Download Open problems in number theory

Open problems in number theory
chris wuthrich
dec 2011
Goldbach’s conjecture
Any even number can be written as a sum of two primes.
Goldbach’s conjecture
Any even number can be written as a sum of two primes.
Examples :
12 = 5 + 7
28 = 5 + 23 = 11 + 17
Goldbach’s conjecture
Any 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 conjecture
Any even number can be written as a sum of two primes.
S OURCE : WIKIPEDIA
Twin primes
There are infinitely many primes p such that p + 2 is also prime.
Twin primes
There 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)
Twin primes
There 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.
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].
Schinzel’s hypothesis
Let f (X) and g(X) be two irreducible polynomials in Z[X].
Then there are infinitely many values of k such that f (k)
and g(k) are both prime numbers.
Schinzel’s hypothesis
Let f (X) and g(X) be two irreducible polynomials in Z[X].
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 prime
conjecture.
Schinzel’s hypothesis
Let f (X) and g(X) be two irreducible polynomials in Z[X].
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 prime
conjecture.
Take f (X) = X − 1 and g(X) = X 2 + 1; we get more than
Landau’s conjecture.
Schinzel’s hypothesis
Let f (X) and g(X) be two irreducible polynomials in Z[X].
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 prime
conjecture.
Take f (X) = X − 1 and g(X) = X 2 + 1; we get more than
Landau’s conjecture.
The hypothesis rules out cases like f (X) = X and
g(X) = X 2 + 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) for
all 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 prime
conjecture.
Take f (X) = X − 1 and g(X) = X 2 + 1; we get more than
Landau’s conjecture.
The hypothesis rules out cases like f (X) = X and
g(X) = X 2 + 1.
Theorem of Terence Tao and Ben Green
There are arbitary long arithmetic progression in the primes.
Theorem of Terence Tao and Ben Green
There are arbitary long arithmetic progression in the primes.
Examples :
(3, 5, 7)
(5, 11, 17)
(7, 13, 19)
Theorem of Terence Tao and Ben Green
There are arbitary long arithmetic progression in the primes.
Examples :
(3, 5, 7)
(5, 11, 17)
(5, 11, 17, 23)
(7, 13, 19)
(7, 37, 67, 97, 127, 157)
Theorem of Terence Tao and Ben Green
There are arbitary long arithmetic progression in the primes.
Examples :
(3, 5, 7)
(5, 11, 17)
(5, 11, 17, 23)
(7, 13, 19)
(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).
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) has
unique factorisation,
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) has
unique factorisation, namely -1, -2, -3, -7, -11, -19, -43, -67,
-163.
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) has
unique factorisation, namely -1, -2, -3, -7, -11, -19, -43, -67,
-163.
Gauss’ conjecture
√
There are infintely many positive D such that Q( D) has
unique factorisation.
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) has
unique factorisation, namely -1, -2, -3, -7, -11, -19, -43, -67,
-163.
Gauss’ conjecture
√
There are infintely many positive D such that Q( D) has
unique 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.
√
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.
√
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 and
only if p is irregular.
√
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 and
only if p is irregular.
If p is regular then Fermat’s last theorem holds for p.
√
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 and
only 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 + ξp y) · · · (x + ξpp−1 y) = zp .
√
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 and
only 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 + ξ¯p ).
Kummer congruences
For all n ≡ m ≡ i (mod (p − 1)), we have, for all r > 0
Bm
Bn 1
m−1
n−1
1−p
< 1
n − m <
⇒
− 1−p
p
pr
m
n p pr+1
Kummer congruences
For all n ≡ m ≡ i (mod (p − 1)), we have, for all ε > 0,
Bm
Bn m−1
n−1
<ε
1−p
n − m < δ
− 1−p
⇒
p
m
n p
Kummer congruences
For all n ≡ m ≡ i (mod (p − 1)), we have, for all ε > 0,
Bm
Bn m−1
n−1
<ε
1−p
n − m < δ
− 1−p
⇒
p
m
n p
The function
Z → Qp
k 7→ 1 − pk−1
is p-adically continuous.
