Solving Diophantine equations using Elliptic curves

Solving Diophantine equations using Elliptic curves Semester Project Report

by

SOURABHASHIS DAS 1311048

to the

School of Mathematical Sciences National Institute of Science Education and Research Bhubaneswar

Abstract The techniques to solve certain Diophantine equations have been studied. The Fermat’s last theorem has been studied in detail and the roles of elliptic curves and modular forms in proving this theorem have been carefully understood. New quantities associated to an elliptic curve such as conductor, L-function, newforms have been introduced. Global minimal Weierstrass equation associated to an elliptic curve has been introduced and the algorithm to find such a equation is expressed in detail. Several different types of Diophantine equations have been solved using the techniques learned.

Contents Page 1 The Geometry of Elliptic Curves 1.1

1.2

1.3

3

Weierstrass Equations . . . . . . . . . . . . . . . . . . . . . . . . . .

3

1.1.1

Different forms of Weierstrass equations . . . . . . . . . . . .

4

1.1.2

Singular points . . . . . . . . . . . . . . . . . . . . . . . . . .

5

The Group Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

6

1.2.1

Composition law . . . . . . . . . . . . . . . . . . . . . . . . .

6

1.2.2

Group law algorithm . . . . . . . . . . . . . . . . . . . . . . .

8

1.2.3

Singular Weierstrass equations . . . . . . . . . . . . . . . . . .

9

Elliptic Curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

10

1.3.1

11

Isogenies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

2 Elliptic Curves Over Field Of Rationals

14

2.1

Minimal model and minimal discriminant . . . . . . . . . . . . . . . .

14

2.2

Reduction of elliptic curves . . . . . . . . . . . . . . . . . . . . . . . .

16

2.2.1

Reduction modulo a prime . . . . . . . . . . . . . . . . . . . .

16

2.2.2

Good and bad reductions . . . . . . . . . . . . . . . . . . . . .

16

2.3

Conductor of an elliptic curve . . . . . . . . . . . . . . . . . . . . . .

18

2.4

L-function associated to an elliptic curve . . . . . . . . . . . . . . . .

18

2.5

Newform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

21

1

CONTENTS

2.5.1 2.6

Dimension of the space of newforms . . . . . . . . . . . . . . .

23

Computing global minimal Weierstrass equation . . . . . . . . . . . .

24

3 Fermat’s Last Theorem

28

3.1

For case n = 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

28

3.2

For case n = 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

33

3.3

For case n ≥ 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

34

3.3.1

Modularity Theorem . . . . . . . . . . . . . . . . . . . . . . .

34

3.3.2

Absence of isogeny . . . . . . . . . . . . . . . . . . . . . . . .

35

3.3.3

Level lowering . . . . . . . . . . . . . . . . . . . . . . . . . . .

36

3.3.4

Frey curves . . . . . . . . . . . . . . . . . . . . . . . . . . . .

37

3.3.5

Proof for the case n ≥ 5 . . . . . . . . . . . . . . . . . . . . .

38

4 Solving Diophantine Equations

41

2

Chapter 1 The Geometry of Elliptic Curves These following notations are going to be used throughout this report. K - a perfect field, i.e., every algebraic extension of K is separable. ¯ - a fixed algebraic closure of K. K

1.1

Weierstrass Equations

Our primary objects of study are elliptic curves, which are curves of genus one having a specified base point. As we will see in later sections, every such curve can be written as the locus in P2 of a cubic equation with only one point, the base point, on the line at ∞. Then, after X and Y are scaled appropriately, an elliptic curve has an equation of the form Y 2 Z + a1 XY Z + a3 Y Z 2 = X 3 + a2 X 2 Z + a4 XZ 2 + a6 Z 3 . ¯ This is a type of Weierstrass Here O = [0, 1, 0] is the base point and a1 , a2 , ..., a6 ∈ K. equation. We will write the Weierstrass equation for our elliptic curve using non homogeneous coordinates x = X/Z and y = Y /Z, E : y 2 + a1 xy + a3 y = x3 + a2 x2 + a4 x + a6 , 3

1 The Geometry of Elliptic Curves

always remembering that there is an extra point O = [0, 1, 0] out at infinity. As usual, if a1 , a2 , ..., a6 ∈ K, then E is said to be defined over K.

1.1.1

Different forms of Weierstrass equations

¯ 6= 2, then we can simplify the equation by completing the square. Thus If char(K) the substitution 1 y 7→ (y − a1 x − a3 ) 2 gives an equation of the form E : y 2 = 4x3 + b2 x2 + 2b4 x + b6 , where b2 = a21 + 4a2 ,

b4 = 2a4 + a1 a3 ,

b6 = a23 + 4a6 .

We also define quantities b8 = a21 a6 + 4a2 a6 − a1 a3 a4 + a2 a23 − a24 , c4 = b22 − 24b4 , c6 = −b32 + 36b2 b4 − 216b6 , ∆ = −b22 b8 − 8b34 − 27b26 + 9b2 b4 b6 , j = c34 /∆, ω=

dx 2y+a1 x+a3

=

dy . 3x2 +2a2 x+a4 −a1 y

One easily verifies that they satisfy the relations 4b8 = b2 b6 − b24

and 1728∆ = c34 − c26 .

¯ 6= 2, 3, then the substitution If further char(K) x − 3b2 y (x, y) 7→ , 36 108 eliminates the x2 term, yielding the simpler equation E : y 2 = x3 − 27c4 x − 54c6 . 4


Definition 1.1.1. The quantity ∆ is the discriminant of the Weierstrass equation, the quantity j is the j-invariant of the elliptic curve, and ω is the invariant differential associated to the Weierstrass equation.

1.1.2

Singular points

Let P = (x0 , y0 ) be a point satisfying Weierstrass equation f (x, y) = y 2 + a1 xy + a3 y − x3 − a2 x2 − a4 x − a6 = 0, and assume that P is a singular point on the curve f (x, y) = 0. Then we have ∂f ∂f (P ) = (P ) = 0. ∂x ∂y ¯ such that the Taylor series expansion of f (x, y) at It follows that there are α, β ∈ K P has the form f (x, y) − f (x0 , y0 ) = ((y − y0 ) − α(x − x0 ))((y − y0 ) − β(x − x0 )) − (x − x0 )3 .

Definition 1.1.2. With notation as above, the singular point P is a node if α 6= β. In this case, the lines y − y0 = α(x − x0 )

and

y − y0 = β(x − x0 )

are the tangent lines at P . If α = β, then we say that P is a cusp, in which case the tangent line at P is given by y − y0 = α(x − x0 ).

Proposition 1.1.1.

(a) The curve given by a Weierstrass equation satisfies:

(i) It is nonsingular if and only if ∆ 6= 0. 5


(ii) It has a node if and only if ∆ = 0 and c4 6= 0. (iii) It has a cusp if and only if ∆ = c4 = 0. In last two cases, there is only one singular point. ¯ if and only if they both have the same (b) Two elliptic curves are isomorphic over K j-invariant. ¯ There exists an elliptic curve defined over K(j ¯ 0 ) whose j-invariant (c) Let j0 ∈ K. is equal to j0 . Proof. Follows from [1, III.1.4].

1.2

The Group Law

Let E be an elliptic curve given by a Weierstrass equation. Thus E ⊂ P2 consists of the points P = (x, y) satisfying the Weierstrass equation, together with the points O = [0, 1, 0] at infinity. Let L ⊂ P2 be a line. Then, since the equation has degree three, the line L intersects E at exactly three points, say P, Q, R. Of course, if L is tangent to E, then P, Q, R need not be distinct. The fact that L ∩ E, taken with multiplicities, consists of exactly three points is a special case of Bézout’s theorem .

1.2.1

Composition law

We define a composition law ⊕ on E by the following rule: Let P, Q ∈ E, let L be the line through P and Q (if P = Q, let L be the tangent line to E at P ), and let R be the third point of intersection of L with E. Let L0 be the line through R and O. Then L0 intersects E at R, O, and a third point. We denote that third point by P ⊕ Q.

