Techniques for Solving Diophantine Equations

Carmen Bruni

November 29th, 2012

Carmen Bruni Techniques for Solving Diophantine Equations

What is a Diophantine Equation?

A Diophantine equation is a polynomial equation over Z in nvariables in which we look for integer solutions (some peopleextend the definition to include any equation where we look forinteger solutions). We ideally wish to classify all integer solutionsto these equations.

Who cares?

Carmen Bruni Techniques for Solving Diophantine Equations

What is a Diophantine Equation?

A Diophantine equation is a polynomial equation over Z in nvariables in which we look for integer solutions (some peopleextend the definition to include any equation where we look forinteger solutions). We ideally wish to classify all integer solutionsto these equations.Who cares?

Carmen Bruni Techniques for Solving Diophantine Equations

Why Study Diophantine Equations?

It’s a fun challenge.

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

It leads to other interesting questions. For example

Pell equations, x2 − dy2 = 1, lead to questions aboutcontinued fractions and fundamental units.Ljunggren’s equation A4 − 2B2 = 8 is related toapproximations of π.Fermat’s Last Theorem xn + yn = zn lead to questions aboutunique factorization domains, cyclotomic fields, elliptic curvesand modular forms.

Carmen Bruni Techniques for Solving Diophantine Equations

Why Study Diophantine Equations?

It’s a fun challenge.

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

It leads to other interesting questions. For example

Pell equations, x2 − dy2 = 1, lead to questions aboutcontinued fractions and fundamental units.Ljunggren’s equation A4 − 2B2 = 8 is related toapproximations of π.Fermat’s Last Theorem xn + yn = zn lead to questions aboutunique factorization domains, cyclotomic fields, elliptic curvesand modular forms.

Carmen Bruni Techniques for Solving Diophantine Equations

Why Study Diophantine Equations?

It’s a fun challenge.

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

It leads to other interesting questions. For example

Pell equations, x2 − dy2 = 1, lead to questions aboutcontinued fractions and fundamental units.Ljunggren’s equation A4 − 2B2 = 8 is related toapproximations of π.Fermat’s Last Theorem xn + yn = zn lead to questions aboutunique factorization domains, cyclotomic fields, elliptic curvesand modular forms.

Carmen Bruni Techniques for Solving Diophantine Equations

Why Study Diophantine Equations?

It’s a fun challenge.

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

It leads to other interesting questions.

For example

Pell equations, x2 − dy2 = 1, lead to questions aboutcontinued fractions and fundamental units.Ljunggren’s equation A4 − 2B2 = 8 is related toapproximations of π.Fermat’s Last Theorem xn + yn = zn lead to questions aboutunique factorization domains, cyclotomic fields, elliptic curvesand modular forms.

Carmen Bruni Techniques for Solving Diophantine Equations

Why Study Diophantine Equations?

It’s a fun challenge.

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

It leads to other interesting questions. For example

Pell equations, x2 − dy2 = 1, lead to questions aboutcontinued fractions and fundamental units.

Ljunggren’s equation A4 − 2B2 = 8 is related toapproximations of π.Fermat’s Last Theorem xn + yn = zn lead to questions aboutunique factorization domains, cyclotomic fields, elliptic curvesand modular forms.

Carmen Bruni Techniques for Solving Diophantine Equations

Why Study Diophantine Equations?

It’s a fun challenge.

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

It leads to other interesting questions. For example

Pell equations, x2 − dy2 = 1, lead to questions aboutcontinued fractions and fundamental units.Ljunggren’s equation A4 − 2B2 = 8 is related toapproximations of π.

Fermat’s Last Theorem xn + yn = zn lead to questions aboutunique factorization domains, cyclotomic fields, elliptic curvesand modular forms.

Carmen Bruni Techniques for Solving Diophantine Equations

Why Study Diophantine Equations?

It’s a fun challenge.

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

It leads to other interesting questions. For example

Pell equations, x2 − dy2 = 1, lead to questions aboutcontinued fractions and fundamental units.Ljunggren’s equation A4 − 2B2 = 8 is related toapproximations of π.Fermat’s Last Theorem xn + yn = zn lead to questions aboutunique factorization domains, cyclotomic fields, elliptic curvesand modular forms.

