Open problems in number theory

chris wuthrich

dec 2011

Goldbach’s conjectureAny even number can be written as a sum of two primes.

Examples :

12 = 5 + 7

28 = 5 + 23 = 11 + 17

168 = 5 + 163 = 11 + 157 = 17 + 151 = 19 + 149

= 29 + 139 = 31 + 137 = 37 + 131 = 41 + 127

= 59 + 109 = 61 + 107 = 67 + 101

= 71 + 97 = 79 + 89

Goldbach’s conjectureAny even number can be written as a sum of two primes.

Examples :

12 = 5 + 7

28 = 5 + 23 = 11 + 17

168 = 5 + 163 = 11 + 157 = 17 + 151 = 19 + 149

= 29 + 139 = 31 + 137 = 37 + 131 = 41 + 127

= 59 + 109 = 61 + 107 = 67 + 101

= 71 + 97 = 79 + 89

Goldbach’s conjectureAny even number can be written as a sum of two primes.

Examples :

12 = 5 + 7

28 = 5 + 23 = 11 + 17

168 = 5 + 163 = 11 + 157 = 17 + 151 = 19 + 149

= 29 + 139 = 31 + 137 = 37 + 131 = 41 + 127

= 59 + 109 = 61 + 107 = 67 + 101

= 71 + 97 = 79 + 89

Goldbach’s conjectureAny even number can be written as a sum of two primes.

SOURCE: WIKIPEDIA

Twin primesThere are infinitely many primes p such that p + 2 is also prime.

Examples :

(3, 5) (5, 7) (11, 13) (17, 19) (29, 31) (41, 43)

(1019, 1021) (2027, 2029) (3119, 3121)

(4001, 4003) (5009, 5011)(6089, 6091)

(7127, 7129) (8009, 8011) (9011, 9013)

Twin primesThere are infinitely many primes p such that p + 2 is also prime.

Examples :

(3, 5) (5, 7) (11, 13) (17, 19) (29, 31) (41, 43)

(1019, 1021) (2027, 2029) (3119, 3121)

(4001, 4003) (5009, 5011)(6089, 6091)

(7127, 7129) (8009, 8011) (9011, 9013)

Twin primesThere are infinitely many primes p such that p + 2 is also prime.

Examples :

(3, 5) (5, 7) (11, 13) (17, 19) (29, 31) (41, 43)

(1019, 1021) (2027, 2029) (3119, 3121)

(4001, 4003) (5009, 5011)(6089, 6091)

(7127, 7129) (8009, 8011) (9011, 9013)

Landau’s conjecture

There are infinitely many primes of the form n2 + 1.

Examples :

2 = 12 + 1

5 = 22 + 1

17 = 42 + 1

37 = 62 + 1

101 = 102 + 1

972197 = 9862 + 1

Landau’s conjecture

There are infinitely many primes of the form n2 + 1.

Examples :

2 = 12 + 1

5 = 22 + 1

17 = 42 + 1

37 = 62 + 1

101 = 102 + 1

972197 = 9862 + 1

Schinzel’s hypothesis

Let f (X) and g(X) be two irreducible polynomials in Z[X].

Suppose there is no integer n such that n divides f (k) · g(k) forall k. Then there are infinitely many values of k such that f (k)and g(k) are both prime numbers.

Take f (X) = X and g(X) = X + 2; we get the twin primeconjecture.Take f (X) = X − 1 and g(X) = X2 + 1; we get more thanLandau’s conjecture.The hypothesis rules out cases like f (X) = X andg(X) = X2 + 1.

Schinzel’s hypothesis

Let f (X) and g(X) be two irreducible polynomials in Z[X].

Suppose there is no integer n such that n divides f (k) · g(k) forall k.

Then there are infinitely many values of k such that f (k)and g(k) are both prime numbers.

Take f (X) = X and g(X) = X + 2; we get the twin primeconjecture.Take f (X) = X − 1 and g(X) = X2 + 1; we get more thanLandau’s conjecture.The hypothesis rules out cases like f (X) = X andg(X) = X2 + 1.

Schinzel’s hypothesis

Let f (X) and g(X) be two irreducible polynomials in Z[X].

Suppose there is no integer n such that n divides f (k) · g(k) forall k.

Then there are infinitely many values of k such that f (k)and g(k) are both prime numbers.