6


Proposition 1.2.1. The composition law has the following properties: (a) If a line L intersects E at the (not necessarily distinct) points P, Q, R, then (P ⊕ Q) ⊕ R = O. (b) P ⊕ O = P for all P ∈ E. (c) P ⊕ Q = Q ⊕ P for all P, Q ∈ E. (d) Let P ∈ E. There is a point of E, denoted by P , satisfying P ⊕ ( P ) = O. (e) Let P, Q, R ∈ E. Then (P ⊕ Q) ⊕ R = P ⊕ (Q ⊕ R). In other words, the composition law makes E into an abelian group with identity element O. (f ) Suppose that E is defined over K. Then E(K) = {(x, y) ∈ K 2 : y 2 + a1 xy + a3 y = x3 + a2 x2 + a4 x + a6 } ∪ {O} is a subgroup of E. Proof. Follows from [1, III.2.2]. Notation. From here on, we drop the special notation ⊕ and and simply write + and − for the group operation on an elliptic curve E. For m ∈ Z and P ∈ E, we let

|m| terms if m0

z }| { [m]P = P + · · · + P ,

z }| { [m]P = −P − · · · − P , 7

[0]P = O.


1.2.2

Group law algorithm

Using the composition law and equation of lines in a plane, we obtain the following group law algorithm for the elliptic curve E given by a Weierstrass equation E : y 2 + a1 xy + a3 y = x3 + a2 x2 + a4 x + a6 . (a) Let P0 = (x0 , y0 ). Then −P0 = (x0 , −y0 − a1 x0 − a3 ). Next let P 1 + P2 = P3

with Pi = (xi , yi ) ∈ E

for i == 1, 2, 3.

(b) If x1 = x2 and y1 + y2 + a1 x2 + a3 = 0, then P1 + P2 = O. Otherwise, define λ and ν by the following formulas: if x1 6= x2 , take λ = and ν =

y1 x2 −y2 x1 . x2 −x1

−x31 +a4 x1 +2a6 −a3 y1 2y1 +a1 x1 +a3

Similarly if x1 = x2 , take λ =

3x21 +2a2 x1 +a4 −a1 y1 2y1 +a1 x1 +a3

y2 −y1 x2 −x1

and ν =

. Then y = λx + ν is the line through P1 and P2 , or tangent to

E if P1 = P2 . (c) With notation as in (b), P3 = P1 + P2 has coordinates x3 = λ2 + a1 λ − a2 − x1 − x2 , y3 = −(λ + a1 )x3 − ν − a3 . (d) As special cases of (c), we have for P1 6= ±P2 , 2 y2 − y1 y2 − y1 x(P1 + P2 ) = + a1 − a2 − x 1 − x 2 , x 2 − x1 x2 − x1 and the duplication formula for P = (x, y) ∈ E, x([2]P ) =

x4 − b4 x2 − 2b6 x − b8 , 4x3 + b2 x2 + 2b4 x + b6

where b2 , b4 , b6 , b8 are the polynomials in the ai ’s given in section 3.1. 8


1.2.3

Singular Weierstrass equations

Definition 1.2.1. Let E be a (possibly singular) curve given by a Weierstrass equation. The nonsingular part of E, denoted by Ens , is the set of nonsingular points of E. Similarly, if E is defined over K, then Ens (K) is the set of nonsingular points of E(K).

Proposition 1.2.2. Let E be a curve given by a Weierstrass equation with ∆ = 0, so E has a singular point S. Then the composition law makes Ens into an abelian group. (a) Suppose that E has a node, so c4 6= 0, and let y = α 1 x + β1

and

y = α2 + β2

be the distinct tangent lines to E at S. Then the map ¯ ∗, Ens 7→ K

(x, y) 7→

y − α 1 x − β1 y − α2 x − β2

is an isomorphism of abelian groups. (b) Suppose that E has a cusp, so c4 = 0, and let y = αx + β be the tangent line to E at S. Then the map ¯ +, Ens 7→ K

(x, y) 7→

is an isomorphism of abelian groups. Proof. Follows from [1, III.2.5]. 9

x − x(S) y − αx − β


1.3

Elliptic Curves

Definition 1.3.1. An elliptic curve is a pair (E, O), where E is a nonsingular curve of genus one and O ∈ E. The elliptic curve E is defined over K, written as E/K, if E is defined over K as a curve and O ∈ E(K).

Proposition 1.3.1. Let E be an elliptic curve defined over K. (a) There exist functions x, y ∈ K(E) such that the map φ : E → P2 ,

φ = [x, y, 1],

gives an isomorphism of E/K onto a curve given by a Weierstrass equation C : Y 2 + a1 XY + a3 Y = X 3 + a2 X 2 + a4 X + a6 where coefficients a1 , ..., a6 ∈ K and satisfying φ(O) = [0, 1, 0]. The functions x and y are called Weierstrass coordinates for the elliptic curve E. (b) Any two Weierstrass equations for E as in (a) are related by a linear change of variables of the form X = u2 X 0 + r,

Y = u3 Y 0 + su2 X 0 + t,

with u ∈ K ∗ and r, s, t ∈ K. (c) Conversely, every smooth cubic curve C given by a Weierstrass equation as in (a) is an elliptic curve defined over K with base point O = [0, 1, 0]. Proof. Follows from [1, III.3.1].

10


1.3.1

Isogenies

Definition 1.3.2. Let E1 and E2 be elliptic curves. An isogeny from E1 to E2 is a morphism φ : E1 → E2

satisfying

φ(O) = O.

Two elliptic curves E1 and E2 are isogenous if there is an isogeny from E1 to E2 with φ(E1 ) 6= {O}. Further, by convention we set deg[0] = 0. This convention ensures that we have φ

ψ

deg(ψ ◦ φ) = deg(ψ)deg(φ) for all chains of isogenies E1 → − E2 − → E3 . Elliptic curves are abelian groups, so the maps between them form groups. We denote the set of isogenies from E1 to E2 by Hom(E1 , E2 ) = {isogenies E1 → E2 }. The sum of two isogenies is defined by (φ + ψ)(P ) = φ(P ) + ψ(P ), and since addition of morphisms is again a morphism, this implies that φ + ψ is a morphism, so it is a isogeny. Hence, Hom(E1 , E2 ) is a group. If E1 = E2 , then we can also compose isogenies. Thus if E is an elliptic curve, we let End(E) = Hom(E, E)

11


be the ring whose addition law is as given above and whose multiplication is composition, (φψ)(P ) = φ(ψ(P )). The ring End(E) is called the endomorphism ring of E. The invertible elements of End(E) form the automorphism group of E, which is denoted by Aut(E).

Example 1.3.1. For each m ∈ Z, we define the multiplication-by-m isogeny [m] : E → E in the natural way. Thus if m > 0, then [m](P ) = P | +P + {z· · · + P}. m terms

For m < 0, we set [m](P ) = [−m](−P ), and we have already defined [0](P ) = O. Since addition of morphisms is a morphism, an easy induction shows that [m] is a morphism, hence an isogeny, since it clearly sends O to O.

Theorem 1.3.1. Let φ : E1 → E2 be an isogeny. Then φ(P + Q) = φ(P ) + φ(Q) for all P, Q ∈ E1 . Proof. Follows from [1, III.4.8].

Corollary 1.3.1. Let φ : E1 → E2 be a nonzero isogeny. Then Ker φ = φ−1 (O) is a finite group. 12


Proof. Follows from [1, III.4.9].

Definition 1.3.3. Let K be a field of characteristic 0 (e.g. Q) and let n ∈ Z>0 . An elliptic curve E/K has a K-rational isogeny of degree n (or n-isogeny) if there exists ¯ of order n with σ(G) = G for all σ ∈ Gal(K/K). ¯ a subgroup G ⊂ E(K) In this case the kernel of the isogeny is the subgroup G itself.

13

Chapter 2 Elliptic Curves Over Field Of Rationals These following notations are going to be used throughout this report. Q - Field of rationals. Z - Ring of integers. ν - A discrete valuation on Q.

