Techniques of Diophantine Equations

8/9/2019 Techniques of Diophantine Equations

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Techniques for Solving Diophantine Equations

Carmen Bruni

November 29th, 2012

Carmen Bruni Techniques for Solving Diophantine Equations

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What is a Diophantine Equation?

A Diophantine equation is a polynomial equation over Z in nvariables in which we look for integer solutions (some people

extend the definition to include any equation where we look forinteger solutions). We ideally wish to classify all integer solutionsto these equations.


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What is a Diophantine Equation?

A Diophantine equation is a polynomial equation over Z in nvariables in which we look for integer solutions (some people

extend the definition to include any equation where we look forinteger solutions). We ideally wish to classify all integer solutionsto these equations.Who cares?


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Why Study Diophantine Equations?


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It’s a fun challenge.


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It gives justification for other studying subjects (for examplealgebraic number theory or algebraic geometry).


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It leads to other interesting questions.


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It leads to other interesting questions. For example

Pell equations, x 2 − dy 2 = 1, lead to questions aboutcontinued fractions and fundamental units.


S ?



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Pell equations, x 2 − dy 2 = 1, lead to questions aboutcontinued fractions and fundamental units.Ljunggren’s equation A4 − 2B 2 = 8 is related toapproximations of π.


Wh S d Di h i E i ?



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Pell equations, x 2 − dy 2 = 1, lead to questions aboutcontinued fractions and fundamental units.Ljunggren’s equation A4 − 2B 2 = 8 is related toapproximations of π.Fermat’s Last Theorem x n + y n = z n lead to questions about

unique factorization domains, cyclotomic fields, elliptic curvesand modular forms.


Phil h f Di h i E i

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Philosophy of Diophantine Equations


Phil h f Di h ti E ti



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Philosophy of Diophantine Equations

It is easier to show that a Diophantine Equations has no solutions

than it is to solve an equation with a solution.


T h i 1 L l b t ti

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Technique 1. Local obstructions


T h i 1 L l bst ti s

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Theorem

The equation x 2

− 3y 2

= 175 has no solutions in integers x and y.


Technique 1 Local obstructions

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Theorem

The equation x 2

− 3y 2


Proof.

Assume we have a solution in x and y .





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Theorem

The equation x 2

− 3y 2


Proof.

Assume we have a solution in x and y . What we will do is looklocally at 4, that is consider the equation over Z4. Then, the

equation becomes x 2 + y 2 = 3 (mod 4).





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Theorem

The equation x 2

− 3y 2


Proof.


equation becomes x 2 + y 2 = 3 (mod 4). Now, notice that

02 ≡ 0 (mod 4) 12 ≡ 1 (mod 4)

22 ≡ 0 (mod 4) 32 ≡ 1 (mod 4)

and hence the only squares in Z4 are 0 and 1.



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Theorem

The equation x 2

− 3y 2


Proof.


equation becomes x 2 + y 2 = 3 (mod 4). Now, notice that

02 ≡ 0 (mod 4) 12 ≡ 1 (mod 4)

22 ≡ 0 (mod 4) 32 ≡ 1 (mod 4)

and hence the only squares in Z4 are 0 and 1. Notice that the sumof any two squares is either 0, 1 or 2. But our equation reduced tothe sum of two squares with value 3, a contradiction.



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Question: How powerful is this technique? Will it always work?



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If the equation has a solution then this technique will neverwork as we will always have local solutions (just use thesolution over Z and project down).



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What if the equation has no solution. Must there be a prime

p such that over Zp our equation has no solution?



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p such that over Zp our equation has no solution?No!



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Definition

An equation is said to satisfy the Hasse Principle (local to global

principle) if whenever it has a solution over R and over every Zp ,then it has one over Z.



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q b





Definition

An equation is said to satisfy the Hasse Principle (local to global

principle) if whenever it has a solution over R and over every Zp ,then it has one over Z.

There exist equations that do not satisfy the Hasse Principle!


Counter Examples to the Hasse Principle

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p p

Theorem (Selmer 1951)

The equation3x 3 + 4y 3 + 5z 3 = 0

has solutions over R and in Zp for every prime p but does not have a solution over Z.