Carmen Bruni Techniques for Solving Diophantine Equations

Philosophy of Diophantine Equations

It is easier to show that a Diophantine Equations has no solutionsthan it is to solve an equation with a solution.

Carmen Bruni Techniques for Solving Diophantine Equations

Philosophy of Diophantine Equations

It is easier to show that a Diophantine Equations has no solutionsthan it is to solve an equation with a solution.

Carmen Bruni Techniques for Solving Diophantine Equations

Technique 1. Local obstructions

Theorem

The equation x2 − 3y2 = 175 has no solutions in integers x and y.

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, theequation becomes x2 + y2 = 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.

Carmen Bruni Techniques for Solving Diophantine Equations

Technique 1. Local obstructions

Theorem

The equation x2 − 3y2 = 175 has no solutions in integers x and y.

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, theequation becomes x2 + y2 = 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.

Carmen Bruni Techniques for Solving Diophantine Equations

Technique 1. Local obstructions

Theorem

The equation x2 − 3y2 = 175 has no solutions in integers x and y.

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, theequation becomes x2 + y2 = 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.

Carmen Bruni Techniques for Solving Diophantine Equations

Technique 1. Local obstructions

Theorem

The equation x2 − 3y2 = 175 has no solutions in integers x and y.

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, theequation becomes x2 + y2 = 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.

Carmen Bruni Techniques for Solving Diophantine Equations

Technique 1. Local obstructions

Theorem

The equation x2 − 3y2 = 175 has no solutions in integers x and y.

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, theequation becomes x2 + y2 = 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.

Carmen Bruni Techniques for Solving Diophantine Equations

Technique 1. Local obstructions

Theorem

The equation x2 − 3y2 = 175 has no solutions in integers x and y.

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, theequation becomes x2 + y2 = 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.

Carmen Bruni Techniques for Solving Diophantine Equations

Technique 1. Local obstructions

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

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).

What if the equation has no solution. Must there be a primep such that over Zp our equation has no solution?

No!

Definition

An equation is said to satisfy the Hasse Principle (local to globalprinciple) 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!

Carmen Bruni Techniques for Solving Diophantine Equations

Technique 1. Local obstructions

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

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).

What if the equation has no solution. Must there be a primep such that over Zp our equation has no solution?

No!

Definition

An equation is said to satisfy the Hasse Principle (local to globalprinciple) 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!

Carmen Bruni Techniques for Solving Diophantine Equations

Technique 1. Local obstructions

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

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).

What if the equation has no solution. Must there be a primep such that over Zp our equation has no solution?

No!

Definition

An equation is said to satisfy the Hasse Principle (local to globalprinciple) 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!

Carmen Bruni Techniques for Solving Diophantine Equations

Technique 1. Local obstructions

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

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).

What if the equation has no solution. Must there be a primep such that over Zp our equation has no solution?

No!

Definition

An equation is said to satisfy the Hasse Principle (local to globalprinciple) 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!

Carmen Bruni Techniques for Solving Diophantine Equations

Technique 1. Local obstructions

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

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).

What if the equation has no solution. Must there be a primep such that over Zp our equation has no solution?

No!

Definition

An equation is said to satisfy the Hasse Principle (local to globalprinciple) 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!

Carmen Bruni Techniques for Solving Diophantine Equations

Technique 1. Local obstructions

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

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).

What if the equation has no solution. Must there be a primep such that over Zp our equation has no solution?

No!

Definition

An equation is said to satisfy the Hasse Principle (local to globalprinciple) 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!

Carmen Bruni Techniques for Solving Diophantine Equations

Counter Examples to the Hasse Principle

Theorem (Selmer 1951)

The equation3x3 + 4y3 + 5z3 = 0

has solutions over R and in Zp for every prime p but does not havea 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 counterexample.