2.1

Minimal model and minimal discriminant

Definition 2.1.1. Let E/Q be an elliptic curve. A Weierstrass equation for E is called a minimal (Weierstrass) equation for E at ν if ν(∆) is minimized subject to condition that a1 , ..., a6 ∈ Z. This minimal value of ν(∆) is called the valuation of the minimal discriminant of E at ν. From now on we will consider the discrete valuation ν as defined below: ν(x) = ordp (x), where p is a prime and ordp (x) = r whenever x = pr q and p - q.

14


Proposition 2.1.1. A Weierstrass model for an elliptic curve with a1 , ..., a6 ∈ Z is minimal at a prime p if ordp (∆) < 12

or

ordp (c4 ) < 4.

Proof. Suppose a Weierstrass equation has discriminant ∆ which is not minimal. Then from Proposition 1.3.1, we obtained another Weierstrass equation using change of coordinates

x0 = u2 x + r

and y 0 = u3 y + su2 x + t,

where r, s, t, u ∈ Q and u 6= 0. This new Weierstrass equation has ∆0 = u−12 ∆ and c04 = u−4 c4 . Thus, ordp (∆) can only be changed by multiple of 12 and ordp (c4 ) can be only changed by multiple of 4. Therefore if ordp (∆) < 12 or ordp (c4 ) < 4, then what we obtain is a minimal Weierstrass equation.

Proposition 2.1.2. For an elliptic curve E/Q there exists a Weierstrass model with a1 , ..., a6 ∈ Z which is minimal at p for all primes p. A Weierstrass model as in the proposition is called a global minimal model, or simply a minimal model, for E. This minimal model is not unique in general, but its discriminant is uniquely determined by E. This invariant of E is called the minimal discriminant of E and denoted by ∆min (E).

Example 2.1.1. Let E : y 2 + xy + y = x3 + x2 + 22x − 9 be an elliptic curve over Q. Then ∆ = −215 · 52 and c4 = −5 · 211 for this curve. Then we see that this is a global minimal model. 15


2.2 2.2.1

Reduction of elliptic curves Reduction modulo a prime

Let E be an elliptic curve over Q and p be any prime. Choose any Weierstrass model for E with a1 , ..., a6 ∈ Z which is minimal at p. This model can be reduced modulo p by mapping ai to a¯i := ai (mod p) ∈ Z/pZ =: Fp . This gives a Weierstrass model with a1 , ..., a6 ∈ Fp and defines a curve E˜ over the finite field Fp . Since we started with a minimal equation for E, then E˜ is unique up to the standard change in coordinates for Weierstrass equation over Q.

2.2.2

Good and bad reductions

Definition 2.2.1. Let E/Q be an elliptic curve, and let E˜ be the reduction modulo p of a minimal Weierstrass equation for E. Then (a) E has good (or stable) reduction if E˜ is nonsingular. (b) E has multiplicative reduction(or semi-stable) if E˜ has a node. (c) E has additive (or bad) reduction if E˜ has a cusp. If E has multiplicative reduction, then the reduction is said to be split if the slope of tangent lines are defined over Fp , otherwise it is said to be non-split.

Proposition 2.2.1. Let E/Q be an elliptic curve given by a minimal Weierstrass equation E : y 2 + a1 xy + a3 y = x3 + a2 x2 + a4 x + a6 . Let ∆ be the discriminant of this equation, and let c4 be the usual expression in a1 , ..., a6 . Then 16

2 Elliptic Curves Over Field Of Rationals (a) E has good reduction modulo p if and only if ordp (∆) = 0. In this case, E˜ is an elliptic curve. (b) E has multiplicative reduction modulo p if and only if ordp (∆) > 0 and ordp (c4 ) = 0. In this case, Eñs is the multiplicative group. ¯ ∼ ¯ ∗. Eñs (Q) = Q (c) E has additive reduction modulo p if and only if ordp (∆) > 0 and ordp (c4 ) > 0. In this case, Eñs is the additive group. ¯ ∼ ¯ +. Eñs (Q) = Q Proof. Proof follows from Proposition 1.2.2 and Proposition 2.1.1.

Example 2.2.1. Let p ≥ 5 be a prime. Then E1 : y 2 = x3 + px2 + 1 has good reduction modulo p, while E 2 = y 2 = x3 + x2 + p has (split) multiplication reduction modulo p, and E3 : y 2 = x3 + p has additive reduction modulo p.

17


2.3

Conductor of an elliptic curve

There is an important representation theoretic invariant associated to any E/Q, called the conductor of E, denoted N (E).

N (E) =

Y

pfp (E) ,

p−prime

where  

0 if E has good reduction modulo p, 1 if E has multiplicative reduction modulo p, fp (E) =  2 + δp if E has additive reduction modulo p. where δp ∈ Z≥0 . Furthermore δ2 ≤ 6, δ3 ≤ 3 and δp = 0 ∀p ≥ 5. In particular, if E has good or multiplicative reduction at 2 and 3, then this completely defines N (E).

2.4

L-function associated to an elliptic curve

Theorem 2.4.1. (Hasse-Weil Bound). The number of points on an elliptic curve E over the finite field Fq of order q, denoted as #E(Fq ) satisfies √ | #E(Fq ) − (q + 1) | ≤ 2 q. Let E be a elliptic curve defined over Q with discriminant ∆. From section 1.1.1, we can assume that E has Weierstrass equation of the form E : y 2 = x3 + Ax + B, which has associated quantities ∆ = −16(4A3 + 27B 2 ) and j = −1728

(4A)3 . ∆

If p - ∆, then this defines an Elliptic curve E˜ over Fp where E˜ is the reduction of E modulo p. 18


Let E/Q be an elliptic curve with conductor N . Fix a prime p and define ˜ pe )|, e ≥ 1, t1 (E) = 2 and tpe (E) = pe + 1 − |E(F ˜ pe )| denotes the number of points of elliptic curve E˜ over Fpe . where |E(F

Theorem 2.4.2. Let 1E be the trivial character modulo N , i.e., 1E (p) =

1 if p - N, 0 if p | N.

Then the normalized solution counts satisfy the recurrence tpe (E) = tp (E)tpe−1 (E) − 1E (p)ptpe−2 (E), e ≥ 2.

Definition 2.4.1. The counting zeta function of E at p is ! X tpe (E) Zp (E, X) = exp Xe . e e≥1 Proposition 2.4.1. The counting zeta function of E at p takes the form Zp (E, X) = (1 − tp (E)X + 1E (p)pX 2 )−1 . Proof. (log Zp (E, X))0 =

1 X 1 tpe (E)X e = · S, X e≥1 X

where (we denote tp in place of tp (E)) S

=

tp X +

X

tpe X e

e≥2

=

tp X + tp X

X

tpe−1 X e−1 − 1E (p)pX 2

e≥2

=

X e≥2

2

tp X + tp X · S − 1E (p)pX (S + 2). 19

tpe−2 X e−2


Now we regroup to get S(1 − tp X + 1E (p)pX 2 ) = X(tp − 1E (p)2pX). So, (log Zp (E, X))0 =

tp − 1E (p)2pX 1 − tp X + 1E (p)pX 2

=

(− log(1 − tp X + 1E (p)pX 2 ))0

=

(log(1 − tp X + 1E (p)pX 2 )−1 )0

Therefore, log Zp (E, X) = log(1 − tp X + 1E (p)pX 2 )−1 + C. We can put X = 0 in this equation and obtain C = 0. So, Zp (E, X) = (1 − tp (E)X + 1E (p)pX 2 )−1 .

Definition 2.4.2. The L-function associated to an elliptic curve is defined as L(E, s) =

Y

Zp (E, p−s ) =

Y (1 − tp (E)p−s + 1E (p)p1−2s )−1 .

p

p

Proposition 2.4.2. The L-function of E expands as L(E, s) =

X n≥1

where a1 = 1, ap = tp (E), ape = ap ape−1 − 1E (p)pape−2 , e ≥ 2; 20

an n−s


amn = am an , gcd(m, n) = 1. Conversely, given an integer sequence {an } satisfying the displayed conditions, the corresponding Dirichlet series has an Euler Factorization X

an n−s =

Y (1 − tp (E)p−s + 1E (p)p1−2s )−1 . p

n≥1

Lemma 2.4.1. Let {fn } be a sequence of function C. Then

Q∞

n=1 (fn (z)

+ 1) is

uniformly and absolutely convergent on any compact subset C of C if and only if P∞ n=1 fn (z) is uniformly and absolutely convergent on C.

