Elliptic Curves
History and Motivation
Background
New Results
Arithmetic Progressions on Mordell Curves
Alejandra Alvarado
Department of Mathematics
Purdue University
18 February 2013 / Eastern Illinois University
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Elliptic Curves
History and Motivation
Background
New Results
Outline
1
Elliptic Curves
2
History and Motivation
3
Background
4
New Results
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Elliptic Curves
History and Motivation
Background
New Results
Louis J. Mordell
1888 – 1972
http://www-history.mcs.st-andrews.ac.uk
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History and Motivation
Background
New Results
Abstract
The study of solutions to diophantine equations has been
around throughout history.
The problem of finding consecutive integer solutions to a
diophantine equation can be traced back to Mohanty.
An r term arithmetic progression (AP) is a collection of
rational numbers {`1 , `2 , ..., `r } such that there is a
common difference d = `i+1 − `i .
When we talk about an r -term AP on the Mordell curve
y 2 = x 3 + k , we mean an AP in the x-coordinates of the
points on the curve {P1 , P2 , ..., Pr }.
The goal of this talk is to give a survey on what is known
about arithmetic progressions on these curves.
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Elliptic Curves
History and Motivation
Background
New Results
What are Elliptic Curves?
Addition of Points on an Elliptic Curve
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Elliptic Curves
History and Motivation
Background
New Results
An elliptic curve E over a field K is given by the set of
solutions to a nonsingular equation of the form
y 2 + a1 xy + a3 y = x 3 + a2 x 2 + a4 x + a6 ,
whose coefficients are in K , plus the point at infinity, denoted by
O.
In our work, we are concerned with elliptic curves over K = Q
and a1 = a2 = a3 = a4 = 0 and a6 6= 0.
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Elliptic Curves
History and Motivation
Background
New Results
To see what we mean by the point at infinity, consider the
homogeneous curve
y 2 z + a1 xyz + a3 yz 2 = x 3 + a2 x 2 z + a4 xz 2 + a6 z 3
whose coefficients are in K . We will need the following
definition.
Let x, y , z ∈ K where K is a field. The projective plane over K ,
denoted P2 (K ), is the set of all equivalence classes of triples
(x : y : z) such that x, y , z are not all zero.
Two points (x : y : z) and (x0 : y0 : z0 ) are said to be equivalent
if there exists a nonzero λ ∈ K such that
x = λx0 , y = λy0 , z = λz0 .
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History and Motivation
Background
New Results
If (x : y : z) is a point on homogeneous curve in P2 such that z
is nonzero, then (x : y : z) = ( xz : yz : 1).
In the homogenous equation, z = 0 implies x = 0 and y is any
nonzero element.
So we say the point at infinity occurs when z is zero.
Scaling, we see that (0 : 1 : 0) is the only point at infinity on E.
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History and Motivation
Background
New Results
If char (K ) 6= 2, 3 then by a change of variables, we end up with
an equation of the form
Y 2 = X 3 + AX + B.
An important feature of elliptic curves is that the set E(K ) forms
an abelian group in a natural way, with O as the identity.
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History and Motivation
Background
New Results
Theorem (Mordell-Weil)
If E is an elliptic curve over a number field K , the group E(K ) is
a finitely generated abelian group.
E(K ) ' Etors (K ) × Zr
with Etors (K ) finite group and r denotes the rank.
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History and Motivation
Background
New Results
mathworld.wolfram.com/MordellCurve.html
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Elliptic Curves
History and Motivation
Background
New Results
History and Motivation
An example of a non-constant 3-term arithmetic progression
whose terms are perfect squares:
{`1 , `2 , `3 } = {1, 25, 49}
where the common difference is d = 24.
There are infinitely many 3-term arithmetic progressions:
{(t 2 − 2t − 1)2 , (t 2 + 1)2 , (t 2 + 2t − 1)2 }
where the common difference is d = 4(t 3 − t).
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Elliptic Curves
History and Motivation
Background
New Results
Fermat showed there are no r -term arithmetic progressions
whose terms are perfect squares if r ≥ 4.
Each 4-term arithmetic progression of square corresponds to a
rational point on the elliptic curve
E:
y 2 = x 3 + 5x 2 + 4x
where E(Q) ' Z2 × Z4 . But these points lead to the trivial case
{`1 , `2 , `3 , `4 } = {1, 1, 1, 1}.
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History and Motivation
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New Results
Over a quadratic field, statement not true.
Rational points on the quadratic twist y 2 = x 3 + 30x 2√+ 144x
correspond to 4-term arithmetic progression over Q( 6),
n
√
√
√
√ o
(9 − 5 6)2 , (15 − 6)2 , (15 + 6)2 , (9 + 5 6)2
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History and Motivation
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New Results
Legendre showed that there does not exist an r-term arithmetic
progression of cubes, r ≥ 3.
3-term arithmetic progression of cubes corresponds to rational
points on the curve
E:
y 2 = x 3 − 27
E(Q) ' Z2 and these points correspond to trivial progression of
cubes, {1, 1, 1} and {−1, 0, 1}.
n
√
√ o
√
Over Q( 2), (4 − 21 2)3 , 223 , (4 − 21 2)3 .
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History and Motivation
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New Results
In general, for r ≥ 3 and n ≥ 3, Denes, Darmon and Merel,
showed there are no r -term arithmetic progression whose
terms are nth powers.
A 3-term progression {an , c n , bn } = {`1 , `2 , `3 } corresponds to
a rational point on an + bn = 2c n .
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History and Motivation
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New Results
Perfect nth powers may be realized as the image of the map
` : C → P1 which sends a rational point P = (x : y : 1) on the
curve C : y = x n to the rational number `(P) = y .
Hence an r -term arithmetic progression of perfect nth powers
`1 , `2 , . . . , `r ⊆ P1 (Q)
is
the image under
the linear polynomial ` of a subset
P1 , P2 , . . . , Pr ⊆ C(Q).
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History and Motivation
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New Results
In general, let C be a curve of degree n = deg(C),