Carmen Bruni Techniques for Solving Diophantine Equations

Counter Examples to the Hasse Principle

Theorem (Selmer 1951)

The equation3x3 + 4y3 + 5z3 = 0

has solutions over R and in Zp for every prime p but does not havea 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 counterexample.

Carmen Bruni Techniques for Solving Diophantine Equations

Counter Examples to the Hasse Principle

Theorem

The equationw2 + x2 + y2 + z2 = −1

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

Proof.

No Solutions over Z: Exercise

Solutions in Zp: Reducing modulo p gives

w2 + x2 + y2 + z2 = −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.

Carmen Bruni Techniques for Solving Diophantine Equations

Counter Examples to the Hasse Principle

Theorem

The equationw2 + x2 + y2 + z2 = −1

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

Proof.

No Solutions over Z:

Exercise

Solutions in Zp: Reducing modulo p gives

w2 + x2 + y2 + z2 = −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.

Carmen Bruni Techniques for Solving Diophantine Equations

Counter Examples to the Hasse Principle

Theorem

The equationw2 + x2 + y2 + z2 = −1

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

Proof.

No Solutions over Z: Exercise

Solutions in Zp: Reducing modulo p gives

w2 + x2 + y2 + z2 = −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.

Carmen Bruni Techniques for Solving Diophantine Equations

Counter Examples to the Hasse Principle

Theorem

The equationw2 + x2 + y2 + z2 = −1

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

Proof.

No Solutions over Z: Exercise

Solutions in Zp: Reducing modulo p gives

w2 + x2 + y2 + z2 = −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.

Carmen Bruni Techniques for Solving Diophantine Equations

Counter Examples to the Hasse Principle

Theorem

The equationw2 + x2 + y2 + z2 = −1

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

Proof.

No Solutions over Z: Exercise

Solutions in Zp: Reducing modulo p gives

w2 + x2 + y2 + z2 = −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.

Carmen Bruni Techniques for Solving Diophantine Equations

Technique 2. Algebraic Number Theory Techniques

We are looking for solutions over Z to Diophantine Equations.What, if any of these properties still hold for other rings?

A early attempt on Fermat’s Last Theorem xn + yn = zn

involved using the ring of integers of the cyclotomic fieldZ[ζn] where ζnn = 1 and attempting to factor the equation.

The biggest obstacle in this attempt was that Z[ζn] 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.

Carmen Bruni Techniques for Solving Diophantine Equations

Technique 2. Algebraic Number Theory Techniques

We are looking for solutions over Z to Diophantine Equations.What, if any of these properties still hold for other rings?

A early attempt on Fermat’s Last Theorem xn + yn = zn

involved using the ring of integers of the cyclotomic fieldZ[ζn] where ζnn = 1 and attempting to factor the equation.

The biggest obstacle in this attempt was that Z[ζn] 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.

Carmen Bruni Techniques for Solving Diophantine Equations

Technique 2. Algebraic Number Theory Techniques

We are looking for solutions over Z to Diophantine Equations.What, if any of these properties still hold for other rings?

A early attempt on Fermat’s Last Theorem xn + yn = zn

involved using the ring of integers of the cyclotomic fieldZ[ζn] where ζnn = 1 and attempting to factor the equation.

The biggest obstacle in this attempt was that Z[ζn] 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.

Carmen Bruni Techniques for Solving Diophantine Equations

Technique 2. Algebraic Number Theory Techniques

We are looking for solutions over Z to Diophantine Equations.What, if any of these properties still hold for other rings?

A early attempt on Fermat’s Last Theorem xn + yn = zn

involved using the ring of integers of the cyclotomic fieldZ[ζn] where ζnn = 1 and attempting to factor the equation.

The biggest obstacle in this attempt was that Z[ζn] 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.

Carmen Bruni Techniques for Solving Diophantine Equations

Technique 2. Algebraic Number Theory Techniques

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

We have that p = x2 + y2 for a prime p if and only ifp ≡ 1 (mod 4).

Proof.