Proposition 2.4.3. L(E, s) converges absolutely for Re(s) > 3/2 where Re(s) means real part of s. √ Proof. Note that by Hasse-Weil bound, |ap | ≤ 2 p. Now we will use Lemma 2.4.1 Q for p (1 − ap p−s + 1E (p)p1−2s )−1 . We have, |ap p−s + p1−2s |

=

|ap p−s | + |p1−2s |

=

2p−Re(s)+ 2 + p1−2Re(s) .

1

P 1−2Re(s) −Re(s)+ 12 2p and converges for Re(s) > 3/2. p pp P Since of sum of two series dominates p (ap p−s + p1−2s ), so the latter converges for Q Re(s) > 3/2. Using Lemma 2.4.1, we obtain that p (1−ap p−s +1E (p)p1−2s ) converges

We can easily see that

P

in the same range and thus L(E, s) converges for Re(s) > 3/2.

2.5

Newform

Let M and N be positive integers such that M |N and let t be a positive integer such N that t| M . So, if f (τ ) ∈ Sk (Γ0 (M )), then f (tτ ) ∈ Sk (Γ0 (N )). This implies we have

21


well-defined map it = iM,N : Mk (Γ0 (M )) t

→

Mk (Γ0 (N ))

f (z)

7→

f (tz).

These maps restrict to spaces of cusp forms.

Definition 2.5.1. Let N ≥ 1 and k ∈ Z. The space of oldforms in the space Sk (Γ0 (N )) of cusp forms, denoted by Sk (Γ0 (N ))old , is the C-linear subspace of Sk (Γ0 (N )) spanned by the images of all the maps

iM,N : Sk (Γ0 (M )) → Sk (Γ0 (N )) t for all M |N with M 6= N and all t|(N/M ), i.e.,

Sk (Γ0 (N ))old =

X

iM,N (Sk (Γ0 (M ))). t

t|M |N M 6=N

The space of newforms in the space Sk (Γ0 (N )) of cusp forms, denoted by Sk (Γ0 (N ))new , is the orthogonal complement of Sk (Γ0 (N ))old with respect to the Petersson inner product, i.e, Sk (Γ0 (N ))new = {f ∈ Sk (Γ0 (N )) | hf, gi = 0 ∀ g ∈ Sk (Γ0 (N ))old }.

Definition 2.5.2. (Newform). A newform is an element f of the subspace Sk (Γ0 (N ))new that is an eigen vector for every Hecke operator, which is normalized, i.e. satisfies a1 (f ) = 1.

22


2.5.1

Dimension of the space of newforms

Theorem 2.5.1. For any even integer k ≥ 2 and any integer N ≥ 1, the dimension of the space of weight-k newforms on Γ0 (N ) is given by the following formula:

g0# (k, N )

k−1 # 1 # = N s0 (N ) − ν∞ (N ) + c2 (k)ν2# (N ) + c3 (k)ν3# (N ) + δ 12 2

k µ(N ), 2

# # # where the quantity g0# (k, N ) and functions s# 0 , ν∞ , c2 , ν2 , c3 , ν3 , δ, µ are defined

as follows (p means a prime): 1. g0# (k, N ) - the dimension of the space of weight-k newforms on Γ0 (N ), 2. s# 0 is the multiplicative function satisfying: s# 0 (p)

1 1 1 # α 2 = 1− , s# 0 (p ) = 1− − 2 , and s0 (p ) = p p p

1 1− p

1 1 − 2 for α ≥ 3. p

# α # (p ) = 0 for odd α and is the multiplicative function satisfying ν∞ 3. ν∞ α

# 2 # α ν∞ (p ) = p − 2, ν∞ (p ) = p 2 −2 (p − 1)2 for even α ≥ 4.

4. ν2# is the multiplicative function satisfying • ν2# (2) = −1, ν2# (4) = −1, ν2# (8) = 1 and ν2# (2α ) = 0 for α ≥ 4; • if p ≡ 1 (mod 4), then ν2# (p) = 0, ν2# (p2 ) = −1 and ν2# (pα ) = 0 for α ≥ 3; • if p ≡ 3 (mod 4), then ν2# (p) = −2, ν2# (p2 ) = 1 and ν2# (pα ) = 0 for α ≥ 3. 5. ν3# is the multiplicative function satisfying • ν3# (3) = −1, ν3# (9) = −1, ν3# (27) = 1 and ν3# (3α ) = 0 for α ≥ 4; • if p ≡ 1 (mod 4), then ν3# (p) = 0, ν3# (p2 ) = −1 and ν3# (pα ) = 0 for α ≥ 3; • if p ≡ 3 (mod 4), then ν3# (p) = −2, ν3# (p2 ) = 1 and ν3# (pα ) = 0 for α ≥ 3. 23


6. c2 is the function defined by c2 (k) =

1 4

− b k4 c − k4 .

7. c3 is the function defined by c3 (k) =

1 3

− b k3 c − k3 .

8. δ function defined as δ(n) =

1 if n = 1, 0 otherwise .

9. µ is the Mobius function. The above result was proved by Greg Martin [6], University of British Columbia in his paper Dimensions of the spaces of cusp forms and newforms on Γ0 (N ) and Γ1 (N ) in 2003.

Theorem 2.5.2. For the following value of N , S2 (Γ0 (N ))new = {0}. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 13, 16, 18, 22, 25, 28, 60.

2.6

Computing global minimal Weierstrass equation

Let E be an elliptic curve over Q. Using Proposition 2.1.2 we can assume that E admits a global minimal Weierstrass equation over Q. We use the method of Tate’s algorithm as defined by Micheal Laska [4] to find such a equation. Let y 2 + a1 xy + a3 y = x3 + a2 x2 + a4 x + a6 be any Weierstrass equation for E over Q. Then y 02 + a01 x0 y 0 + a03 y 0 = x03 + a02 x02 + a04 x0 + a06 is also a Weierstrass equation for E over Q if and only if x = u2 x0 + r, y = u3 y 0 + u2 sx0 + t, r, s, t, u ∈ Q, u 6= 0. Also, 24

2 Elliptic Curves Over Field Of Rationals ua01 = a1 + 2s, u2 a02 = a2 − sa1 + 3r − s2 , u3 a03 = a3 + ra1 + 2t, u4 a04 = a4 − sa3 + 2ra2 − (t + rs)a1 + 3r2 − 2st, u6 a06 = a6 + ra4 + r2 a2 + r3 − ta3 − rta1 − t2 , u4 c04 = c4 , u6 c06 = c6 , u12 ∆0 = ∆. and a41 ≡ c4 (mod 8), a32 ≡ −a61 − c6 (mod 3), a1 a3 ≡ a21 a2 +

c4 −a41 8

(mod 2),

and moreover if a1 ≡ 0 (mod 2), then c6 ≡ 0 (mod 8) and a23 ≡

c6 8

(mod 4).