X


Cijk x i y j z k = 0 ,
C = (x : y : z) ∈ P2 

i+j+k =n
and let ` : C → P1 be a linear polynomial
` : (x : y : z) 7→ a x + b y + c z.
The set P1 , P2 , . . . , Pr is an r -term
arithmeticprogression
on C with respect to ` if its image `1 , `2 , . . . , `r under ` is an
r -term arithmetic progression of rational numbers.
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Motivating Question
Given a projective curve C and a linear polynomial ` : C → P1 ,
what
is the longest
non-constant r -term arithmetic progression
P1 , P2 , . . . , Pr on C with respect to `?
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Background
Consider the case where deg(C) = 2. Allison found an infinite
family of conic sections
C:
y 2 = c2 x 2 + c1 x + c0
that contain an 8-term AP with respect to x. These curves have
an axis of symmetry halfway between two integers.
Points in AP correspond to points on
E:
Y 2 = (X − 3)(X − 6)(X − 9)
where E(Q) ' Z2 × Z2 × Z.
For example, y 2 = −420x 2 + 2100x + 2809 contains the
x-coordinates in AP:
{(−1, 17), (0, 53), (1, 67), (2, 73), (3, 73), (4, 67), (5, 53), (6, 17)}.
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More recently, we considered the non-symmetric case. We
showed there exists an infinite family of conic sections that
contain 5-term x-AP.
A search found 5 examples which contain AP of length 7 (two
previously found by Allison).
For example, y 2 = −4980x 2 + 32100x + 2809 contains
{(0, 53), (1, 173), (2, 217), (3, 233), (4, 227), (5, 197), (6, 127)}.
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New Results
Consider the case where deg(C) = 3. Bremner, then Campbell,
found distinct families of elliptic curves with an 8-term AP with
respect to x.
Bremner showed that points on the rank 1 elliptic curve
y 2 = x 3 − x 2 − 36x + 36 give rise to an 8-term AP on
y 2 = x 3 + Ax + B where A and B are in two parameters.
For example, Bremner found the rank 3 elliptic curve
y 2 = x 3 − 112x + 400
{(−12, 4), (−8, 28), (−4, 28), (0, 20), (4, 4), (8, 4), (12, 28), (16, 52)}
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New Results
Mohanty’s Conjectures, 1980
Conjecture (Mohanty)
There does not exist length five or greater x-AP on y 2 = x 3 + k .
Conjecture (Mohanty)
There does not exist length five or greater y -AP on y 2 = x 3 + k .
In the y -coordinates, we found a counter example, the rank two
elliptic curve, y 2 = x 3 + 274625 with the following length six
y -AP,
{(70, ±735), (−14, ±441), (−56, ±147)}
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History and Motivation
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New Results
Mohanty also conjectured that there are finitely many k ’s,
possibly only k = 1025, such that there exists consecutive
length four y -AP on y 2 = x 3 + k .
This curve has rank 3 and the points in arithmetic progression
are:
{(−5, 30), (−4, 31), (−1, 32), (4, 33)}
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New Results
Goal
For each integer r ≥ 3, classify those cubic curves
C : y 2 = x 3 + k which contain an r −term AP {P1 , P2 , ..., Pr }
with respect to ` : (x : y : 1) 7→ x.
For each integer r ≥ 3, define AP(r ) as the collection of
sequences,
{(P1 , k1 ), (P2 , k2 ), · · · , (Pr , kr )} ⊆ P2 × Gm
such that Pi = (xi : yi : 1), ki−1 = ki = ki+1 and
xi+1 + xi−1 = 2xi .
There is an action ◦ : Gm × AP(r ) → AP(r ) defined by
λ ◦ ((x : y : 1), k ) = (λ2 x : λ3 y : 1), λ6 k
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Theorem (Alvarado, Goins)
Let S (5) ⊆ P2 × P2 × Gm consist of ((X : Y : 1), (Z : W : 1), t)
where
Y 2 = X (X − e1 (t))(X − e2 (t))
W 2 = Z (Z − e3 (t))(Z − e4 (t)) as well as
p(X , Y , t) = q(Z , W , t) and J(X , Y , t) 6= 0, 1, ∞.
Then S (5) ' AP(r )/Gm is a surface such that if
((X : Y : 1), (Z : W : 1), t) ∈ S (5) , then the set {Pi |1 ≤ i ≤ 5} is
a 5-term AP on C : y 2 = x 3 + λ6 J−1
with respect to
J3
` : (x : y : 1) 7→ x; and
There are infinitely many cubic curves C : y 2 = x 3 + k which
contain 3- and 4-term AP’s with respect to ` : (x : y : 1) 7→ x.
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History and Motivation
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New Results
S
(5)

