open problems in number theory · pdf filetake f(x) = x and g(x) = x + 2; we get the twin...
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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.