Lyness Cycles, Elliptic Curves,
and Hikorski Triples
Jonny Griffiths, Maths Dept
Paston Sixth Form College
Open University, June 2012
MSc by Research, UEA, 2009-12
(Two years part-time)
Supervisors:
Professor Tom Ward
Professor Graham Everest
Professor Shaun Stevens
Mathematics
The Structure of this Talk
1. Lyness cycles (periodic recurrence relations)
2. An introduction to elliptic curves
3. The link between Lyness cycles and elliptic curves
4. Hikorski triples
5. Cross-ratio-type functions
6. Conclusions
Part 1: Lyness Cycles
un+1 = un + un-1,
u0 = 0, u1 = 1
The Fibonacci sequence
z=x+y
x, y, x + y, x + 2y, 2x + 3y, 3x + 5y, ...
x
y
Can this be periodic?
x = 2x + 3y, y = 3x + 5y
x = 0, y = 0.
Order-2, periodic for these starting values,
(locally periodic).
Can we have a recurrence relation
that is periodic
for (almost) all starting values?
Globally periodic
x1, x2,..., xn, f(x1,...,xn), f(x2,...,f(x1,...xn))..., x1, x2...
xn+1,
xn+2
.........
xm+1, xm+2
Order-n, period-m
Globally periodic behaviour is very atypical of
difference equations, and accordingly only a
very restrictive class of functions f(x1, x2, ...)
exhibit this behaviour. Mestel.
Robert Cranston Lyness, 1909-1997
Mathematical Gazette, 1942
Globally periodic for x and y non-zero,
order-2, period-5.
Imagine you have series of numbers such that if you add 1 to any
number, you get the product of its left and right neighbours.
Then this series will repeat itself at every fifth step!
The difference between a mathematician and a nonmathematician is not just being able to discover something like
this, but to care about it and to be curious why it's true, what it
means and what other things it might be connected with.
In this particular case, the statement itself turns out to be
connected with a myriad of deep topics in advanced
mathematics: hyperbolic geometry, algebraic K-theory, and the
Schrodinger equation of quantum mechanics. I find this kind of
connection between very elementary and very deep
mathematics overwhelmingly beautiful. Zagier
Regular Lyness Cycles
Order 1
Order 2
Order 3
Order-1 regular Lyness cycles:
what periods are possible?
For what values of n does
have solutions for a, b, c, d and k in Q?
Answer: n = 1, 2, 3, 4, 6.
Related question:
what are the finite subgroups of GL2(Q)/N,
Where N = {
}?
But that is it for rational coefficients...
Proof: Cull, Flahive, Robson
..... so  is a root of K(x),
AND a rational quadratic equation.
Since K(x) is irreducible,
(K) = 1 or 2 (where  is the totient function).
(1) = 1, (2) = 1, (3) = 2, (4) =2,
(5) = 4, (6) =2, (n) > 2 for n > 6.
Regular Lyness cycles order-2:
what periods are possible?
All coefficients rational.
Possible periods: 2, 3, 4, 5, 6, 8, 12
2
3
4
5
6
Can add
constants easily
to these
Pseudo-cycle
Pseudo-cycle
Pseudo-cycle
But that is it for rational coefficients...
Symmetric QRT maps
(Quispel, Roberts and Thompson).
Tsuda has given a theorem that
restricts the periods for periodic
symmetric QRT maps to 2, 3, 4, 5, 6.
Note: x, y, ky - x,...
can have any period if you are
choosing k from R.
k = 1/, period 5,
k = √2, period 8,
k = , period 10.
x, y, |y| - x, ... is period 9.
Part 2: An Introduction
to Elliptic Curves
ax + by + c = 0
Straight line
ax2 + bxy + cy2 + dx + ey + f = 0
Conics
Circle, ellipse, parabola, hyperbola,
pair of straight lines
ax3 + bx2y + cxy2 + dy3
+ ex2 + fxy + gy2 + hx + iy + j = 0
Elliptic curves
UNLESS
The curve has singularities;
a cusp or a loop
or it factorises into straight lines...
y2 = x4 + ax3 + bx2 + cx + d
can be elliptic too...
Any elliptic curve can be transformed
into Weierstrass Normal Form
2
Y
=
3
X
+ aX + b
using a birational map;
that is, you can get from the original curve
to this normal form
and back again using rational maps;
The curves are said to be ISOMORPHIC.