Bk
k
p-adic zeta-function
There is an analytic function ζp : Zp → Qp such that
1
ζp (−k) = 1 − k ζ(−k)
p
for all integers k > 0.
p-adic zeta-function
There is an analytic function ζp : Zp → Qp such that
1
ζp (−k) = 1 − k ζ(−k)
p
for all integers k > 0.
ζp (s) has a simple pole at s = 1 ∈ Zp .
p-adic zeta-function
There is an analytic function ζp : Zp → Qp such that
1
ζp (−k) = 1 − k ζ(−k)
p
for all integers k > 0.
ζp (s) has a simple pole at s = 1 ∈ Zp .
Conjecture
ζp (k) 6= 0 for k ∈ 2Z.
p-adic zeta-function
There is an analytic function ζp : Zp → Qp such that
1
ζp (−k) = 1 − k ζ(−k)
p
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 number
of Q(ξpn ).
We found that
x = a2 − b2
y = 2ab
z = a2 + b2
parametrises all solutions of x2 + y2 = z2 over Z.
We found that
x = a2 − b2
y = 2ab
z = a2 + b2
parametrises all solutions of x2 + y2 = z2 over Z. Or Q.
We found that
x = a2 − b2
y = 2ab
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 that
xy = 2n ?
We found that
x = a2 − b2
y = 2ab
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 that
xy = 2n ?
We are looking for triangles with rational sides, with a 90 degree
angle and area equal to n.
We found that
x = a2 − b2
y = 2ab
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 that
xy = 2n ?
We are looking for triangles with rational sides, with a 90 degree
angle and area equal to n. If there is one we say n is a
congruent number.
Examples : 6 is a congruent number as 32 + 42 = 52 and
3 · 4 = 2 · 6.
Examples : 6 is a congruent number as 32 + 42 = 52 and
3 · 4 = 2 · 6.
5
6
7
13
14
15
(20/3, 3/2, 41/6)
(4, 3, 5)
(24/5, 35/12, 337/60)
(323/30, 780/323, . . . )
(21/2, 8/3, 65/6)
(15/2, 4, 17/2)
21
22
23
29
30
31
(12, 7/2, 25/2)
(140/3, . . .
(41496/3485, . . .
(52780/99, . . .
(12, 5, 13)
(8897/360, . . .
Conjecture
If n ≡ 5, 6 or 7 (mod 8) then n is a congruent number.
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
Y 2 = X 3 − n2 X .
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
Y 2 = X 3 − n2 X .
L(E, s) =
1
Y
p-2n
1−
ap
ps
+
p
p2s
for s > 23 .
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
Y 2 = X 3 − n2 X .
L(E, s) =
1
Y
p-2n
1−
ap
ps
+
p
p2s
for s > 23 .
Conjecture (part of Birch and Swinnerton-Dyer conjecture)
n is congruent if and only if L(E, 1) = 0.
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
Y 2 = X 3 − n2 X .
L(E, s) =
1
Y
p-2n
1−
ap
ps
+
p
p2s
for s > 23 .
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 .
Riemann hypothesis
The non-trivial zeroes of ζ(s) lie on Re(s) = 12 .
Continuation of ζ :
s
ζ(s) =
−s·
s−1
Z
1
∞
x − [x]
dx
xs+1
for Re(s) > 0
Riemann hypothesis
The non-trivial zeroes of ζ(s) lie on Re(s) = 12 .
Continuation of ζ :
s
ζ(s) =
−s·
s−1
Z
1
∞
x − [x]
dx
xs+1
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 .
S OURCE : H T T P :// S E C A M L O C A L . E X . A C . U K / P E O P L E / S T A F F / M R W A T K I N / Z E T A / E N C O D I N G 1. H T M
Riemann hypothesis
P
1
The series ζ(s)
= n>1
µ(n)
ns
converges for all Re(s) > 12 .