Take f (X) = X and g(X) = X + 2; we get the twin primeconjecture.

Take f (X) = X − 1 and g(X) = X2 + 1; we get more thanLandau’s conjecture.The hypothesis rules out cases like f (X) = X andg(X) = X2 + 1.

Schinzel’s hypothesis

Let f (X) and g(X) be two irreducible polynomials in Z[X].

Suppose there is no integer n such that n divides f (k) · g(k) forall k.

Then there are infinitely many values of k such that f (k)and g(k) are both prime numbers.

Take f (X) = X and g(X) = X + 2; we get the twin primeconjecture.Take f (X) = X − 1 and g(X) = X2 + 1; we get more thanLandau’s conjecture.

The hypothesis rules out cases like f (X) = X andg(X) = X2 + 1.

Schinzel’s hypothesis

Let f (X) and g(X) be two irreducible polynomials in Z[X].

Suppose there is no integer n such that n divides f (k) · g(k) forall k.

Then there are infinitely many values of k such that f (k)and g(k) are both prime numbers.

Take f (X) = X and g(X) = X + 2; we get the twin primeconjecture.Take f (X) = X − 1 and g(X) = X2 + 1; we get more thanLandau’s conjecture.The hypothesis rules out cases like f (X) = X andg(X) = X2 + 1.

Schinzel’s hypothesis

Let f (X) and g(X) be two irreducible polynomials in Z[X].Suppose there is no integer n such that n divides f (k) · g(k) forall k. Then there are infinitely many values of k such that f (k)and g(k) are both prime numbers.

Take f (X) = X and g(X) = X + 2; we get the twin primeconjecture.Take f (X) = X − 1 and g(X) = X2 + 1; we get more thanLandau’s conjecture.The hypothesis rules out cases like f (X) = X andg(X) = X2 + 1.

Theorem of Terence Tao and Ben GreenThere are arbitary long arithmetic progression in the primes.

Examples :

(3, 5, 7) (5, 11, 17) (7, 13, 19)

(5, 11, 17, 23) (7, 37, 67, 97, 127, 157)

(1564588127269043,

1564588127269043 + 278810314282500,

1564588127269043 + 2 · 278810314282500, . . .

1564588127269043 + 23 · 278810314282500)

Theorem of Terence Tao and Ben GreenThere are arbitary long arithmetic progression in the primes.

Examples :

(3, 5, 7) (5, 11, 17) (7, 13, 19)

(5, 11, 17, 23) (7, 37, 67, 97, 127, 157)

(1564588127269043,

1564588127269043 + 278810314282500,

1564588127269043 + 2 · 278810314282500, . . .

1564588127269043 + 23 · 278810314282500)

Theorem of Terence Tao and Ben GreenThere are arbitary long arithmetic progression in the primes.

Examples :

(3, 5, 7) (5, 11, 17) (7, 13, 19)

(5, 11, 17, 23) (7, 37, 67, 97, 127, 157)

(1564588127269043,

1564588127269043 + 278810314282500,

1564588127269043 + 2 · 278810314282500, . . .

1564588127269043 + 23 · 278810314282500)

Theorem of Terence Tao and Ben GreenThere are arbitary long arithmetic progression in the primes.

Examples :

(3, 5, 7) (5, 11, 17) (7, 13, 19)

(5, 11, 17, 23) (7, 37, 67, 97, 127, 157)

(1564588127269043,

1564588127269043 + 278810314282500,

1564588127269043 + 2 · 278810314282500, . . .

1564588127269043 + 23 · 278810314282500)

Unique factoristaion hold in the integers Z[i] of Q(i).

Theorem (Heegner, Stark, Baker)

There exists only finitely many negative D such that Q(√

D) hasunique factorisation, namely -1, -2, -3, -7, -11, -19, -43, -67,-163.

Gauss’ conjecture

There are infintely many positive D such that Q(√

D) hasunique factorisation.

Examples :D = 2, 3, 5, 6, 7, 11, 13, 14, 17, 21, 29, 33, 37, 41, 53, 57, 61, . . . .

Unique factoristaion hold in the integers Z[i] of Q(i).

Theorem (Heegner, Stark, Baker)

There exists only finitely many negative D such that Q(√

D) hasunique factorisation,

namely -1, -2, -3, -7, -11, -19, -43, -67,-163.