Y2 = X3 + aX + b
a = -2.5
b=1
a = 2.5
b=1
Transforming to Normal Form
For example...
This
becomes...
Where does a straight line
cross our elliptic curve
in normal form?
We are solving simultaneously
y = mx + c, y2 = x3 + ax + b
which gives
x3 - m2x2 + x(a - 2cm) + b - c2 = 0
This is a cubic equation
with at most three real roots.
Note; if it has two real roots,
it must have a third real root.
So if we pick two points on the curve,
the line joining them
MUST cut the curve in a third point.
P+Q+R=0
P+Q=-R
We can form multiples of a point
by taking the tangent at that point.
Sometimes we find that kP = 0.
In this case we say that P is a torsion point.
2
3
y =x +1
6P=0
P is of
order 6
Amazing fact...
The set of points on the curve
together with this addition operation
form a group.
Closed – certainly.
We want P and –P to be inverses.
So P + -P = 0, and we define 0,
the identity here, as the point at infinity.
Associativity?
Geogebra
demonstration
Notice also that if a, b are rational,
then the set of rational points
on the curve form a group.
Closed – certainly.
y = mx + c connects two rational points,
so m and c must be rational.
x3 - m2x2 + x(a - 2cm) + b - c2 = 0
If two roots are rational, the third must be.
Inverses and identity
as before
Mordell Theorem (1922)
Let E be an elliptic curve
defined over Q.
Then E(Q) is a finitely generated
Abelian group.
(Mordell-Weil Theorem [1928]
generalises this.)
Siegel’s Theorem (1929)
If a, b and c are rational,
(and if x3 + ax2 + bx + c = 0
has no repeated solutions),
then there are finitely many
integer points on
y2 = x3 + ax2 + bx + c.
Mordell’s Theorem implies that
E(Q) is isomorphic to
Etorsion(Q)  Zr
The number r is called
the RANK of the elliptic curve.
How big can the rank be?
Nobody knows.
Largest rank so far found; 18 by Elkies (2006)
y2 + xy = x3
− 2617596002705884096311701787701203903556438969515
+ 5106938147613148648974217710037377208977
9103253890567848326775119094885041.
Curves of rank at least 28 exist.
x
Mazur’s Theorem (1977)
The torsion subgroup of E(Q) is
isomorphic to Z/nZ
for some n in
{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12}
or to Z/2nZ  Z/2Z
for some n in {1, 2, 3, 4}.
Part 3: The link between
Lyness cycles and elliptic curves
x = 3, k = -14
Other roots are -7, -2, -1/3.
x = 3, k = -19/3
Other roots are -7, -2, -1/3.
An elliptic curve!
Note:
simplifies to exactly the same set of curves.
Additional note:
This curve has 5-torsion
Choose A = (a, b) to be on the curve,
which gives k.
X = (0,-1) is a torsion point of order 5.
What happens if we repeatedly add X to A?
We expect to get back to A, and we do.
But we get this sequence of points...
The recurrence that built the curve is linked
geometrically to adding a torsion point on it.
Does this work for other periods?
Note:
This doesn’t work for
period 8 and 12
pseudocycles.
Mazur’s Theorem link?
Wouldn’t we expect the possibilities in Mazur’s Theorem
to link with the possible periods for Lyness cycles?
Mazur: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12
Lyness order-2 cycles: 2, 3, 4, 5, 6, 8, 12
Part 4: Hikorski Triples
What does
mean to you?
What do
mean to you?
Adding Speeds Relativistically
Suppose we say the speed of light is 1.
How do we add two parallel speeds?
Try the recurrence relation:
x, y, (xy+1)/(x+y)…
Not much to report...
Try the recurrence relation:
x, y, (x+y)/(xy+1)…
still nothing to report...
But try the recurrence relation:
x, y, -(xy+1)/(x+y)…
Periodic, period 3
Now try the recurrence relation:
x, y, -(x+y)/(xy+1)…
Also periodic, period 3
Are both periodic, period 6
GCSE Resit Worksheet, 2002
How many different equations can you make by putting
the numbers into the circles?
Solve them!
Suppose a, b, c, and d are in the bag.
If ax + b = cx + d,
then the solution to this equation is x =
There are 24 possible equations,
but they occur in pairs, for example:
ax + b = cx + d and cx + d = ax + b
will have the same solution.
So there are a maximum of twelve distinct solutions.