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p p

Theorem (Selmer 1951)

The equation3x 3 + 4y 3 + 5z 3 = 0

has solutions over R and in Zp for every prime p but does not have a solution over Z.

The proof of this is a bit tricky. If we remove the requirement thatwe have a real solution, we can come up with a very nice counter

example.



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p p

Theorem

The equation

w 2 + x 2 + y 2 + z 2 = −1

has solutions in Zp for every prime p but does not have a solutionover Z.



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Theorem

The equation

w 2 + x 2 + y 2 + z 2 = −1


Proof.

No Solutions over Z:



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Theorem

The equation

w 2

+ x 2

+ y 2

+ z 2

= −1


Proof.

No Solutions over Z: Exercise



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Theorem

The equation

w 2

+ x 2

+ y 2

+ z 2

= −1


Proof.


Solutions in Zp : Reducing modulo p gives

w 2 + x 2 + y 2 + z 2 = −1 ≡ p − 1 (mod p )



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Theorem

The equation

w 2

+ x 2

+ y 2

+ z 2

= −1


Proof.


Solutions in Zp : Reducing modulo p gives

w 2 + x 2 + y 2 + z 2 = −1 ≡ p − 1 (mod p )

Since p − 1 ≥ 0, we invoke Lagrange’s theorem on sums of foursquares which says that every non-negative integer can be writtenas a sum of four squares of integers and conclude the result.


Technique 2. Algebraic Number Theory Techniques

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We are looking for solutions over Z to Diophantine Equations.

What, if any of these properties still hold for other rings?



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A early attempt on Fermat’s Last Theorem x n + y n = z n

involved using the ring of integers of the cyclotomic fieldZ[ζ n] where ζ

nn = 1 and attempting to factor the equation.



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The biggest obstacle in this attempt was that Z[ζ n] is notalways a unique factorization domain! (In fact erroneousproofs of FLT appeared in the literature which assumed thisfact).



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The biggest obstacle in this attempt was that Z[ζ n] is notalways a unique factorization domain! (In fact erroneousproofs of FLT appeared in the literature which assumed thisfact).

Perhaps if we look at rings with unique factorization (or evenbetter - if we look at Euclidean domains) we can extend someof factorization techniques.



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Theorem (Fermat’s theorem on sums of two squares)

We have that p = x 2 + y 2 for a prime p if and only if

p ≡ 1 (mod 4).



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p ≡ 1 (mod 4).

Proof.

If p ≡ 3 (mod 4), then our argument from before shows thatp = x 2 + y 2 has no solutions. So we suppose that p ≡ 1 (mod 4).



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p ≡ 1 (mod 4).

Proof.

If p ≡ 3 (mod 4), then our argument from before shows thatp = x 2 + y 2 has no solutions. So we suppose that p ≡ 1 (mod 4).Since Z∗p is cyclic, it has a generator say g with

g p −1 ≡ 1 (mod p ). Thus, g (p −1)/2 ≡ −1 (mod p ) and so thereexists an integer, namely m = g (p −1)/4, such that p | m2 + 1(notice this fails if p ≡ 3 (mod 4)).



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p ≡ 1 (mod 4).

Proof.

If p ≡ 3 (mod 4), then our argument from before shows thatp = x 2 + y 2 has no solutions. So we suppose that p ≡ 1 (mod 4).Since Z∗p is cyclic, it has a generator say g with

g p −1 ≡ 1 (mod p ). Thus, g (p −1)/2 ≡ −1 (mod p ) and so thereexists an integer, namely m = g (p −1)/4, such that p | m2 + 1(notice this fails if p ≡ 3 (mod 4)). Look at this over Z[i ], which

says that p divides m2

+ 1 = (m + i )(m − i ). If p were prime, thenp would divide one of m ± i over Z[i ]. Then p (a + bi ) = m ± i andso pb = ±1 a contradiction since b ∈ Z. Thus, p is reducible inZ[i ]. (Continued on next slide)



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Proof.

Recap: p ≡ 1 (mod 4) implies that p is reducible in Z[i ].