Gauss’ conjecture

There are infintely many positive D such that Q(√

D) hasunique factorisation.

Examples :D = 2, 3, 5, 6, 7, 11, 13, 14, 17, 21, 29, 33, 37, 41, 53, 57, 61, . . . .

Unique factoristaion hold in the integers Z[i] of Q(i).

Theorem (Heegner, Stark, Baker)

There exists only finitely many negative D such that Q(√

D) hasunique factorisation, namely -1, -2, -3, -7, -11, -19, -43, -67,-163.

Gauss’ conjecture

There are infintely many positive D such that Q(√

D) hasunique factorisation.

Examples :D = 2, 3, 5, 6, 7, 11, 13, 14, 17, 21, 29, 33, 37, 41, 53, 57, 61, . . . .

Unique factoristaion hold in the integers Z[i] of Q(i).

Theorem (Heegner, Stark, Baker)

There exists only finitely many negative D such that Q(√

D) hasunique factorisation, namely -1, -2, -3, -7, -11, -19, -43, -67,-163.

Gauss’ conjecture

There are infintely many positive D such that Q(√

D) hasunique factorisation.

Examples :D = 2, 3, 5, 6, 7, 11, 13, 14, 17, 21, 29, 33, 37, 41, 53, 57, 61, . . . .

Unique factoristaion hold in the integers Z[i] of Q(i).

Theorem (Heegner, Stark, Baker)

There exists only finitely many negative D such that Q(√

D) hasunique factorisation, namely -1, -2, -3, -7, -11, -19, -43, -67,-163.

Gauss’ conjecture

There are infintely many positive D such that Q(√

D) hasunique factorisation.

Examples :D = 2, 3, 5, 6, 7, 11, 13, 14, 17, 21, 29, 33, 37, 41, 53, 57, 61, . . . .

For each number field K, e.g. Q(i), Q( 7√−3), . . . , there is a

class humber h.

Unique factorisation holds if and only if h = 1.

Theorem (Kummer)

Let ξp = e2πi/p. The class number of Q(ξp) is divisible by p if andonly if p is irregular.

If p is regular then Fermat’s last theorem holds for p.

Vandiver’s conjecture

p does not divide the class number of Q(ξp) ∩ R = Q(ξp + ξ̄p).

For each number field K, e.g. Q(i), Q( 7√−3), . . . , there is a

class humber h.Unique factorisation holds if and only if h = 1.

Theorem (Kummer)

Let ξp = e2πi/p. The class number of Q(ξp) is divisible by p if andonly if p is irregular.

If p is regular then Fermat’s last theorem holds for p.

Vandiver’s conjecture

p does not divide the class number of Q(ξp) ∩ R = Q(ξp + ξ̄p).

For each number field K, e.g. Q(i), Q( 7√−3), . . . , there is a

class humber h.Unique factorisation holds if and only if h = 1.

Theorem (Kummer)

Let ξp = e2πi/p. The class number of Q(ξp) is divisible by p if andonly if p is irregular.

If p is regular then Fermat’s last theorem holds for p.

Vandiver’s conjecture

p does not divide the class number of Q(ξp) ∩ R = Q(ξp + ξ̄p).

For each number field K, e.g. Q(i), Q( 7√−3), . . . , there is a

class humber h.Unique factorisation holds if and only if h = 1.

Theorem (Kummer)

Let ξp = e2πi/p. The class number of Q(ξp) is divisible by p if andonly if p is irregular.

If p is regular then Fermat’s last theorem holds for p.

Vandiver’s conjecture

p does not divide the class number of Q(ξp) ∩ R = Q(ξp + ξ̄p).

For each number field K, e.g. Q(i), Q( 7√−3), . . . , there is a

class humber h.Unique factorisation holds if and only if h = 1.

Theorem (Kummer)

Let ξp = e2πi/p. The class number of Q(ξp) is divisible by p if andonly if p is irregular.

If p is regular then Fermat’s last theorem holds for p.Because one can factor

xp + yp = (x + y)(x + ξpy) · · · (x + ξp−1p y) = zp.

Vandiver’s conjecture

p does not divide the class number of Q(ξp) ∩ R = Q(ξp + ξ̄p).