This maximum is possible:
for example, if 7, -2, 3 and 4 are in the bag,
then the solutions are:
If x is a solution, then –x, 1/x and -1/x
will also be solutions.
ax + b = cx + d
a + b(1/x) = c + d(1/x)
c(-x) + b = a(-x) + d
a + d(-1/x) = c + b(-1/x)
The solutions in general will be:
{p, -p, 1/p, -1/p}
{q, -q, 1/q, -1/q}
and {r, -r, 1/r, -1/r}
where p, q and r are all ≥ 1
It is possible for p, q and r
to be positive integers.
For example, 1, 2, 3 and 8
in the bag give (p, q, r) = (7, 5, 3).
In this case,
they form a Hikorski Triple (or HT).
Are (7, 5, 3) linked in any way?
Will this always work?
a, b, c, d in the bag
gives the same as
a + k, b + k, c + k, d + k
in the bag.
Translation Law
a, b, c, d in the bag
gives the same as
ka, kb, kc, kd
in the bag.
Dilation Law
So we can start with 0, 1, a and b
(a, b rational numbers
with 0 < 1 < a < b)
in the bag without loss of
generality.
a, b, c, d in the bag
gives the same as
-a, -b, -c, -d
in the bag.
Reflection Law
Suppose we have 0, 1, a, b
in the bag, with 0 < 1 < a < b
and with b – a < 1
then this gives the same as –b, – a, – 1, 0
which gives the same as 0, b – a, b – 1, b
which gives the same as 0, 1, (b – 1)/(b – a), b/(b – a)
Now b/(b – a) – (b – 1)/(b – a) = 1/(b – a) > 1
If the four numbers in the bag are given as
{0, 1, a, b}
with 1< a < b and b – a > 1,
then we can say the bag is in Standard Form.
So our four-numbers-in-a-bag situation
obeys three laws:
the Translation Law,
the Reflection Law and the Dilation Law.
Given a bag of numbers in
Standard Form,
where might the whole numbers
for our HT come from?
The only possible whole numbers here are:
(b – 1)/a must be the smallest here.
Twelve solutions to bag problem are:
Pythagorean Triples
Parametrisation?
(2rmn, r(m2 - n2), r(m2 + n2))
Hikorski Triples
Parametrisation?
How many HTs are there?
Plenty...
H(n) = # of HTs in which n appears = d(n2-1).
All n > 2 feature in an least 4 HTs.
If n separates a pair of primes, then d(n2-1) = 4.
For how many n does n feature in exactly 4 HTs?
Is abc unique
for each HT?
The Uniqueness
Conjecture
If (a, b, c) and (p, q, r)
are non-trivial HTs
with abc = pqr,
then (a, b, c) = (p, q, r).
Part 5: Cross-ratio-type functions
The Cross-ratio
Takes six values as A, B and C permute:
Form a group isomorphic to S3 under composition
Cross-ratio-type functions
So the cross-ratio
and these cross-ratio-type functions
all obey the three laws:
the Translation Law,
the Reflection Law and the Dilation Law.
Cross-ratio-type functions
and Lyness Cycles
x
y
?
Find a in terms of x and e in terms of y and then substitute...
?
=
(x+y-1)/(x-1)
x
y
Find a in terms of x and e in terms of y and then substitute...
What if we try the same trick here?
( a  b)
(c  d )
x
(b  c)
( d  e)
y
(c  d )
(e  a )
z
( d  e)
( a  b)
?
And here?
( a  b)
(c  b )
x
(b  c)
(d  c)
y
(c  d )
(a  d )
?
So this works with the other cross-ratio type functions too...
Note:
all the Lyness cycles
generated by the
various cross-ratio methods
are regular.
Part 6: Conclusions
Consider the example we had earlier:
is a periodic recurrence relation.
has a torsion point of order 3 when X = 0.
X
If we take a point (p, q) on
map to our cubic, add X and then
map back, we get:
(p, q) maps to (q, -(pq+1)/(p+q))
The recurrence that built the curve is linked
geometrically to adding a torsion point on it.
is an elliptic curve
has integral points (5,3), (3,-2), (-2,5)
and (3,5), (-2,3), (5,-2)
and (30,1), (1,-1), (-1,30)
and (1,30), (30,-1), (-1,1)
If the uniqueness conjecture is true...
Why the name?
www.jonny-griffiths.net
Google ‘Jonny Griffiths’