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Proof.


Now, Let N (a + bi ) = a2 + b 2 = |a + bi |2 and suppose that p = wz

with w , z ∈Z

[i ] and w , z = ±1, ±i (these are the units of Z

[i ]).Taking the norms of both sides gives p 2 = N (w )N (z ).



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Proof.


Now, Let N (a + bi ) = a2 + b 2 = |a + bi |2 and suppose that p = wz

with w , z ∈Z

[i ] and w , z = ±1, ±i (these are the units of Z

[i ]).Taking the norms of both sides gives p 2 = N (w )N (z ). This meansthat either one of the elements has norm 1 which means that it isone of ±1, ±i , a contradiction or that each element has norm p .Writing w = x + iy , we see that p = N (w ) = x 2 + y 2 as required.


Technique 3. Linear Forms in Logarithms



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Theorem (Baker 1966)

Let α

1,..,α

n be algebraic integers (roots of monic polynomials)and let H be a number larger than the absolute value of the coefficients of the minimum polynomials of the αi . Let b i ∈ Z

∗

and let B = maxi |b i |. Let

Λ := |

ni =1

b i log αi |

and suppose that Λ = 0.



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Theorem (Baker 1966)

Let α

1,..,α

n be algebraic integers (roots of monic polynomials)and let H be a number larger than the absolute value of the coefficients of the minimum polynomials of the αi . Let b i ∈ Z

∗

and let B = maxi |b i |. Let

Λ := |

ni =1

b i log αi |

and suppose that Λ = 0. Then there exist computable constants C and D so that

Λ > exp (−L(log H )n log(log H )n−1 log B )

where L = (Cnd )Dn and d = [Q(α1,..,αn) : Q].



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Today, under slight modifications to the above statement, Matveev

has shown that we can take L to be

L := 2 · 30n+4(n + 1)6d n+2(d + 1).



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L := 2 · 30n+4(n + 1)6d n+2(d + 1).

A similar statement even holds for Λ = ni =1

αb i

i

− 1.



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L := 2 · 30n+4(n + 1)6d n+2(d + 1).


αb i

i

− 1.What can we prove with this theorem?



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L := 2 · 30n+4(n + 1)6d n+2(d + 1).


αb i

i

− 1.What can we prove with this theorem?

Theorem

Let a, b , k ∈ Z with a, b > 1 and k nonzero. If am − b n = k for

m, n ∈ Z, then m and n are bounded by a computable number depending only on k and the greatest prime divisor of ab.



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The proof to the previous theorem follows from this corollary:



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The proof to the previous theorem follows from this corollary:

Theorem

Let P be the set of all the primes up to M ∈ N. Define

S := p i ∈P

p ai i : ai ∈ N.

Order the elements of S by 0


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Proof.

Let ni =

p i ∈P p ai i and ni +1 =

p i ∈P p

b i i . Look at

ni +1ni

− 1 =p i ∈P

p b i −ai i − 1



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Proof.

Let ni =

p i ∈P p ai i and ni +1 =

p i ∈P p

b i i . Look at

ni +1ni

− 1 =p i ∈P

p b i −ai i − 1

The exponents are bounded by 4 log ni , the primes are bounded byM . Applying our bounds on linear forms in logarithms shows that

ni +1ni

− 1 ≥ (log ni )−C

for some computable C depending only on M .


Technique 4. The Modular Method

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This technique is best showed by an example. Let’s solve FLT,

that is, there are no solutions to the equation x n + y n = z n inpairwise coprime integers x , y , z and n ≥ 3.



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that is, there are no solutions to the equation x n + y n = z n inpairwise coprime integers x , y , z and n ≥ 3. Without loss of generality, suppose n is prime and n ≥ 5.



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that is, there are no solutions to the equation x n + y n = z n inpairwise coprime integers x , y , z and n ≥ 3. Without loss of generality, suppose n is prime and n ≥ 5. The techniques here useelliptic curves, modular forms, and a whole lot of machinery.