For each number field K, e.g. Q(i), Q( 7√−3), . . . , there is a

class humber h.Unique factorisation holds if and only if h = 1.

Theorem (Kummer)

Let ξp = e2πi/p. The class number of Q(ξp) is divisible by p if andonly if p is irregular.

If p is regular then Fermat’s last theorem holds for p.

Vandiver’s conjecture

p does not divide the class number of Q(ξp) ∩ R = Q(ξp + ξ̄p).

Kummer congruences

For all n ≡ m ≡ i (mod (p− 1)), we have, for all r > 0

∣∣n−m∣∣p <

1pr ⇒

∣∣∣∣(1−pm−1)Bm

m−(1−pn−1)Bn

n

∣∣∣∣p<

1pr+1

Kummer congruences

For all n ≡ m ≡ i (mod (p− 1)), we have, for all ε > 0,

∣∣n− m∣∣p < δ ⇒

∣∣∣∣(1− pm−1)Bm

m−(1− pn−1)Bn

n

∣∣∣∣p< ε

The function

Z→ Qp

k 7→(1− pk−1)Bk

k

is p-adically continuous.

Kummer congruences

For all n ≡ m ≡ i (mod (p− 1)), we have, for all ε > 0,

∣∣n− m∣∣p < δ ⇒

∣∣∣∣(1− pm−1)Bm

m−(1− pn−1)Bn

n

∣∣∣∣p< ε

The function

Z→ Qp

k 7→(1− pk−1)Bk

k

is p-adically continuous.

Kummer congruences

For all n ≡ m ≡ i (mod (p− 1)), we have, for all ε > 0,

∣∣n− m∣∣p < δ ⇒

∣∣∣∣(1− pm−1)Bm

m−(1− pn−1)Bn

n

∣∣∣∣p< ε

The function

Z→ Qp

k 7→(1− pk−1)Bk

k

is p-adically continuous.

p-adic zeta-functionThere is an analytic function ζp : Zp → Qp such that

ζp(−k) =(

1− 1pk

)ζ(−k)

for all integers k > 0.

ζp(s) has a simple pole at s = 1 ∈ Zp.

Conjecture

ζp(k) 6= 0 for k ∈ 2Z.

Each zero of ζp(s) has something to do with the class numberof Q(ξpn).

p-adic zeta-functionThere is an analytic function ζp : Zp → Qp such that

ζp(−k) =(

1− 1pk

)ζ(−k)

for all integers k > 0.

ζp(s) has a simple pole at s = 1 ∈ Zp.

Conjecture

ζp(k) 6= 0 for k ∈ 2Z.

Each zero of ζp(s) has something to do with the class numberof Q(ξpn).

p-adic zeta-functionThere is an analytic function ζp : Zp → Qp such that

ζp(−k) =(

1− 1pk

)ζ(−k)

for all integers k > 0.

ζp(s) has a simple pole at s = 1 ∈ Zp.

Conjecture

ζp(k) 6= 0 for k ∈ 2Z.

Each zero of ζp(s) has something to do with the class numberof Q(ξpn).

p-adic zeta-functionThere is an analytic function ζp : Zp → Qp such that

ζp(−k) =(

1− 1pk

)ζ(−k)

for all integers k > 0.

ζp(s) has a simple pole at s = 1 ∈ Zp.

Conjecture

ζp(k) 6= 0 for k ∈ 2Z.

Each zero of ζp(s) has something to do with the class numberof Q(ξpn).

We found that

x = a2 − b2 y = 2 a b z = a2 + b2

parametrises all solutions of x2 + y2 = z2 over Z.

Or Q.

Congruent number problem

Given an integer n, is there a rational solution such thatx y = 2 n ?

We are looking for triangles with rational sides, with a 90 degreeangle and area equal to n. If there is one we say n is acongruent number.

We found that

x = a2 − b2 y = 2 a b z = a2 + b2

parametrises all solutions of x2 + y2 = z2 over Z. Or Q.

Congruent number problem

Given an integer n, is there a rational solution such thatx y = 2 n ?

We are looking for triangles with rational sides, with a 90 degreeangle and area equal to n. If there is one we say n is acongruent number.

We found that

x = a2 − b2 y = 2 a b z = a2 + b2

parametrises all solutions of x2 + y2 = z2 over Z. Or Q.