If p ≡ 3 (mod 4), then our argument from before shows thatp = x2 + y2 has no solutions. So we suppose that p ≡ 1 (mod 4).Since Z∗p is cyclic, it has a generator say g with

gp−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 ], whichsays 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)

Carmen Bruni Techniques for Solving Diophantine Equations

Technique 2. Algebraic Number Theory Techniques

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

We have that p = x2 + y2 for a prime p if and only ifp ≡ 1 (mod 4).

Proof.

If p ≡ 3 (mod 4), then our argument from before shows thatp = x2 + y2 has no solutions. So we suppose that p ≡ 1 (mod 4).

Since Z∗p is cyclic, it has a generator say g with

gp−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 ], whichsays 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)

Carmen Bruni Techniques for Solving Diophantine Equations

Technique 2. Algebraic Number Theory Techniques

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

We have that p = x2 + y2 for a prime p if and only ifp ≡ 1 (mod 4).

Proof.

If p ≡ 3 (mod 4), then our argument from before shows thatp = x2 + y2 has no solutions. So we suppose that p ≡ 1 (mod 4).Since Z∗p is cyclic, it has a generator say g with

gp−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 ], whichsays 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)

Carmen Bruni Techniques for Solving Diophantine Equations

Technique 2. Algebraic Number Theory Techniques

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

We have that p = x2 + y2 for a prime p if and only ifp ≡ 1 (mod 4).

Proof.

If p ≡ 3 (mod 4), then our argument from before shows thatp = x2 + y2 has no solutions. So we suppose that p ≡ 1 (mod 4).Since Z∗p is cyclic, it has a generator say g with

gp−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 ], whichsays 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)

Carmen Bruni Techniques for Solving Diophantine Equations

Technique 2. Algebraic Number Theory Techniques

Proof.

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

Now, Let N(a + bi) = a2 + b2 = |a + bi |2 and suppose that p = wzwith w , z ∈ Z[i ] and w , z 6= ±1,±i (these are the units of Z[i ]).Taking the norms of both sides gives p2 = 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) = x2 + y2 as required.

Carmen Bruni Techniques for Solving Diophantine Equations

Technique 2. Algebraic Number Theory Techniques

Proof.

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

Now, Let N(a + bi) = a2 + b2 = |a + bi |2 and suppose that p = wzwith w , z ∈ Z[i ] and w , z 6= ±1,±i (these are the units of Z[i ]).Taking the norms of both sides gives p2 = 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) = x2 + y2 as required.

Carmen Bruni Techniques for Solving Diophantine Equations

Technique 2. Algebraic Number Theory Techniques

Proof.

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

Now, Let N(a + bi) = a2 + b2 = |a + bi |2 and suppose that p = wzwith w , z ∈ Z[i ] and w , z 6= ±1,±i (these are the units of Z[i ]).Taking the norms of both sides gives p2 = 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) = x2 + y2 as required.

Carmen Bruni Techniques for Solving Diophantine Equations

Technique 3. Linear Forms in Logarithms

Theorem (Baker 1966)

Let α1, .., αn be algebraic integers (roots of monic polynomials)and let H be a number larger than the absolute value of thecoefficients of the minimum polynomials of the αi . Let bi ∈ Z∗and let B = maxi |bi |. Let

Λ := |n∑

i=1

bi logαi |

and suppose that Λ 6= 0. Then there exist computable constants Cand 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].

Carmen Bruni Techniques for Solving Diophantine Equations

Technique 3. Linear Forms in Logarithms

Theorem (Baker 1966)

Let α1, .., αn be algebraic integers (roots of monic polynomials)and let H be a number larger than the absolute value of thecoefficients of the minimum polynomials of the αi . Let bi ∈ Z∗and let B = maxi |bi |. Let

Λ := |n∑

i=1

bi logαi |

and suppose that Λ 6= 0.

Then there exist computable constants Cand 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].

Carmen Bruni Techniques for Solving Diophantine Equations

Technique 3. Linear Forms in Logarithms

Theorem (Baker 1966)

Let α1, .., αn be algebraic integers (roots of monic polynomials)and let H be a number larger than the absolute value of thecoefficients of the minimum polynomials of the αi . Let bi ∈ Z∗and let B = maxi |bi |. Let

Λ := |n∑

i=1

bi logαi |

and suppose that Λ 6= 0. Then there exist computable constants Cand 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].