Riemann hypothesis
P
1
The series ζ(s)
= n>1
µ(n)
ns
converges for all Re(s) > 12 .
Link to prime numbers
Z
log ζ(s) = s ·
2
∞
π(x) dx
·
xs − 1 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:
for some A and B in Q.
y2 = x 3 + A x + B
An elliptic curve is an equation of the form
E:
y2 = x 3 + A x + B
for some A and B in Q.
Question
Are there infinitely many rational solutions to E ?
An elliptic curve is an equation of the form
E:
y2 = x 3 + A x + B
for some A and B in Q.
Question
Are there infinitely many rational solutions to E ?
Examples :
Question
Are there infinitely many rational solutions to E ?
Examples :
E2 :
y2 = x3 + x + 2
Question
Are there infinitely many rational solutions to E ?
Examples :
E2 :
y2 = x3 + x + 2
has only three solutions (−1, 0), (1, −2), and (1, 2).
Question
Are there infinitely many rational solutions to E ?
Examples :
E1 :
y2 = x3 + x + 1
Question
Are there infinitely many rational solutions to E ?
Examples :
E1 :
y2 = x3 + x + 1
has infinitely many solutions. (0, ±1), ( 14 , ± 94 ), (72, ±611), . . .
Question
Are 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
287
− 1296
,
43992
26862913
82369 ,
1493284 ,
7549090222465
51865013741670864
8662944250944 ,
6504992707996225 ,
139455877527
3596697936
− 8760772801
,
1824793048 ,
173161424238594532415
− 310515636774481238884 , . . .
Let Np be the number of solutions of E modulo p plus 1.
Let Np be the number of solutions of E modulo p plus 1.
Consider the function
X
Np
f (X) =
log
p
p∈P, p6X
Let Np be the number of solutions of E modulo p plus 1.
Consider the function
X
Np
f (X) =
log
p
p∈P, p6X
Birch and Swinnerton-Dyer conjecture
f (X) stays bounded if and only if there are only finitely many
solutions in Q.
Let Np be the number of solutions of E modulo p plus 1.
Consider the function
X
Np
f (X) =
log
p
p∈P, p6X
Birch and Swinnerton-Dyer conjecture
f (X) stays bounded if and only if there are only finitely many
solutions 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 bound
√
The value of |ap | = |Np − p − 1| is bounded by 2 p.
Hasse-Weil bound
√
The value of |ap | = |Np − p − 1| is bounded by 2 p.
Theorem (R. Taylor), Sato-Tate conjecture
The value of
√
2
2
π 1 − t dt.
ap
√
2 p
is distributed in [−1, 1] like the measure
Hasse-Weil bound
√
The value of |ap | = |Np − p − 1| is bounded by 2 p.
Theorem (R. Taylor), Sato-Tate conjecture
The value of
√
2
2
π 1 − t dt.
ap
√
2 p
is distributed in [−1, 1] like the measure
Hasse-Weil bound
√
The value of |ap | = |Np − p − 1| is bounded by 2 p.
Theorem (R. Taylor), Sato-Tate conjecture
The value of
√
2
2
π 1 − t dt.
ap
√
2 p
is distributed in [−1, 1] like the measure
Hasse-Weil bound
√
The value of |ap | = |Np − p − 1| is bounded by 2 p.
Theorem (R. Taylor), Sato-Tate conjecture
The value of
√
2
2
π 1 − t dt.
ap
√
2 p
is distributed in [−1, 1] like the measure
Hasse-Weil bound
√
The value of |ap | = |Np − p − 1| is bounded by 2 p.
Theorem (R. Taylor), Sato-Tate conjecture
The value of
√
2
2
π 1 − t dt.
ap
√
2 p
is distributed in [−1, 1] like the measure