Congruent number problem

Given an integer n, is there a rational solution such thatx y = 2 n ?

We are looking for triangles with rational sides, with a 90 degreeangle and area equal to n. If there is one we say n is acongruent number.

We found that

x = a2 − b2 y = 2 a b z = a2 + b2

parametrises all solutions of x2 + y2 = z2 over Z. Or Q.

Congruent number problem

Given an integer n, is there a rational solution such thatx y = 2 n ?

We are looking for triangles with rational sides, with a 90 degreeangle and area equal to n.

If there is one we say n is acongruent number.

We found that

x = a2 − b2 y = 2 a b z = a2 + b2

parametrises all solutions of x2 + y2 = z2 over Z. Or Q.

Congruent number problem

Given an integer n, is there a rational solution such thatx y = 2 n ?

We are looking for triangles with rational sides, with a 90 degreeangle and area equal to n. If there is one we say n is acongruent number.

Examples : 6 is a congruent number as 32 + 42 = 52 and3 · 4 = 2 · 6.

5 (20/3, 3/2, 41/6) 21 (12, 7/2, 25/2)6 (4, 3, 5) 22 (140/3, . . .7 (24/5, 35/12, 337/60) 23 (41496/3485, . . .

13 (323/30, 780/323, . . . ) 29 (52780/99, . . .14 (21/2, 8/3, 65/6) 30 (12, 5, 13)15 (15/2, 4, 17/2) 31 (8897/360, . . .

Examples : 6 is a congruent number as 32 + 42 = 52 and3 · 4 = 2 · 6.

5 (20/3, 3/2, 41/6) 21 (12, 7/2, 25/2)6 (4, 3, 5) 22 (140/3, . . .7 (24/5, 35/12, 337/60) 23 (41496/3485, . . .

13 (323/30, 780/323, . . . ) 29 (52780/99, . . .14 (21/2, 8/3, 65/6) 30 (12, 5, 13)15 (15/2, 4, 17/2) 31 (8897/360, . . .

Conjecture

If n ≡ 5, 6 or 7 (mod 8) then n is a congruent number.

For each prime p put p− ap the number of solutions over Z/pZ to

Y2 = X3 − n2X .

L(E, s) =∏p-2n

11− ap

ps + pp2s

for s > 32 .

Conjecture (part of Birch and Swinnerton-Dyer conjecture)

n is congruent if and only if L(E, 1) = 0.

Theorem (Kolyvagin)

If L(E, s) has a simple zero at s = 1, then n is congruent.

Conjecture

If n ≡ 5, 6 or 7 (mod 8) then n is a congruent number.

For each prime p put p− ap the number of solutions over Z/pZ to

Y2 = X3 − n2X .

L(E, s) =∏p-2n

11− ap

ps + pp2s

for s > 32 .

Conjecture (part of Birch and Swinnerton-Dyer conjecture)

n is congruent if and only if L(E, 1) = 0.

Theorem (Kolyvagin)

If L(E, s) has a simple zero at s = 1, then n is congruent.

Conjecture

If n ≡ 5, 6 or 7 (mod 8) then n is a congruent number.

For each prime p put p− ap the number of solutions over Z/pZ to

Y2 = X3 − n2X .

L(E, s) =∏p-2n

11− ap

ps + pp2s

for s > 32 .

Conjecture (part of Birch and Swinnerton-Dyer conjecture)

n is congruent if and only if L(E, 1) = 0.

Theorem (Kolyvagin)

If L(E, s) has a simple zero at s = 1, then n is congruent.

Conjecture

If n ≡ 5, 6 or 7 (mod 8) then n is a congruent number.

For each prime p put p− ap the number of solutions over Z/pZ to

Y2 = X3 − n2X .

L(E, s) =∏p-2n

11− ap

ps + pp2s

for s > 32 .

Conjecture (part of Birch and Swinnerton-Dyer conjecture)

n is congruent if and only if L(E, 1) = 0.

Theorem (Kolyvagin)

If L(E, s) has a simple zero at s = 1, then n is congruent.

Conjecture

If n ≡ 5, 6 or 7 (mod 8) then n is a congruent number.

For each prime p put p− ap the number of solutions over Z/pZ to

Y2 = X3 − n2X .

L(E, s) =∏p-2n

11− ap

ps + pp2s

for s > 32 .