Carmen Bruni Techniques for Solving Diophantine Equations

Technique 3. Linear Forms in Logarithms

Today, under slight modifications to the above statement, Matveevhas shown that we can take L to be

L := 2 · 30n+4(n + 1)6dn+2(d + 1).

A similar statement even holds for Λ =

∣∣∣∣∏ni=1 α

bii − 1

∣∣∣∣.What can we prove with this theorem?

Theorem

Let a, b, k ∈ Z with a, b > 1 and k nonzero. If am − bn = k form, n ∈ Z, then m and n are bounded by a computable numberdepending only on k and the greatest prime divisor of ab.

Carmen Bruni Techniques for Solving Diophantine Equations

Technique 3. Linear Forms in Logarithms

Today, under slight modifications to the above statement, Matveevhas shown that we can take L to be

L := 2 · 30n+4(n + 1)6dn+2(d + 1).

A similar statement even holds for Λ =

∣∣∣∣∏ni=1 α

bii − 1

∣∣∣∣.What can we prove with this theorem?

Theorem

Let a, b, k ∈ Z with a, b > 1 and k nonzero. If am − bn = k form, n ∈ Z, then m and n are bounded by a computable numberdepending only on k and the greatest prime divisor of ab.

Carmen Bruni Techniques for Solving Diophantine Equations

Technique 3. Linear Forms in Logarithms

Today, under slight modifications to the above statement, Matveevhas shown that we can take L to be

L := 2 · 30n+4(n + 1)6dn+2(d + 1).

A similar statement even holds for Λ =

∣∣∣∣∏ni=1 α

bii − 1

∣∣∣∣.

What can we prove with this theorem?

Theorem

Let a, b, k ∈ Z with a, b > 1 and k nonzero. If am − bn = k form, n ∈ Z, then m and n are bounded by a computable numberdepending only on k and the greatest prime divisor of ab.

Carmen Bruni Techniques for Solving Diophantine Equations

Technique 3. Linear Forms in Logarithms

Today, under slight modifications to the above statement, Matveevhas shown that we can take L to be

L := 2 · 30n+4(n + 1)6dn+2(d + 1).

A similar statement even holds for Λ =

∣∣∣∣∏ni=1 α

bii − 1

∣∣∣∣.What can we prove with this theorem?

Theorem

Let a, b, k ∈ Z with a, b > 1 and k nonzero. If am − bn = k form, n ∈ Z, then m and n are bounded by a computable numberdepending only on k and the greatest prime divisor of ab.

Carmen Bruni Techniques for Solving Diophantine Equations

Technique 3. Linear Forms in Logarithms

Today, under slight modifications to the above statement, Matveevhas shown that we can take L to be

L := 2 · 30n+4(n + 1)6dn+2(d + 1).

A similar statement even holds for Λ =

∣∣∣∣∏ni=1 α

bii − 1

∣∣∣∣.What can we prove with this theorem?

Theorem

Let a, b, k ∈ Z with a, b > 1 and k nonzero. If am − bn = k form, n ∈ Z, then m and n are bounded by a computable numberdepending only on k and the greatest prime divisor of ab.

Carmen Bruni Techniques for Solving Diophantine Equations

Technique 3. Linear Forms in Logarithms

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 :=

{ ∏pi∈P

paii : ai ∈ N

}.

Order the elements of S by 0 < n1 < n2 < .... Then there exists acomputable number C depending on M only so that

ni+1 − ni ≥ni

(log ni )C

Carmen Bruni Techniques for Solving Diophantine Equations

Technique 3. Linear Forms in Logarithms

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 :=

{ ∏pi∈P

paii : ai ∈ N

}.

Order the elements of S by 0 < n1 < n2 < .... Then there exists acomputable number C depending on M only so that

ni+1 − ni ≥ni

(log ni )C

Carmen Bruni Techniques for Solving Diophantine Equations

Technique 3. Linear Forms in Logarithms

Proof.

