Diophantine equations - Mathematics and Statistics · Diophantine equations Cubic equations FLT Pell’s equation Elliptic curves Back to FLT Some questions a number theorist might

Diophantine equations Cubic equations FLT Pell’s equation Elliptic curves Back to FLT

Diophantine equations

Henri Darmon

McGill University

CRM-ISM Colloquium, UQAM, January 8, 2010


1 Diophantine equations

2 Cubic equations

3 FLT

4 Pell’s equation

5 Elliptic curves

6 Back to FLT


Definition

A Diophantine equation is a system of polynomial equations withinteger coefficients:

f1(x1, . . . , xm) = · · · = fn(x1, . . . , xm) = 0,

in which one is solely interested in the integer solutions.

Some examples:

1 Cubic equations, like y2 = x3 + 1;

2 The Fermat-Pell equation: x2 − Dy2 = 1;

3 Fermat’s equation: xn + yn = zn;

4 x19y3 − 198z713 + 15xyz3 = 1098w2001.

A large part of number theory is concerned with the study ofDiophantine equations.


Definition


f1(x1, . . . , xm) = · · · = fn(x1, . . . , xm) = 0,


Some examples:




4 x19y3 − 198z713 + 15xyz3 = 1098w2001.



Definition


f1(x1, . . . , xm) = · · · = fn(x1, . . . , xm) = 0,


Some examples:




4 x19y3 − 198z713 + 15xyz3 = 1098w2001.



Definition


f1(x1, . . . , xm) = · · · = fn(x1, . . . , xm) = 0,


Some examples:




4 x19y3 − 198z713 + 15xyz3 = 1098w2001.



Definition


f1(x1, . . . , xm) = · · · = fn(x1, . . . , xm) = 0,


Some examples:




4 x19y3 − 198z713 + 15xyz3 = 1098w2001.



Definition


f1(x1, . . . , xm) = · · · = fn(x1, . . . , xm) = 0,


Some examples:




4 x19y3 − 198z713 + 15xyz3 = 1098w2001.



Definition


f1(x1, . . . , xm) = · · · = fn(x1, . . . , xm) = 0,


Some examples:




4 x19y3 − 198z713 + 15xyz3 = 1098w2001.



Some questions a number theorist might be asked

This is the 21st century. Are there still questions about wholenumbers that we don’t know how to answer?

Isn’t the study of Diophantine equations just a recreationalpursuit?

Claim: Diophantine equations lie beyond the realm of recreationalmathematics, because their study draws on a rich panoply ofmathematical ideas. These ideas, and the new questions they leadto, are just as interesting (perhaps more!) than the equationswhich might have led to their discovery.


















2 Cubic equations

3 FLT

4 Pell’s equation

5 Elliptic curves

6 Back to FLT


First example: the equation y 2 + y = x3

Theorem

The equation y2 + y = x3 has only two solutions, namely(x , y) = (0, 0) and (0,−1).

Proof.

Factor the left-hand side: y(y + 1) = x3.

Unique factorisation in Z:

If gcd(a, b) = 1 and ab = x3, then a = x31 , b = x3

2 .

Hence y and y + 1 are perfect cubes,

{y , y + 1} ⊂ {. . . ,−27,−8,−1, 0, 1, 8, 27, . . .}.

It follows that y = −1 or 0.



Theorem


Proof.




2 .


{y , y + 1} ⊂ {. . . ,−27,−8,−1, 0, 1, 8, 27, . . .}.




Theorem


Proof.




2 .


{y , y + 1} ⊂ {. . . ,−27,−8,−1, 0, 1, 8, 27, . . .}.




Theorem


Proof.




2 .


{y , y + 1} ⊂ {. . . ,−27,−8,−1, 0, 1, 8, 27, . . .}.




Theorem


Proof.




2 .


{y , y + 1} ⊂ {. . . ,−27,−8,−1, 0, 1, 8, 27, . . .}.