Conjecture (part of Birch and Swinnerton-Dyer conjecture)

n is congruent if and only if L(E, 1) = 0.

Theorem (Kolyvagin)

If L(E, s) has a simple zero at s = 1, then n is congruent.

Riemann hypothesis

The non-trivial zeroes of ζ(s) lie on Re(s) = 12 .

Continuation of ζ :

ζ(s) =s

s− 1− s ·

∫ ∞1

x− [x]

xs+1 dx for Re(s) > 0

Since ζ(1− n) = −Bnn , there are “trivial” zeroes at odd negative

integers.

Riemann hypothesis

The non-trivial zeroes of ζ(s) lie on Re(s) = 12 .

Continuation of ζ :

ζ(s) =s

s− 1− s ·

∫ ∞1

x− [x]

xs+1 dx for Re(s) > 0

Since ζ(1− n) = −Bnn , there are “trivial” zeroes at odd negative

integers.

Riemann hypothesis

The non-trivial zeroes of ζ(s) lie on Re(s) = 12 .

Continuation of ζ :

ζ(s) =s

s− 1− s ·

∫ ∞1

x− [x]

xs+1 dx for Re(s) > 0

Since ζ(1− n) = −Bnn , there are “trivial” zeroes at odd negative

integers.

Conjecture (Riemann hypothesis)

The non-trivial zeroes of ζ(s) lie on Re(s) = 12 .

SOURCE: HTTP://SECAMLOCAL.EX.AC.UK/PEOPLE/STAFF/MRWATKIN/ZETA/ENCODING1.HTM

Riemann hypothesis

The series 1ζ(s) =

∑n>1

µ(n)ns converges for all Re(s) > 1

2 .

Link to prime numbers

log ζ(s) = s ·∫ ∞

2

π(x)

xs − 1· dx

x

Riemann hypothesis

The series 1ζ(s) =

∑n>1

µ(n)ns converges for all Re(s) > 1

2 .

Link to prime numbers

log ζ(s) = s ·∫ ∞

2

π(x)

xs − 1· dx

x

Approximation of π(x) using the first n pairs of zeroes of ζ(s).

Approximation of π(x) using the first n pairs of zeroes of ζ(s).

Approximation of π(x) using the first n pairs of zeroes of ζ(s).

Approximation of π(x) using the first n pairs of zeroes of ζ(s).

Approximation of π(x) using the first n pairs of zeroes of ζ(s).

Approximation of π(x) using the first n pairs of zeroes of ζ(s).

Approximation of π(x) using the first n pairs of zeroes of ζ(s).

Approximation of π(x) using the first n pairs of zeroes of ζ(s).

Approximation of π(x) using the first n pairs of zeroes of ζ(s).

Approximation of π(x) using the first n pairs of zeroes of ζ(s).

Approximation of π(x) using the first n pairs of zeroes of ζ(s).

Approximation of π(x) using the first n pairs of zeroes of ζ(s).

Approximation of π(x) using the first n pairs of zeroes of ζ(s).

Approximation of π(x) using the first n pairs of zeroes of ζ(s).

Approximation of π(x) using the first n pairs of zeroes of ζ(s).

Approximation of π(x) using the first n pairs of zeroes of ζ(s).

Approximation of π(x) using the first n pairs of zeroes of ζ(s).

Approximation of π(x) using the first n pairs of zeroes of ζ(s).

Approximation of π(x) using the first n pairs of zeroes of ζ(s).

Approximation of π(x) using the first n pairs of zeroes of ζ(s).

Approximation of π(x) using the first n pairs of zeroes of ζ(s).

Approximation of π(x) using the first n pairs of zeroes of ζ(s).

Approximation of π(x) using the first n pairs of zeroes of ζ(s).

Approximation of π(x) using the first n pairs of zeroes of ζ(s).

Approximation of π(x) using the first n pairs of zeroes of ζ(s).

Approximation of π(x) using the first n pairs of zeroes of ζ(s).

An elliptic curve is an equation of the form

E : y2 = x3 + A x + B

for some A and B in Q.

QuestionAre there infinitely many rational solutions to E ?

Examples :

An elliptic curve is an equation of the form

E : y2 = x3 + A x + B

for some A and B in Q.

QuestionAre there infinitely many rational solutions to E ?