Let ni =∏

pi∈P paii and ni+1 =

∏pi∈P pbi

i . Look at

ni+1

ni− 1 =

∏pi∈P

pbi−aii − 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+1

ni− 1 ≥ (log ni )

−C

for some computable C depending only on M.

Carmen Bruni Techniques for Solving Diophantine Equations

Technique 3. Linear Forms in Logarithms

Proof.

Let ni =∏

pi∈P paii and ni+1 =

∏pi∈P pbi

i . Look at

ni+1

ni− 1 =

∏pi∈P

pbi−aii − 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+1

ni− 1 ≥ (log ni )

−C

for some computable C depending only on M.

Carmen Bruni Techniques for Solving Diophantine Equations

Technique 4. The Modular Method

This technique is best showed by an example. Let’s solve FLT,that is, there are no solutions to the equation xn + yn = zn inpairwise coprime integers x , y , z and n ≥ 3.

Without loss ofgenerality, suppose n is prime and n ≥ 5. The techniques here useelliptic curves, modular forms, and a whole lot of machinery.

Definition

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

y2 = x3 + a2x2 + a4x + a6

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

Carmen Bruni Techniques for Solving Diophantine Equations

Technique 4. The Modular Method

This technique is best showed by an example. Let’s solve FLT,that is, there are no solutions to the equation xn + yn = zn inpairwise coprime integers x , y , z and n ≥ 3. Without loss ofgenerality, suppose n is prime and n ≥ 5.

The techniques here useelliptic curves, modular forms, and a whole lot of machinery.

Definition

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

y2 = x3 + a2x2 + a4x + a6

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

Carmen Bruni Techniques for Solving Diophantine Equations

Technique 4. The Modular Method

This technique is best showed by an example. Let’s solve FLT,that is, there are no solutions to the equation xn + yn = zn inpairwise coprime integers x , y , z and n ≥ 3. Without loss ofgenerality, suppose n is prime and n ≥ 5. The techniques here useelliptic curves, modular forms, and a whole lot of machinery.

Definition

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

y2 = x3 + a2x2 + a4x + a6

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

Carmen Bruni Techniques for Solving Diophantine Equations

Technique 4. The Modular Method

Associated to each elliptic curve y2 = x3 + a2x2 + a4x + a6 is anumber 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, y2 = x3 + x2 + 4x + 4 has conductor N = 20 = 22 · 5.This elliptic curve is defined in all Fp where p 6= 2, 5. Reducing thecurve modulo 5 for example gives

y2 = x3 + x2 + 4x + 4 ≡ x3 + x2− x −1 ≡ (x + 1)2(x −1) (mod 5)

which has a double root and hence is not smooth.

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

Technique 4. The Modular Method

Associated to each elliptic curve y2 = x3 + a2x2 + a4x + a6 is anumber 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, y2 = x3 + x2 + 4x + 4 has conductor N = 20 = 22 · 5.This elliptic curve is defined in all Fp where p 6= 2, 5. Reducing thecurve modulo 5 for example gives

y2 = x3 + x2 + 4x + 4 ≡ x3 + x2− x −1 ≡ (x + 1)2(x −1) (mod 5)

which has a double root and hence is not smooth.

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

Technique 4. The Modular Method

Associated to each elliptic curve y2 = x3 + a2x2 + a4x + a6 is anumber 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, y2 = x3 + x2 + 4x + 4 has conductor N = 20 = 22 · 5.This elliptic curve is defined in all Fp where p 6= 2, 5. Reducing thecurve modulo 5 for example gives

y2 = x3 + x2 + 4x + 4 ≡ x3 + x2− x −1 ≡ (x + 1)2(x −1) (mod 5)

which has a double root and hence is not smooth.