Second example: the equation y 2 + 2 = x3

Theorem (Euler)

The equation y2 + 2 = x3 has only two solutions, namely(x , y) = (3,±5).

Proof.

Factor the left hand side in the larger ring Z[√−2]:

(y +√−2)(y −

√−2) = x3.

Observe that y is odd, so gcd(y +√−2, y −

√−2) = 1.

Unique factorisation in Z[√−2] =⇒

y +√−2 = (a + b

√−2)3 = a(a2 − 6b2) + b(3a2 − 2b2)

√−2.

Elementary manipulations =⇒ b = 1, a = ±1, so y = ±5.



Theorem (Euler)


Proof.


(y +√−2)(y −

√−2) = x3.


√−2) = 1.


y +√−2 = (a + b

√−2)3 = a(a2 − 6b2) + b(3a2 − 2b2)

√−2.




Theorem (Euler)


Proof.


(y +√−2)(y −

√−2) = x3.


√−2) = 1.


y +√−2 = (a + b

√−2)3 = a(a2 − 6b2) + b(3a2 − 2b2)

√−2.




Theorem (Euler)


Proof.


(y +√−2)(y −

√−2) = x3.


√−2) = 1.


y +√−2 = (a + b

√−2)3 = a(a2 − 6b2) + b(3a2 − 2b2)

√−2.




Theorem (Euler)


Proof.


(y +√−2)(y −

√−2) = x3.


√−2) = 1.


y +√−2 = (a + b

√−2)3 = a(a2 − 6b2) + b(3a2 − 2b2)

√−2.




Theorem (Euler)


Proof.


(y +√−2)(y −

√−2) = x3.


√−2) = 1.


y +√−2 = (a + b

√−2)3 = a(a2 − 6b2) + b(3a2 − 2b2)

√−2.



The gap in Euler’s proof

Euler’s proof is interesting because it invokes a non-trivialstructural property – unique factorisation – of the the ring Z[

√−2].

Legend has it that Euler did not attempt to justify this claim(although he would have been able to do so, if challenged).

At the time, the prevailing (perhaps unconscious) belief may havebeen that “simple” rings like Z[

√m] possess unique factorisation,

just like the regular integers.




√−2].








√−2].






Third example: the equation y 2 + 118 = x3

Theorem

The equation y2 + 118 = x3 has no integer solutions.

Proof.


(y +√−118)(y −

√−118) = x3.

Proceed exactly as before, using unique factorisation inZ[√−118].

But... 152 + 118 = 73, so the theorem is wrong!



Theorem


Proof.


(y +√−118)(y −

√−118) = x3.





Theorem


Proof.


(y +√−118)(y −

√−118) = x3.





Theorem


Proof.


(y +√−118)(y −

√−118) = x3.





Theorem


Proof.


(y +√−118)(y −

√−118) = x3.




Unique factorisation

Conclusion: Unique factorisation fails in Z[√−118].

The possible failure of unique factorisation is a highly interestingphenomenon, which often poses an obstruction to analysingdiophantine equations. Number theorists have devoted a lot ofefforts to better understanding and controlling this phenomenon,spurring the development of algebraic number theory andcommutative algebra.

A key notion: The class number of a (Dedekind) ring.


















2 Cubic equations

3 FLT

4 Pell’s equation

5 Elliptic curves

6 Back to FLT


Fermat’s Last Theorem

Theorem (Fermat, 1635?)

If n ≥ 3, then the equation xn + yn = zn has no integer solutionwith xyz 6= 0.

Natural opening gambit:

(x + y)(x + ζny) · · · (x + ζn−1n y) = zn,

where ζn = e2πi/n is an nth root of unity.

Theorem (Lame)

Suppose p > 2 is prime. If Z[ζp] has unique factorisation, thenxp + yp = zp has no non-trivial solution.