Definition 2.6.1. We call a Weierstrass equation for E over Q of restricted type if a1 , a3 ∈ {0, 1} and a2 ∈ {−1, 0, 1}. Algorithm to compute global minimal Weierstrass equation [4] :Let a Weierstrass equation be given by y 2 + a1 xy + a3 y = x3 + a2 x2 + a4 x + a6 with discriminant ∆. 1. Compute the quantities c4 and c6 for the Weierstrass equation: c4 = (a21 + 4a2 )2 − 24(a1 a3 + 2a4 ), c6 = −(a21 + 4a2 )3 + 36(a21 + 4a2 )(a1 a3 + 2a4 ) − 216(a23 + 4a6 ). 2. Determine umax , the largest integer u satisfying the following conditions There exist xu , yu ∈ Z such that u4 xu = c4 , u6 yu = c6 . 25

2 Elliptic Curves Over Field Of Rationals Factor umax = 2e2 3e3 v with v prime to 6. 3. Choose u = 2f2 3f3 v ≤ umax = 2e2 3e3 v such that u = 2f2 tests successfully in step 6 for some a01 , a2 , a03 and u = 3f3 tests successfully in step 6 for some a01 , a2 , a03 . 4. Choose a01 , a02 ∈ {0, 1}, a02 ∈ {−1, 0, 1} subject to the conditions a04 1 ≡ xu

06 a03 2 ≡ −a1 − yu

(mod 8),

(mod 3),

and additionally in the case a01 = 0: yu ≡ 0

(mod 8),

a02 3 ≡

yu 8

(mod 4).

In the case a01 = 1: a03 ≡ a02 +

xu − 1 8

(mod 2).

5. Solve the following equations for a04 and a06 successively: 0 2 0 0 0 xu = (a02 1 + 4a2 ) − 24(a1 a3 + 2a4 ), 0 3 02 0 0 0 0 02 0 yu = −(a02 1 + 4a2 ) + 36(a1 + 4a2 )(a1 a3 + 2a4 ) − 216(a3 + 4a6 ).

If a04 or a06 is not in Z, then continue with (8), otherwise continue with (6). 6. Solve the following equations for s, r, t successively: ua01 = a1 + 2s, u2 a02 = a2 − sa1 + 3r − s2 , u3 a03 = a3 + ra1 + 2t. If s, r, t are not in Z, then continue with (8); otherwise continue as follows:: If the values for s, r, t are not related by the equations: u4 a04 = a4 − sa3 + 2ra2 − (t + rs)a1 + 3r2 − 2st, 26

2 Elliptic Curves Over Field Of Rationals u6 a06 = a6 + ra4 + r2 a2 + r3 − ta3 − t2 − rta1 , then continue with (8), otherwise go to (7). 7. Store this new integral Weierstrass equation Γu,a01 ,a02 ,a03 and its discriminant u−12 ∆, provided the store is empty. Then continue with (9). If store is not empty, then continue as follows: Compare the discriminant in the store with this new discriminant. If the discriminant of equation in the store divides this new discriminant, then continue with (9); otherwise continue as follows: Store the equation Γu,a01 ,a02 ,a03 and its discriminant. Then continue with (9). 8. If all the possible a01 , a02 and a03 subject to the conditions have already been chosen, then continue with (9), otherwise choose new a01 , a02 and a03 and return to (5). 9. Look at the store. It is not empty and contains a global minimal equation of restricted type for E. Since every global minimal equation can be transformed into a restricted type, thus what we obtain in the store finally is the required global minimal equation for E.

27

Chapter 3 Fermat’s Last Theorem Fermat’s Last Theorem (FLT) states that no three positive integers a, b, and c satisfy the equation an + bn = cn for any integer value of n greater than 2. Now suppose n = kl. If an + bn = cn has an integer solution (x, y, z), then al + bl = cl has an integer solution (xk , y k , z k ). Therefore to prove FLT for all n, it is enough to prove FLT for all l ≤ n. Note that any n > 2 is of the following form. 1. Either n = 2m , m > 2, which is same as n = 4 · 2k−2 . Choose l = 4 in this case. 2. or an odd prime. Choose l = n. 3. or a multiple of an odd prime. Choose l to be that odd prime. Thus, it is enough to prove FLT for the case n = 4 and n = odd prime .

3.1

For case n = 3

Lemma 3.1.1. Given that there exist p, q with the following properties: (a) gcd(p, q) = 1, (b) p, q have opposite parities (one is odd, one is even), (c) p2 + 3q 2 is a cube. 28


Then there exists a, b such that: (a) p = a3 − 9ab2 , (b) q = 3a2 b − 3b3 , (c) gcd(a, b) = 1, (d) a, b have opposite parities.

Theorem 3.1.1. There does not exist a positive integer solution for the equation a3 + b 3 = c 3 . Proof. Let us suppose x, y, z is a solution to this equation. Then xyz 6= 0. WLOG we can assume that x, y, z are co-prime. The proof is carried out in following steps. 1. First we observe that there exist p, q such that • gcd(p, q) = 1 • p, q have opposite parities. • p, q are positive integers. • 2p(p2 + 3q 2 ) is a cube. We can easily see that at least one of x, y, z is even. Case 1 :- (z is even). Then x, y are odd. So x+y, x−y are even. Take p = (x+y)/2 and q = (x−y)/2. Then x = p + q and y = p − q. Now, since x and y are co-prime, we get p and q to be co-prime. We also suppose p, q to be positive. Because p = (x + y)/2, q = (x − y)/2 and x 6= y as p, q are co-prime. If x + y is negative, substitute −x, −y and −z since (−x)3 + (−y)3 − (−z)3 . If y > x, flip x and y since they are symmetric. 29


Also. p and q have opposite parities because x = p + q and y = p − q and both x, y are odd. Since z 3 = x3 + y 3 = (x + y)(x2 − xy + y 2 ) = 2p(p2 + 3q 2 ). Thus, 2p(p2 + 3q 2 ) is a cube. Case 2 :- (x is even). The other case for y is even follows similarly. Then z, y are odd and co-prime to x. Hence, z + y and z − y are both even. Let p = (z − y)/2 and q = (z + y)/2. Then z = p + q and y = q − p. Also, p, q have opposite parity since z, y are odd. The arguments for p, q to be co-prime and positive is same as the arguments in case-1. Lastly, x3 = z 3 − y 3 = (z − y)(z 2 + zy + y 2 ) = 2p(p2 + 3q 2 ). So 2p(p2 + 3q 2 ) is a cube. 2. Second, we see that gcd(2p, p2 + 3q 2 ) is either 1 or 3. Suppose we assume that there exist prime f such that f |2p and f |(p2 + 3q 2 ). Clearly f 6= 2 because f |(p2 + 3q 2 ) and p2 + 3q 2 is odd. Now suppose f > 3. Then there exist integers P, Q such that 2p = f P and p2 + 3q 2 = f Q. Since f 6= 2, thus 2|P . Let P = 2H. So, p = f H. Now 3q 2 = f Q − p2 = f Q − f 2 H 2 = f (Q − f H 2 ). Also f - 3 as f > 3 which gives f |q. Therefore, f |p and f |q which is a contradiction since p, q are co-prime. Hence f = 1, 3 are the only possibilities. 3. Third, we show that if gcd(2p, p2 + 3q 2 ) = 1, then there must be a smaller solution to FLT for n = 3.

30

3 Fermat’s Last Theorem Since gcd(2p, p2 + 3q 2 ) = 1, and 2p(p2 + 3q 2 ) is a cube, we get that they both are cube. Applying Lemma 3.1.1, we get there exist a, b as defined in the lemma. Hence 2p = 2a3 − 18ab2 = 2a(a − 3b)(a + 3b). We claim that 2a, a − 3b, a + 3b are all co-primes. (a) First we show that 2a is co-prime to a − 3b and a + 3b. We know both a − 3b and a + 3b are odd because a, b are of opposite parity. If a had common factor with either of them, then that factor divides b which is a contradiction to the fact that a, b are co-prime. (b) If any odd prime > 3 divides both a − 3b and a + 3b, then it divides both a and b as 2a = a − 3b + a + 3b and 6b = a + 3b − (a − 3b) which is again a contradiction. (c)We already know that 2 does not divide a − 3b and a + 3b as both are odd. We need to check that 3 does not divide both. If suppose 3 divide both of them. Then 3|a as 2a = a − 3b + a + 3b and so 3|p as p = a3 − 9ab2 . This is a contradiction because gcd(2p, p2 + 3q 2 ) = 1. Hence 2a, a − 3b, a + 3b are all co-primes and there product equals 2p which is a cube. So, they all are cubes. So we take 2a = A3 , a − 3b = B 3 , a + 3b = C 3 which is another solution to FLT for n = 3 since A3 = 2a = a − 3b + a + 3b = B3 + C 3. This solution is clearly a smaller solution since A3 B 3 C 3 = 2p and by (1), either x3 = 2p(p2 + 3q 2 ) or z 3 = 2p(p2 + 3q 2 ). Hence ABC < x or ABC < z which in both cases gives smaller values. 31