S

(4) S (3)
P1

σ5
σ4
σ3
/ AP(5)
/ AP(4)
/ AP(3)
(x : y : 1), (X : Y : 1), t
_
(x : y : 1), t
(z : 1), t
_
/
(P, k ), (Q, k ),
(R, k ), (S, k ), (T , k )
_
_
/
(P, k ), (Q, k ),
(R, k ), (S, k )
_
/ (P, k ), (Q, k ), (R, k )
_
π
P1
t
k
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Example
Still searching for a 5-term AP in the x-coordinates. Found a
point on the intersection w = W but corresponds to J = 0.
There is an infinite family of curves with “almost” length five
x-AP.
The rank 3 curve y 2 = x 3 + 1025 has the following points in
“almost” length five x-AP:
√
{(10, 45), (20, 95), (30, 5 1121), (40, 255), (50, 355)}
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Cijk x1i x2j x0k = 0
X
` : C → P1
Longest Length r for a Family
Longest Length r for an Example
y = x2
(x : y : 1) 7→ y
3 (Fermat, Euler)
3 (Fermat, Euler)
y = x3
(x : y : 1) 7→ y
2 (Legendre)
2 (Legendre)
y = xn
(x : y : 1) 7→ y
2 (Dénes, Darmon-Merel)
2 (Dénes, Darmon-Merel)
y 2 = f (x) with deg(f ) = 2
(x : y : 1) 7→ x
8 (Allison)
8 (Allison)
y 2 = f (x) with deg(f ) = 3
(x : y : 1) 7→ x
8 (Bremner, Campbell)
8 (Bremner, Campbell)
y 2 = f (x) with deg(f ) = 4
(x : y : 1) 7→ x
12 (Ulas)
14 (MacLeod, Alvarado)
y 2 = f (x) with deg(f ) = 5
(x : y : 1) 7→ x
12 (Alvarado)
12 (Alvarado)
y 2 = f (x) with deg(f ) = 6
(x : y : 1) 7→ x
16 (Ulas)
18 (Ulas)
y2 = x3 + k
(x : y : 1) 7→ x
4 (Alvarado)
4 (Alvarado)
y2 = x3 + k
(x : y : 1) 7→ y
6 (Alvarado)
6 (Alvarado)
C:
i+j+k =n
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New Results
THANK YOU!
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