(x + y)(x + ζny) · · · (x + ζn−1n y) = zn,


Theorem (Lame)







(x + y)(x + ζny) · · · (x + ζn−1n y) = zn,


Theorem (Lame)



Kummer’s theorem

Theorem (Kummer)

Suppose p > 2 is prime. If p does not divide the class number ofZ[ζp], then xp + yp = zp has no non-trivial solution. In particular,Fermat’s Last theorem is true for p < 100.

Kummer’s theorem leads to fascinating questions about cyclotomicrings (rings of the form Z[ζn]). Many of these are still open!

Nonetheless, Fermat’s Last Theorem was eventually proved in1995, by Andrew Wiles, relying on a very different circle of ideas,which I will touch upon later.


Kummer’s theorem

Theorem (Kummer)





Kummer’s theorem

Theorem (Kummer)






2 Cubic equations

3 FLT

4 Pell’s equation

5 Elliptic curves

6 Back to FLT


Pell’s equation

The Fermat-Pell equation is the equation

x2 − dy2 = 1,

where d > 0 is a non-square integer.

The group law.

(x1, y1) ∗ (x2, y2) = (x1x2 + dy1y2, x1y2 + y1x2).

Theorem (Fermat)

For any non-square d > 0, the Pell equation x2 − dy2 has anon-trivial fundamental solution (x0, y0) such that all othersolutions are of the form

(±x ,±y) = (x0, y0)∗n.


Pell’s equation


x2 − dy2 = 1,


The group law.

(x1, y1) ∗ (x2, y2) = (x1x2 + dy1y2, x1y2 + y1x2).

Theorem (Fermat)


(±x ,±y) = (x0, y0)∗n.


Pell’s equation


x2 − dy2 = 1,


The group law.

(x1, y1) ∗ (x2, y2) = (x1x2 + dy1y2, x1y2 + y1x2).

Theorem (Fermat)


(±x ,±y) = (x0, y0)∗n.


Explanation of the group law

Key remark: If (x , y) is a solution to Pell’s equation, thenx + y

√d is a unit (invertible element) of the ring Z[

√d ].

One can rewrite

(x1, y1) ∗ (x2, y2) = (x3, y3)

as(x1 + y1

√d)(x2 + y2

√d) = (x3 + y3

√d).

Solving Pell’s equation can now be recast as:

Problem: Calculate the group of units in the ring Z[√

d ].





√d ].

One can rewrite

(x1, y1) ∗ (x2, y2) = (x3, y3)

as(x1 + y1

√d)(x2 + y2

√d) = (x3 + y3

√d).



d ].





√d ].

One can rewrite

(x1, y1) ∗ (x2, y2) = (x3, y3)

as(x1 + y1

√d)(x2 + y2

√d) = (x3 + y3

√d).



d ].


Some examples of fundamental solutions

If d = 2, then (x0, y0) = (3, 2).

If d = 61, then (x0, y0) = (1766319049, 226153980).

If d = 313, then

(x0, y0) = (32188120829134849, 1819380158564160).

The standard (and still the best) method to find the fundamentalsolution is the method based on continued fractions. It wasdiscovered by the Indian mathematicians of the 12th century, andrediscovered by Fermat in the 17th century.



If d = 2, then (x0, y0) = (3, 2).

If d = 61, then (x0, y0) = (1766319049, 226153980).

If d = 313, then

(x0, y0) = (32188120829134849, 1819380158564160).




If d = 2, then (x0, y0) = (3, 2).

If d = 61, then (x0, y0) = (1766319049, 226153980).

If d = 313, then

(x0, y0) = (32188120829134849, 1819380158564160).




If d = 2, then (x0, y0) = (3, 2).

If d = 61, then (x0, y0) = (1766319049, 226153980).

If d = 313, then

(x0, y0) = (32188120829134849, 1819380158564160).



A cyclotomic approach to Pell’s equation

Theorem (Gauss)

Suppose (for simplicity) that d ≡ 1 (mod 4). Then the ring Z[√

d ]is contained in the cyclotomic ring Z[ζd ], where ζd = e2πi/d .