3 Fermat’s Last Theorem 4. Fourth, we show that if gcd(2p, p2 + 3q 2 ) = 3, then there must be a smaller solution to FLT for n = 3. Let gcd(2p, p2 + 3q 2 ) = 3. So, 3|p and 3 - q as p, q are co-prime. So, there exist s such that p = 3s and 2p(p2 + 3q 2 ) = 32 · 2s(3s2 + q 2 ). We can easily check that 32 ·2s, 3s2 +q 2 are relatively prime because any common factor of these two would be a common factor of p, q and p, q are co-prime which gives us the desired result. Since 3 - q, this gives 3 - 3s2 + q 2 . Since p = 3s. s has same parity as p and thus opposite parity as q. So, 2 - 3s2 + q 2 as it is odd. Moreover, gcd(s, q) = 1 as gcd(p, q) = 1. So, 32 · 2s, 3s2 + q 2 are cubes as they are cop-rime and their product is a cube. Using Lemma 3.1.1 we obtain that there exist a, b such that q = a3 − 9ab2 , s = 3a2 b − 3b3 , gcd(a, b) = 1. So, we see that 32 · 2s = 33 · 2b(a − b)(a+) is a cube. Similar work as we did in the last part shows us that 2b, a − b, a + b are co-prime and since their product is a cube, so they are individually cubes. Let us take A = 2b, B = a − b, C = a + b. Clearly A3 = 2b = (a + b) − (a − b) = C 3 − B 3 . Since C 3 = a + b which is less than s = 3b(a − b)(a + b) which is less than p = 3s which is less than either x3 = 2p(p2 + 3q 2 ) or z 3 = 2p(p2 + 3q 2 ), we get a smaller solution to FLT for n = 3. 5. Lastly, every integer solution to the FLT for n = 3 gives rise to another smaller solution of same type. Thus we obtain a infinite chain of positive integer solution 32


starting from a given positive integer solution. This is a contradiction according to Fermat’s method of infinite descent. Thus we have no such positive integer solution for the equation a3 + b3 = c3 where abc 6= 0.

3.2

For case n = 4

Theorem 3.2.1. There exists no positive integer solution to the equation x4 + y 4 = z 2 . Proof. Now WLOG we may assume that x2 , y 2 , z are co-prime. From the solution to Pythagorean triples, we obtain that there exist p, q such that x2 = 2pq y 2 = p2 − q 2 z = p2 + q 2 , p, q are relatively prime and of opposite parity. From this we obtain another Pythagorean triple since y 2 + q 2 = p2 . Therefore there exist a, b such that q = 2ab y = a2 − b 2 p = a2 + b 2 , gcd(a, b) = 1 and a, b are of opposite parity. Combining equations we get, x2 = 2pq = 2(a2 + b2 )(2ab) = 4(ab)(a2 + b2 ) We can easily check that ab and a2 + b2 are relatively prime because any prime dividing these two would also divide both a and b which is not possible as gcd(a, b) = 1. Therefore both ab and a2 + b2 are squares because their product is a square. 33

3 Fermat’s Last Theorem Therefore, there exist P such that P 2 = a2 + b2 . But P 2 = a2 + b2 = p which is less than p2 + q 2 = z which is less than z 2 . So, P is less than z. Hence existence of initial solution leads to a new smaller solution and thus we obtain a infinite sequence of positive integer solutions staring from the initial positive integer solution. This is a contradiction by Fermat’s method of infinite descent. Thus we have no positive integer solution to the equation x4 + y 4 = z 2 where xyz 6= 0.

Corollary 3.2.1. There exists no positive integer solution to the equation x4 + y 4 = z 4 . Proof. Suppose we have a solution to x4 + y 4 = z 4 , then we have a solution to x4 + y 4 = t2 where t = z 2 . But by Theorem 3.2.1, we see that there does not exist any such solution to x4 + y 4 = t2 . Thus our assumption is wrong. So, we have no positive solution to the equation x4 + y 4 = z 4 where xyz 6= 0.

3.3 3.3.1

For case n ≥ 5 Modularity Theorem

Definition 3.3.1. Let E/Q be an elliptic curve with conductor N (E). Then E is said to be modular if there exists a newform f ∈ S2 (Γ0 (N (E))) such that ap (E) = ap (f ) for all primes p where ap (E) is the p-th coefficient of the L-series associated to E and ap (f ) is the p-th coefficient of the q-series expansion of f . We denote E = Ef . A equivalent definition for modularity says that if there exists some M ∈ Z>0 and a newform f ∈ S2 (Γ0 (M )) such that ap (E) = ap (f ) for all but possibly finitely many primes p, then M = N (E) and ap (E) = ap (f ) for all primes p. The conjecture that all elliptic curve curve over Q are in fact modular is known as the Shimura-Taniyama-Weil conjecture. 34


Theorem 3.3.1. (Modularity [16]). Every elliptic curve over Q is modular. For any given positive integer N , the association f 7→ Ef is a bijection between rational newforms of level N and isogeny classes of elliptic curves of conductor N (E). This was proved for semi-stable elliptic curves in 1994 by Wiles [11], [13], with help from Taylor. This sufficed to complete the proof of Fermat’s Last Theorem. Subsequently, the methods of Wiles and Taylor were generalized until finally in 1999 the full modularity theorem above was proved by Breuil, Conrad, Diamond, and Taylor [15], [16].

3.3.2

Absence of isogeny

Theorem 3.3.2. (Mazur [7]). Suppose E/Q is an elliptic curve and that at least one of the following conditions hold: • p ≥ 17 and j(E) ∈ / Z[ 12 ], • or p ≥ 11 and E is a semi-stable elliptic curve, • or p ≥ 5, #E(Q)[2] = 4, and E is a semi-stable elliptic curve. Then E does not have any p-isogenies.

Theorem 3.3.3. (Diamond and Kramer [8]). Suppose that E/Q is an elliptic curve with conductor N . If ord2 (N ) = 3, 5, 7, then E does not have any isogenies of odd degree.

35


3.3.3

Level lowering

Definition 3.3.2. An element a in a ring containing Z (e.g. C or Q) is called an algebraic integer if f (a) = 0 for some monic polynomial f ∈ Z[x]. For an algebraic integer a there is an unique monic f ∈ Z[x] of minimal degree with f (a) = 0, called the minimal polynomial of a.

Theorem 3.3.4. Let f =

P∞

n=1

an (f )q n be a newform. Then an (f ) is an algebraic

integer for all n ∈ Z>0 . Proof. From modular forms, we know that the eigen values of a normalized eigen form are algebraic integers. From the definition of newforms, we see that a newform is a normalized eigen form with eigen values an (f ) for the Hecke operator Tn . Thus an (f ) is an algebraic integer for all n ∈ Z>0 .

Theorem 3.3.5. Let E/Q be an elliptic curve and let l be an odd prime. Define Sl := {p prime : ordp (N (E)) = 1 and ordp (∆min (E)) ≡ 0

(mod l)}

and Nl =

N (E) Q p∈Sl p

! .

If E is does not have a Q-rational l-isogeny, then there exists a newform f ∈ S2 (Γ0 (Nl )), such that for every prime p - N (E) · l, we have l | mp (ap (E)), where mp denotes the minimal polynomial of ap (f ). 36


In case ap (f ) ∈ Z, we note that mp (f ) = x − ap (f ). So, the last theorem reduces to l | ap (E) − ap (f ), which means the same as ap (E) ≡ ap (f )

(mod l).

Notation :- We denote the relation between E and f in the above theorem by E ∼l f . If f is a rational newform, then we know that f corresponds to some elliptic curve F (this is Ef in the notation of Theorem 3.3.1 ). If E arises modulo p from f , then we shall also say that E arises modulo p from F (and write E ∼p F ).