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

Technique 4. The Modular Method

Associated to each elliptic curve y2 = x3 + a2x2 + a4x + a6 is anumber 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, y2 = x3 + x2 + 4x + 4 has conductor N = 20 = 22 · 5.This elliptic curve is defined in all Fp where p 6= 2, 5. Reducing thecurve modulo 5 for example gives

y2 = x3 + x2 + 4x + 4 ≡ x3 + x2− x −1 ≡ (x + 1)2(x −1) (mod 5)

which has a double root and hence is not smooth.

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

Technique 4. The Modular Method

Associated to each elliptic curve y2 = x3 + a2x2 + a4x + a6 is anumber 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, y2 = x3 + x2 + 4x + 4 has conductor N = 20 = 22 · 5.This elliptic curve is defined in all Fp where p 6= 2, 5. Reducing thecurve modulo 5 for example gives

y2 = x3 + x2 + 4x + 4 ≡ x3 + x2− x −1 ≡ (x + 1)2(x −1) (mod 5)

which has a double root and hence is not smooth.

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

Technique 4. The Modular Method

We ‘prove’ FLT. Suppose that ap + bp = cp for a, b, c ∈ Zpairwise coprime and nonzero and p ≥ 5 prime. WLOG b iseven.

Associate to this solution an elliptic curve

Y 2 = X (X − ap)(X + bp)

The [minimal] discriminant and conductor are of the form

∆ = 2−8(abc)2p N = 2∏

q 6=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 oflevel 2! This completes the proof.

Carmen Bruni Techniques for Solving Diophantine Equations

Technique 4. The Modular Method

We ‘prove’ FLT. Suppose that ap + bp = cp for a, b, c ∈ Zpairwise coprime and nonzero and p ≥ 5 prime. WLOG b iseven.

Associate to this solution an elliptic curve

Y 2 = X (X − ap)(X + bp)

The [minimal] discriminant and conductor are of the form

∆ = 2−8(abc)2p N = 2∏

q 6=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 oflevel 2! This completes the proof.

Carmen Bruni Techniques for Solving Diophantine Equations

Technique 4. The Modular Method

We ‘prove’ FLT. Suppose that ap + bp = cp for a, b, c ∈ Zpairwise coprime and nonzero and p ≥ 5 prime. WLOG b iseven.

Associate to this solution an elliptic curve

Y 2 = X (X − ap)(X + bp)

The [minimal] discriminant and conductor are of the form

∆ = 2−8(abc)2p N = 2∏

q 6=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 oflevel 2! This completes the proof.

Carmen Bruni Techniques for Solving Diophantine Equations

Technique 4. The Modular Method

We ‘prove’ FLT. Suppose that ap + bp = cp for a, b, c ∈ Zpairwise coprime and nonzero and p ≥ 5 prime. WLOG b iseven.

Associate to this solution an elliptic curve

Y 2 = X (X − ap)(X + bp)

The [minimal] discriminant and conductor are of the form

∆ = 2−8(abc)2p N = 2∏

q 6=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 oflevel 2! This completes the proof.

Carmen Bruni Techniques for Solving Diophantine Equations

Other Techniques

Skolem’s Method

Chabauty’s Method

Hyper Geometric Method

The Brauer-Manin Obstruction

The Descent Obstruction

Carmen Bruni Techniques for Solving Diophantine Equations

Other Techniques

Skolem’s Method

Chabauty’s Method

Hyper Geometric Method

The Brauer-Manin Obstruction

The Descent Obstruction

Carmen Bruni Techniques for Solving Diophantine Equations

Other Techniques

Skolem’s Method

Chabauty’s Method

Hyper Geometric Method

The Brauer-Manin Obstruction

The Descent Obstruction

Carmen Bruni Techniques for Solving Diophantine Equations

Other Techniques

Skolem’s Method

Chabauty’s Method

Hyper Geometric Method

The Brauer-Manin Obstruction

The Descent Obstruction

Carmen Bruni Techniques for Solving Diophantine Equations

Other Techniques

Skolem’s Method

Chabauty’s Method

Hyper Geometric Method

The Brauer-Manin Obstruction

The Descent Obstruction

Carmen Bruni Techniques for Solving Diophantine Equations

Thank you!

Carmen Bruni Techniques for Solving Diophantine Equations