Proof.

Gauss sums:

g =d−1∑j=0

(j

d

)ζ jd .

Direct calculation:g2 = d .



Theorem (Gauss)



Proof.

Gauss sums:

g =d−1∑j=0

(j

d

)ζ jd .




Theorem (Gauss)



Proof.

Gauss sums:

g =d−1∑j=0

(j

d

)ζ jd .



The cyclotomic approach to Pell’s equation, cont’d

The usefulness of Gauss’s theorem for Pell’s equation arises fromthe fact that Z[ζd ] contains some obvious units: the circular units.

u = ζd + 1 =ζ2d − 1

ζd − 1.

Now letx + y

√d := norm

Z[ζd ]

Z[√

d ](u).

Then (x , y) is a (not necessarily fundamental!) solution to Pell’sequation.




u = ζd + 1 =ζ2d − 1

ζd − 1.

Now letx + y

√d := norm

Z[ζd ]

Z[√

d ](u).





u = ζd + 1 =ζ2d − 1

ζd − 1.

Now letx + y

√d := norm

Z[ζd ]

Z[√

d ](u).




2 Cubic equations

3 FLT

4 Pell’s equation

5 Elliptic curves

6 Back to FLT


Elliptic Curves

An elliptic curve is an equation in two variables x , y of the form

y2 = x3 + ax + b, with a, b ∈ Q.

We are interested in the rational rather than integer solutions tosuch an equation.

Elliptic curve equations exhibit many of the features of Pell’sequation:

1 The set of (rational) solutions to an elliptic curve equation isequipped with a natural group law;

2 The cyclotomic approach to solving Pell’s equation has aninteresting (and quite deep) counterpart for elliptic curves.


Elliptic Curves


y2 = x3 + ax + b, with a, b ∈ Q.






Elliptic Curves


y2 = x3 + ax + b, with a, b ∈ Q.






Elliptic Curves


y2 = x3 + ax + b, with a, b ∈ Q.






The group law for elliptic curves

x

y y = x + a x + b2 3

P

Q

R

P+Q


Ring theoretic formulation of the problem

To the elliptic curve E : y2 = x3 + ax + b, we attach the ring

QE := Q[x , y ]/(y2 − (x3 + ax + b)).

Elementary (but important) remark: Rational solutions of E arein natural bijection with homomorphisms from QE to Q: given asolution (x , y) = (r , s) , let ϕ : QE −→ Q be given by

ϕ(x) = r , ϕ(y) = s.

Problem: Construct homomorphisms from QE to Q (or at least toQ) in a non-trivial way.




QE := Q[x , y ]/(y2 − (x3 + ax + b)).


ϕ(x) = r , ϕ(y) = s.





QE := Q[x , y ]/(y2 − (x3 + ax + b)).


ϕ(x) = r , ϕ(y) = s.





QE := Q[x , y ]/(y2 − (x3 + ax + b)).


ϕ(x) = r , ϕ(y) = s.



Modular functions

Let H be the Poincare upper half plane.

Theorem

There is a unique holomorphic function j : H −→ C satisfying

j

(az + b

cz + d

)= j(z), for all

a b

c d

∈ SL2(Z),

j(z) = q−1 + O(q), where q = e2πiz .

The j-function is the prototypical example of a modular function.It has been said that number theory is largely the study of suchobjects.


Modular functions


Theorem


j

(az + b

cz + d

)= j(z), for all

a b

c d

∈ SL2(Z),




Modular functions


Theorem


j

(az + b

cz + d

)= j(z), for all

a b

c d

∈ SL2(Z),




Modular functions


Theorem


j

(az + b

cz + d

)= j(z), for all

a b

c d

∈ SL2(Z),




Why number theorists like the j-function

1 Moonshine: Its q expansion, or Fourier expansion, hasinteger coefficients:

j(q) = q−1 + 196884q + 21493760q2 + · · ·

The coefficients in this expansion encode information aboutfinite-dimensional representations of certain sporadic simplegroups. (John McKay’s “monstrous moonshine”).