3.3.4

Frey curves

Given a Diophantine equation, suppose that it has a solution, we associate with it an elliptic curve E called a Frey curve, if possible. The key properties of the Frey curve are: • Coefficients of elliptic curve somehow depend on the solution to the Diophantine equation. • Minimal discriminant can be written in the form ∆ = C · Dp where D depends on the solutions and C only depend on the Diophantine equation. • E has multiplicative reduction at the primes dividing D (i.e. if p | D, then p||N ). To our interested Diophantine equation given by al + b l = c l , we associate the following Frey curve E : y 2 = x(x − al )(x + bl ) 37


such that ∆(E) = 16a2l b2l c2l ,

N (E) =

Y

l,

c4 = 24 (a2l + al bl + b2l ).

l|abc

3.3.5

Proof for the case n ≥ 5

Proposition 3.3.1. Consider the Weierstrass model E : y 2 = x 3 + a2 x 2 + a4 x + a6 with a2 , a4 , a6 ∈ Z and ∆(E) 6= 0. Let f (x) denote the polynomial x3 +a2 x2 +a4 x+a6 . Let p be an odd prime. If for some a, b ∈ Z with a 6≡ b (mod p), we have f (x) ≡ (x − a)2 (x − b)

(mod p),

then the model is minimal at p and E has multiplicative reduction at p. Proof. Since y 2 ≡ (x − a)2 (x − b) (mod p), we obtain ∆ ≡ 0 (mod p) and c4 = 24 (a−b)2 6≡ o (mod p) as a 6≡ b (mod p). So, ordp (c4 ) = 0 < 4. By Proposition 2.1.1, we get that the model is minimal. Furthermore, the model has p | ∆ and p - c4 , which ˜ p has a node, i.e., E has multiplicative reduction at by Proposition 1.1.1 implies E/F p.

Theorem 3.3.6. There exists no positive integer solution to the equation an + b n = c n , where n ≥ 5. Proof. It suffices to prove for prime l ≥ 5. Let us assume that there are nonzero a, b, c ∈ Z and a prime l ≥ 5 such that al + bl = cl . The aim is to arrive at a contradiction which will make our assumption false and thus prove FLT for n ≥ 5. 38


Now without loss of generality, we may assume that gcd(a, b, c) = 1,

2 | b,

a ≡ −1

(mod 4).

The Frey curve associated to the above Diophantine equation is the elliptic curve given by E : y 2 = x(x − al )(x + bl ). We compute ∆ = 24 (abc)2l ,

c4 = 24 (a2l + al bl + b2l ).

Using Proposition 3.3.1, we obtain that for odd prime p | abc, the model is minimal at p and ordp (N (E)) = 1. This is because when we reduce the curve modulo p, we obtain the same form as we have in the proposition. If p - abc, then p - ∆, so in this case the model is minimal at p by Proposition 2.1.1 as well and ordp (N (E)) = 0. So in order to compute ∆min (E) and N (E), all we need is to compute ord2 (N (E)) and ord2 (∆min (E)). Now consider the change of coordinates x = 4x0 ,

y = 8y 0 + 4x0 .

This gives E : y 02 + x0 y 0 = x03 +

bl − al − 1 02 al bl 0 x − x 4 16

with ∆0 =

∆ (abc)2l = , 212 28

c04 =

c4 = a2l + al bl + b2l . 24

Since by our assumption a is odd and b is even, this gives that c04 is odd and thus 2 - c04 . Therefore ord2 (c04 ) = 0 < 4. By Proposition 2.1.1, this new model is minimal at 2 and therefore we obtain a global minimal model.

39

3 Fermat’s Last Theorem Finally since l ≥ 5, we obtain 2 | ∆0 . Since 2 | ∆0 and 2 - c04 , by Proposition 1.1.1, we get a multiplicative reduction at 2. Thus, ord2 (N (E)) = 1. Therefore we obtain ∆min (E) =

(abc)2l , 28

N (E) =

Y

p.

p|abc

Thus by Theorem 3.3.2, since l ≥ 5, #E(Q)[2] = 4 and conductor of E is square free, we obtain that E does not have any Q-rational l-isogenies. Now we know Sl := {p prime : ordp (N (E)) = 1 and ordp (∆min (E)) ≡ 0

(mod l)}

and Nl =

N (E) Q p∈Sl p

! .

In this case, we obtain Sl = {p prime : p | abc, p 6= 2} and Nl = 2. Thus by Theorem 3.3.5, we get that there exist a newform f of level 2 satisfying the properties as described in the theorem. But by Theorem 2.5.2, we get that there is no newforms at level 2. Therefore we get a contradiction to our assumption that there are nonzero a, b, c ∈ Z and a prime l ≥ 5 such that al + bl = cl . Thus we proved that there exists no positive integer solution to the equation an + bn = cn where n ≥ 5.

40

Chapter 4 Solving Diophantine Equations In this chapter, we try to solve certain types of Diophantine equations using the method discussed in the last chapter.

Proposition 4.0.1. The Diophantine equation ap + Lr bp + cp = 0,

abc 6= 0,

p ≥ 5 is prime,

has no positive integer solution for all r ∈ Z≥0 where L = 3, 5, 11. Proof. Let us assume that there exists a solution (a, b, c) to the above equation for L = 3, 5, 11 and any r ∈ Z≥0 . Without loss of generality, we may assume that gcd(a, b, c) = 1 and 0 < r < p. Let A, B, C be a permutation of ap , Lr bp , cp such that 2 | B,

A ≡ −1

(mod 4).

The Frey curve associated to this Diophantine equation is given by E : y 2 = x(x − A)(x + B)

41


. Now as in last section of last chapter, we use Proposition 3.3.1, Proposition 2.1.1 and Proposition 1.1.1 to obtain (l means prime) ∆min =

L2r (abc)2p , 28

N (E) =

Y

l.

l|Labc

For this case we obtain Sp = {l prime : l | abc, l 6= 2},

N (E) = 2L. and Np = Q l∈Sp l

Thus by Theorem 3.3.5, we get that there exist a newform f of level 2L satisfying the properties as described in the theorem. But by Theorem 2.5.2, we get that there is no newforms at level 6, 10 and 22, i.e., there exists no newforms for L = 3, 5, 11 which is a contradiction to our assumption. Thus we proved that there exists no positive integer solution to the equation ap + Lr bp + cp = 0,

abc 6= 0,

p ≥ 5 is prime,

where r ∈ Z≥0 where L = 3, 5, 11.

Definition 4.0.1. An elliptic curve E has complex multiplication if End(E) ∼ 6= Z.

¯ be a primitive fourth root Example 4.0.1. Assume that char(K) 6= 2 and let i ∈ K of unity, i.e., i2 = −1. Then the elliptic curve given by the equation E : y 2 = x3 − x has endomorphism ring End(E) strictly larger than Z, since it contains a map, which we denote by [i], given by [i] : (x, y) 7→ (−x, iy).

42


Thus, E has complex multiplication. We also observe that, [i] ◦ [i](x, y) = [i](−x, iy) = (x, −y) = −(x, y). So [i] ◦ [i] = [−1]. There is thus a ring homomorphism Z[i] → End(E),

m + ni 7→ [m] + [n] ◦ [i].

For K = Q, this map is an isomorphism, Z[i] ∼ = End(E).

Theorem 4.0.1. Suppose E and F are elliptic curves over Q and F has complex multiplication in a number field L. Suppose E ∼p F for some prime p. Then 1. (Halberstadt and Kraus [12]) if p = 11 or p ≥ 17 and p splits in L, then the conductor E and F are equal. 2. (Darmon and Merel [9]) if p ≥ 5 and p is inert in L, and E has Q- rational subgroup of order 2 OR 3, then j(E) ∈ Z[ p1 ]. Moreover, if we assume p2 | N and p - N 0 where N and N 0 are respectively the conductor of E an F , then j(E) ∈ Z.