DefinitionAn elliptic curve over Q is any smooth curve (no double roots)satisfying

y 2 = x 3 + a2x 2 + a4x + a6

with ai ∈ Q. By clearing denominators and relabeling, we mayassume ai ∈ Z.



Associated to each elliptic curve y 2 = x 3 + a2x 2 + a4x + a6 is a

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p y + 2 + 4 + 6number N called the conductor and another number ∆ called thediscriminant (much like the discriminant of a quadratic or a cubic).

These values tell you when the curve is also defined over Fp .




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p y + 2 + 4 + 6number N called the conductor and another number ∆ called thediscriminant (much like the discriminant of a quadratic or a cubic).

These values tell you when the curve is also defined over Fp . Forexample, y 2 = x 3 + x 2 + 4x + 4 has conductor N = 20 = 22 · 5.This elliptic curve is defined in all Fp where p = 2, 5. Reducing thecurve modulo 5 for example gives

y 2 = x 3 + x 2 + 4x + 4 ≡ x 3 + x 2 − x − 1 ≡ (x + 1)2(x − 1) (mod 5)

which has a double root and hence is not smooth.




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p y 2 4 6number N called the conductor and another number ∆ called thediscriminant (much like the discriminant of a quadratic or a cubic).


y 2 = x 3 + x 2 + 4x + 4 ≡ x 3 + x 2 − x − 1 ≡ (x + 1)2(x − 1) (mod 5)


Wiles (et al.) showed that newforms are a generalization of elliptic curves. The conductor of an elliptic curve corresponds

to something called the level of a modular form.




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ynumber N called the conductor and another number ∆ called thediscriminant (much like the discriminant of a quadratic or a cubic).


y 2 = x 3 + x 2 + 4x + 4 ≡ x 3 + x 2 − x − 1 ≡ (x + 1)2(x − 1) (mod 5)



to something called the level of a modular form.Ribet showed that newforms with [minimal] discriminants thathave very high prime powers in their exponents can beassociated to newforms with lower levels.




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number N called the conductor and another number ∆ called thediscriminant (much like the discriminant of a quadratic or a cubic).


y 2 = x 3 + x 2 + 4x + 4 ≡ x 3 + x 2 − x − 1 ≡ (x + 1)2(x − 1) (mod 5)



to something called the level of a modular form.Ribet showed that newforms with [minimal] discriminants thathave very high prime powers in their exponents can beassociated to newforms with lower levels.

The smallest level for a newform is N = 11.Carmen Bruni Techniques for Solving Diophantine Equations


We ‘prove’ FLT. Suppose that ap + b p = c p for a, b , c ∈ Z

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p S pp + , , ∈pairwise coprime and nonzero and p ≥ 5 prime. WLOG b iseven.




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p pp + , ,pairwise coprime and nonzero and p ≥ 5 prime. WLOG b iseven.

Associate to this solution an elliptic curve

Y 2 = X (X − ap )(X + b p )




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p pp , ,pairwise coprime and nonzero and p ≥ 5 prime. WLOG b iseven.


Y 2 = X (X − ap )(X + b p )

The [minimal] discriminant and conductor are of the form

∆ = 2−8(abc )2p N = 2

q =2,q |abc

q






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pairwise coprime and nonzero and p ≥ 5 prime. WLOG b iseven.


Y 2 = X (X − ap )(X + b p )

The [minimal] discriminant and conductor are of the form

∆ = 2−8(abc )2p N = 2

q =2,q |abc

q

Since all elliptic curves are modular (that is, they arenewforms), and our discriminant has integers to very highexponential powers, we use Ribet’s level lowering theorem tofind a newform at level N = 2. But there are no newforms of level 2! This completes the proof.


Other Techniques

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Skolem’s Method


Other Techniques

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Skolem’s Method

Chabauty’s Method


Other Techniques



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Skolem’s Method

Chabauty’s Method

Hyper Geometric Method


Other Techniques

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Skolem’s Method

Chabauty’s Method


The Brauer-Manin Obstruction


Other Techniques

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Skolem’s Method

Chabauty’s Method


The Brauer-Manin Obstruction

The Descent Obstruction


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Thank you!

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Techniques of Diophantine Equations