2 Modular polynomials: Let N be an integer. The functionsj(z) and j(Nz) satisfy a polynomial equation ΦN(x , y) in twovariables with integer coefficients. The polynomial ΦN(x , y)is called the N-th modular polynomial.

3 Complex multiplication: If z ∈ H satisfies a quadraticequation with rational coefficients, then j(z) is an algebraicnumber.




j(q) = q−1 + 196884q + 21493760q2 + · · ·







j(q) = q−1 + 196884q + 21493760q2 + · · ·







j(q) = q−1 + 196884q + 21493760q2 + · · ·





Modular rings

Using the modular polynomial ΦN(x , y), we can associate to eachN a ring of modular functions

QN := Q[x , y ]/(ΦN(x , y)) = Q(j(z), j(Nz)).

The ring QN will be called the modular ring of level N. Modularrings play the same role in the study of elliptic curves ascyclotomic rings in the study of Pell’s equation.


Modular rings





Modular rings





Modular rings





Wiles’ Theorem

Theorem (Wiles, Breuil, Conrad, Diamond, Taylor)

Let E : y2 = x3 + ax + b be an elliptic curve (with a, b ∈ Q).Then the ring QE is contained in (the fraction field of) themodular ring QN , for some integer N ≥ 1 (which can be explicitlycalculated from an equation of E ).

Proof.

Wiles, Andrew. Modular elliptic curves and Fermat’s LastTheorem. Annals of Mathematics 141: 443–551.

Taylor R, Wiles A. Ring theoretic properties of certain Heckealgebras. Annals of Mathematics 141: 553–572.


Wiles’ Theorem

Theorem (Wiles, Breuil, Conrad, Diamond, Taylor)

Let E : y2 = x3 + ax + b be an elliptic curve (with a, b ∈ Q).Then the ring QE is contained in (the fraction field of) themodular ring QN , for some integer N ≥ 1 (which can be explicitlycalculated from an equation of E ).

Proof.

Wiles, Andrew. Modular elliptic curves and Fermat’s LastTheorem. Annals of Mathematics 141: 443–551.

Taylor R, Wiles A. Ring theoretic properties of certain Heckealgebras. Annals of Mathematics 141: 553–572.


Using Wiles’ theorem to solve elliptic curve equations

Let τ = a + b√−d ∈ H be any quadratic number.

1 By the theory of complex multiplication, we have ahomomorphism

evτ : QN −→ Q,

sending j(z) to j(τ) and j(Nz) to j(Nτ).

2 By Wiles’ theorem, QE is a subring of the modular ring QN .

3 Restricting evτ to QE gives a homomorphism

ϕτ : QE −→ Q;

this homomorphism corresponds to an algebraic solution of E .





evτ : QN −→ Q,




ϕτ : QE −→ Q;






evτ : QN −→ Q,




ϕτ : QE −→ Q;






evτ : QN −→ Q,




ϕτ : QE −→ Q;




2 Cubic equations

3 FLT

4 Pell’s equation

5 Elliptic curves

6 Back to FLT


Wiles’ theorem and FLT

By a startling twist of fate, the very same theorem of Wiles thatleads to a method for solving elliptic curve equations can also beused to prove Fermat’s Last Theorem!

Unfortunately, there is no time to explain how this comes about.

If you want to learn more about this, you can consult the longintroduction of Wiles’ article, or the exposition in

H. Darmon, F. Diamond, R. Taylor, Fermat’s Last Theorem,Current Developments in Math. Vol. 1, pp. 1–157, InternationalPress, 1996.














Better yet, come to Montreal and study number theory!

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Diophantine equations - Mathematics and Statistics · Diophantine equations Cubic equations FLT Pell’s equation Elliptic curves Back to FLT Some questions a number theorist might