Examples :

An elliptic curve is an equation of the form

E : y2 = x3 + A x + B

for some A and B in Q.

QuestionAre there infinitely many rational solutions to E ?

Examples :

QuestionAre there infinitely many rational solutions to E ?

Examples :

E2 : y2 = x3 + x + 2

QuestionAre there infinitely many rational solutions to E ?

Examples :

E2 : y2 = x3 + x + 2

has only three solutions (−1, 0), (1,−2), and (1, 2).

QuestionAre there infinitely many rational solutions to E ?

Examples :

E1 : y2 = x3 + x + 1

QuestionAre there infinitely many rational solutions to E ?

Examples :

E1 : y2 = x3 + x + 1

has infinitely many solutions. (0,±1), (14 ,±

94), (72,±611), . . .

QuestionAre there infinitely many rational solutions to E ?

Examples :

E1 : y2 = x3 + x + 1

has infinitely many solutions. (0,±1), (14 ,±

94), (72,±611), . . .

The following x-coordinates are

− 2871296 ,

4399282369 ,

268629131493284 ,

1394558775271824793048 , −3596697936

8760772801 ,75490902224658662944250944 ,

518650137416708646504992707996225 , − 173161424238594532415

310515636774481238884 , . . .

Let Np be the number of solutions of E modulo p plus 1.

Consider the function

f (X) =∑

p∈P, p6X

log(

Np

p

)

Birch and Swinnerton-Dyer conjecture

f (X) stays bounded if and only if there are only finitely manysolutions in Q.

Let Np be the number of solutions of E modulo p plus 1.Consider the function

f (X) =∑

p∈P, p6X

log(

Np

p

)

Birch and Swinnerton-Dyer conjecture

f (X) stays bounded if and only if there are only finitely manysolutions in Q.

Let Np be the number of solutions of E modulo p plus 1.Consider the function

f (X) =∑

p∈P, p6X

log(

Np

p

)

Birch and Swinnerton-Dyer conjecture

f (X) stays bounded if and only if there are only finitely manysolutions in Q.

Let Np be the number of solutions of E modulo p plus 1.Consider the function

f (X) =∑

p∈P, p6X

log(

Np

p

)

Birch and Swinnerton-Dyer conjecture

f (X) stays bounded if and only if there are only finitely manysolutions in Q.

Birch and Swinnerton-Dyer conjecture

f (X) grows like r · log(log(X)), where r is the so-called rank of E.

Birch and Swinnerton-Dyer conjecture

f (X) grows like r · log(log(X)), where r is the so-called rank of E.

E1 : y2 = x3 + x + 1.E2 : y2 = x3 + x + 2.

Birch and Swinnerton-Dyer conjecture

f (X) grows like r · log(log(X)), where r is the so-called rank of E.

Hasse-Weil boundThe value of |ap| = |Np − p− 1| is bounded by 2

√p.

Theorem (R. Taylor), Sato-Tate conjecture

The value of ap2√

p is distributed in [−1, 1] like the measure2π

√1− t2 dt.

Hasse-Weil boundThe value of |ap| = |Np − p− 1| is bounded by 2

√p.

Theorem (R. Taylor), Sato-Tate conjecture

The value of ap2√

p is distributed in [−1, 1] like the measure2π

√1− t2 dt.

Hasse-Weil boundThe value of |ap| = |Np − p− 1| is bounded by 2

√p.

Theorem (R. Taylor), Sato-Tate conjecture

The value of ap2√

p is distributed in [−1, 1] like the measure2π

√1− t2 dt.

Hasse-Weil boundThe value of |ap| = |Np − p− 1| is bounded by 2

√p.

Theorem (R. Taylor), Sato-Tate conjecture

The value of ap2√

p is distributed in [−1, 1] like the measure2π

√1− t2 dt.

Hasse-Weil boundThe value of |ap| = |Np − p− 1| is bounded by 2

√p.

Theorem (R. Taylor), Sato-Tate conjecture

The value of ap2√

p is distributed in [−1, 1] like the measure2π

√1− t2 dt.

Hasse-Weil boundThe value of |ap| = |Np − p− 1| is bounded by 2

√p.

Theorem (R. Taylor), Sato-Tate conjecture

The value of ap2√

p is distributed in [−1, 1] like the measure2π

√1− t2 dt.