(m + n)+ r - 9
3
so we must have
2
3(3n2 + 2)(m + n)
1
m 2 (m + n) +?9
which is false.
by Corollary 4.29, 2 is a fair upper approximation to 46
Part 3: - =3 and 2 m2> L.9 Then
am2
The cases where n 2 = 1 and M2 < 7 can be checked by computer, so let us pass to the case where
n2
_
m2
46k - 5
9k - I
for some k > 1. Let j = 30k + 20. The jth triangulating condition implies that, no matter how
small n is (as long as it is positive), we must have
100k - m2 + 693 < (30k + 20)D,
or else the set Sj
(D;
n,
)
,
will simply be too small.
Combining this with the result in
Proposition 4.33, we see that
1Gm 2
2
3(30k + 20)
4-
3
100k(9k - 1) + 693 < D <
30k + 20
3
3
1
+ 3(30k - 5)'
2 .m
which implies that
2
.1
30k +20
30k - 5'
0
which is evidently false even for k = 1.
Corollary 4.41. If (D; ", n, n) is weakly triangulating with mi, M 2 , M 3 > 2, then there exist
consecutive odd-indexed Fibonacci numbers b, c and f E Z such that (m 1 , n 1 ) = f(b2 , c 2 ) + (a 2 , b2 +2).
Proof. In light of the Lemma, it remains to eliminate a finitely many cases, enumerated in Lemma
4.34. But this is easily verified by computer.
Lemma 4.42. The sequence (3;
m' >1.
M/,
9
+ e) is strongly pseudo-triangulatingfor all
Proof. Let
Ti := {(XYZ) E N3 I X <9m'+1and X+9Y+(9m'+1)Z (i}.
Note by Lemma 4.21 and Proposition 4.16 that (3; 2, 1) is a strongly triangulating sequence. Therefore, for each j E N, we have
j+2
2 )
2 13)
=
#Sj (3;
=
#{(X, Y) E N 2 I X
3' 1
#
1 and X + 9Y < 3j}
=#{(X, Y, Z) E N2 IX < 9m'+ 1 and (9m'+ 1)Z + X 5 1 and (9m'+ 1)Z + X + 9Y < 3j}
I
1 or Z > 1}
=#{(X, Y, Z) E T3j X
= #T3j - #{(X, Y, Z) E T 3j X = 1 and Z = 0}
#T3
(4-6.1)
-
142
Meanwhile, the semigroup W' generated by 2m', 3m', and 9m'+ 1 is contained in the semigroup
generated by m' and 9m' + 1; more precisely, each element of W' has a unique representation as
m'X + (9m'+ 1)Y, where (X, Y) E N 2 and the following two conditions both hold: (i) X < 9m' + 1,
is in bijection with the set of
, 9m'+1 + E
and (ii) X y 1 or Y > m'. Thus, Sj 9m'+1. 2 3m
triples (X, Y, Z) E N3 with (X, Y) satisfying the conditions (i) and (ii), such that
m'X + (9m'+ 1)Y +
Z <
1 +E
9
m'(9m'+ 1) +
3
'
3
or in other words
9m'X
+ 9Y + (9m'+ 1 + e)Z < 3j,
9m' + I
(4.6.2)
for all sufficiently small E > 0. But as X < 9m' + 1, the inequality 4.6.2 is easily seen to be
equivalent to the conjunction of the following two conditions: (iii) X + 9Y + (9m' + 1)Z ' 3j, and
(iv) if X + 9Y + (9m' + 1)Z = 3j, then X > 1 or Z = 0. The upshot is that
9m' +1
3
2 3m'+1 9m'+1
3'
m/
9
=
(4.6.3)
#T 3j - #{(X,Y, Z) E T3j IX = l and Y < m'}
- #{(X, Y, Z) E T3j I X = 0 and Z > 1 and X + 9Y + (9m' + 1)Z = 3j}
=
#T3j - #{(Y, Z) E N 2 I Y < m' and 1 + 9Y + (9m' + 1)Z < 3j}
- #{(Y, Z) E N2 I Z >lI and 9Y + (9m'+ 1)Z = 3j}.
(4.6.4)
Now we claim that, for (Y, Z) E N 2 with Y < m', we have
1 + 9Y + (9m'+ 1)Z < 3j Indeed, on the one hand, if Y + m'Z <
3m'j
3m'j
Y + m'Z < 9m'+1
(4.6.5)
, then in particular
Z <
3j
9m' + '
so
9(y + m'Z) <
3 -9m'j
9m'
=3j
3j
9m' 1
3j - Z,
then in particular
and 1+9Y+(9m'+1)Z < 3j, as desired. On the other hand, if Y+m'Z > 93m'
3m'j
3m'+3
9m' + 1'
m(Z + 1) = m' + mZ >
so
Z>
Thus, Z
3j
9m'+ 1
1.
L>9F1 , so
1 + 9Y + (9m' + 1)Z> 1 + 9Y + 9m'Z+Z+
143
3___
3
19+11
9m' + 1
-9m'j
+
3j
9m'I+1I
3j
= 3j,
as needed. We conclude that (4.6.5) holds, so
#{(Y, Z) E N2 I Y < m' and 1 + 9Y + (9m' + 1)Z
#
3j} =
[93m' ~
9m'+1K
3m'j]
Y+m'ZEN IY+m'Z < 9m/1
9m' + I
(4.6.6)
Meanwhile,
Z E N>o I Z
#{(Y, Z) E T3 j I Z ;> 1 and 9Y+(9m'+1)Z = 3j} = #
Now
3j
(mod 9) and Z < 9m'+1}
(4.6.7)
]
3j+9 -
is the least strictly positive integer congruent to 3j modulo 9, so any given Z counted in the righthand side of equation (4.6.7) may be written as
Z
=
1
9-
9Z' +3j+
3j
for a unique Z' E N. Thus,
#
ZE N>O IZ=3j
=#
(mod 9) and Z
Z' C N IZ'
9m' + 1
}
=
3
+] 3(9m'+1)
#
Z'
E N I 9Z'+ 9 + 3
1}=F+3_
-3m'j3
9m' +i
=Fj]
3 ( 9 m'
[
-
-
9
F1
3
<9m+1}
+ 1)
9m'+1 I
(4.6.8)
Putting equations (4.6.4), (4.6.6), (4.6.7), (4.6.8), and (4.6.1) together, we have
9m'
#Sj (-3
+1 2
'13'
F31
-FI
,
=
3m'+ 1 9m' + 1
m'
'
9
=#T
=
33 -
#Tj -
as desired.
144
9m + 11
]= #Sj
3;
3m+
9m' + 1
j + 2
1)
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