Theorem 4.0.2. Suppose p ≥ 5 is a prime. The equation ap + 2 r b p + c p = 0 has no integer solutions with abc 6= 0 and a, b, c are pairwise co-prime except when r = 1 and (a, b, c) = ±(−1, 1, −1). Proof. This Theorem is due to Wiles [11] for r = 0, Ribet [10] for r ≥ 2, and Darmon and Merel [9] for r = 1. Let a, b, c be a solution to the above equation. Without loss of generality, we can assume that ap ≡ −1

(mod 4),

2r bp ≡ 0 43

(mod 2), 0 ≤ r < p.


The associated Frey curve is the elliptic curve given by E : y 2 = x(x − ap )(x + 2r bp ) with the associated quantities c4 = 16(c2p − 2r ap bp ), ∆ = 22r+4 (abc)2p , j =

(c2p − 2r ap bp )3 . 22r−8 (abc)2p

We use the method of finding minimal discriminant discussed in Section 2.6 to get 2r+4 2 (abc)2p if 16 - 2r bp , ∆min = 22r−8 (abc)2p if 16 | 2r bp . Now for a positive integer R and a prime q, we define Radq (R) =

Y

l.

l|R prime l6=q

We compute the conductor of E to obtain  2 Rad2 (abc) if r    Rad2 (abc) if r N (E) = 8 Rad2 (abc) if r    32 Rad2 (abc) if r

≥ 5 or b is even, = 4 and b is odd, = 2, 3 and b is odd, = 1 and b is odd.

Thus we obtain  2      1 2 Np =   8    32

if if if if if

r ≥ 5 or r = 0, r=4 1 ≤ r ≤ 3 and b is even, r = 2, 3 and b is odd, r = 1 and b is odd.

Now we know that E(Q)[2] = 4. Thus by Theorem 3.3.2, if the conductor is square free then E doesn’t have any p-isogenies. We can also see that if N (E) is not square free, then ord2 (N (E)) = 3 or 5. Now by Theorem 3.3.3, we see that if ord2 (N (E)) = 3 or 5, then E has no odd degree isogenies, hence no p-isogenies. Now by Theorem 3.3.5, we get that there is a newform f of level Np such that E ∼p f . But Theorem 2.5.2 says that there is no newform at level 1, 2, 8. Thus we 44


get r = 1 and b is odd as the only possible values of r and b left for which we can obtain a solution for the given Diophantine equation. Unlike all other cases, since we have newforms at level 32, so we don’t get a direct contradiction and some work is required to prove the theorem. Using Theorem 2.5.1 for k = 2 and N = 32 gives us that the space of newforms of weight 2 and level 32 has dimension 1 and in fact there is exactly one newform at level 32 corresponding to the elliptic curve F : y 2 = x(x + 1)(x + 2) with Cremona reference 32A2 [17]. Notice that we can get the elliptic curve F by letting a = −1, b = 1, c = 1, in the model for E. That is, we are substituting a solution to the Diophantine equation that satisfies the additional constraints placed above. At this point all that we can conclude is that E arises modulo p from F . The curve F is special in that it has complex multiplication. Now F has complex multiplication by Z[i]. By Theorem 4.0.1, if p ≡ 1 (mod 4) (p splits in Q(i)), then the conductor of E is equal to conductor of F which is 32. From the formula for the conductor of E we know that a, b and c are not divisible by any odd primes. But a, b, c are odd, and it follows (a, b, c) = ±(−1, 1, −1). Suppose now that p ≡ 3 (mod 4) (p is inert in Q(i)). Note also that E has a point of order 2. Then j(E) ∈ Z. From the formula for the j-invariant above and the pairwise co-primality of a, b, c above, we see again that a, b, c is not divisible by odd primes and so (a, b, c) = ±(−1, 1, −1). Hence the theorem is proved.

Proposition 4.0.2. Suppose p ≥ 5 is a prime and r ≥ 4. The equation ap + 6 r b p + c p = 0 45


has no integer solutions with abc 6= 0, where a, b, c are pairwise co-prime and a, b, c are not divisible by 2 and 3. Proof. Let a, b, c be a solution to the above equation. Without loss of generality, we can assume that ap ≡ −1

(mod 4),

6r bp ≡ 0

(mod 2), 0 ≤ r < p.

The associated Frey curve is the elliptic curve given by E : y 2 = x(x − ap )(x + 6r bp ). We use the method of finding minimal discriminant discussed in Section 2.6 to get 4 2r 2 6 (abc)2p if 16 - 6r bp , ∆min = 2−8 62r (abc)2p if 16 | 6r bp . We compute the conductor of E to obtain  2 Rad2 (6r abc)    Rad2 (6r abc) N (E) = 8 Rad2 (6r abc)    32 Rad2 (6r abc) Thus we obtain

 6    3 Np = 24    96

if if if if

r r r r

if if if if

r r r r

≥ 5, = 4, = 2, 3, = 1.

≥ 5 or r = 0, =4 = 2, 3, = 1.

Now we know that E(Q)[2] = 4. Thus by Theorem 3.3.2, if the conductor is square free then E doesn’t have any p-isogenies. We can also see that if N (E) is not square free, then ord2 (N (E)) = 3 or 5. Now by Theorem 3.3.3, we see that if ord2 (N (E)) = 3 or 5, then E has no odd degree isogenies, hence no p-isogenies. Now by Theorem 3.3.5, we get that there is a newform f of level Np such that E ∼p f . But Theorem 2.5.2 says that there is no newform at level 3 and 6. Thus we get that the solution to the above equation exist nowhere except maybe when r = 1, 2, 3. 46

References [1] J. H. Silverman. The arithmetic of elliptic curves, volume 106 of Graduate Texts in Mathematics. Springer-Verlag, New York, second edition, 2009. [2] J. H. Silverman. The advance topics in the arithmetic of elliptic curves, volume 106 of Graduate Texts in Mathematics. Springer-Verlag, New York, second edition, 1994. [3] N. Koblitz, Introduction to elliptic curves and modular forms. Springer-Verlag New York, Inc, 1993. [4] M. Laska, An Algorithm for finding a minimal Weierstrass equation for an elliptic curve, Mathematics of Computation, Vol. 38, No. 157, pp. 257-260. American Mathematical Society, http://www.jstor.org/stable/2007483, Jan 1982. [5] S. Siksek, The modular approach to Diophantine equations, Mathematics Institute, University of Warwick, Coventry, July 2016. [6] G. Martin, Dimensions of the spaces of cusp forms and newforms on Γ0 (N ) and Γ1 (N ), Journal of Number Theory 112 (2005) 298–331. Dec 2004. [7] B. Mazur, Rational isogenies of prime degree, Invent. Math. 44, 129–162, 1978. [8] F. Diamond and K. Kramer, Modularity of a family of elliptic curves, Math. Res. Lett. 2, No. 3, 299–304, 1995. 47

REFERENCES

[9] H. Darmon and L. Merel, Winding quotients and some variants of Fermat’s Last Theorem, J. reine angew. Math. 490, 81–100 1997. [10] K. Ribet, On the equation ap + 2bp + cp = 0, Acta Arith. LXXIX.1, 7–15, 1997. [11] A. Wiles, Modular elliptic curves and Fermat’s Last Theorem, Annals of Math. 141, 443–551, 1995. [12] E. Halberstadt and A. Kraus, Sur les modules de torsion des courbes elliptiques, Math. Ann. 310, 47–54, 1998. [13] R. Taylor and A. Wiles, Ring theoretic properties of certain Hecke algebras, Annals of Math. 141, 553–572, 1995. [14] F. Diamond, On deformation rings and Hecke rings, Ann. of Math. 144, no. 1, 137–166, 1996. [15] B. Conrad, F. Diamond, R. Taylor, Modularity of certain potentially BarsottiTate Galois representations, J. Amer. Math. Soc. 12, no. 2, 521–567, 1999. [16] C. Breuil, B. Conrad, F. Diamond and R. Taylor, On the modularity of elliptic curves over Q: wild 3-adic exercises, J. Amer. Math. Soc. 14 No.4, 843–939, 2001. [17] J. Cremona, Algorithms for modular elliptic curves, 2nd edition, Cambridge University Press, 1996.

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