## Jacobsthal identity for Q( √

## −2)

Yifan Yang (with Ki-Ichiro Hashimoto and Ling Long)

National Chiao Tung University, Taiwan

5 November 2010, NCKU Colloquium

## Jacobsthal’s identity

Theorem (Fermat)

An odd prime p is a sum of two integer squares if and only if p ≡ 1 mod 4.

Theorem (Jacobsthal)

Let p be a prime congruent to 1 modulo 4 and n be a quadratic nonresidue modulo p. Set

A = 1 2

p−1

X

x =0

x^{3}− x
p

, B = 1

2

p−1

X

x =0

x^{3}− nx
p

.

Then A, B ∈ Z and satisfies p = A^{2}+B^{2}.

## Jacobsthal’s identity

Theorem (Fermat)

An odd prime p is a sum of two integer squares if and only if p ≡ 1 mod 4.

Theorem (Jacobsthal)

Let p be a prime congruent to 1 modulo 4 and n be a quadratic nonresidue modulo p. Set

A = 1 2

p−1

X

x =0

x^{3}− x
p

, B = 1

2

p−1

X

x =0

x^{3}− nx
p

.

Then A, B ∈ Z and satisfies p = A^{2}+B^{2}.

## Legendre symbols

Definition

Let p be an odd prime. An integer a relatively prime to p is aquadratic residue(resp.quadratic nonresidue) modulo p if the congruence equation

x^{2}≡ a mod p
is solvable (resp. unsolvable) in integers.

Definition

Let p be an odd prime. Then theLegendre symbol

· p

is defined by

a p

=

0, if p|a,

1, if a is a quadratic residue modulo p,

−1, if a is a quadratic nonresidue modulo p.

## Legendre symbols

Definition

Let p be an odd prime. An integer a relatively prime to p is aquadratic residue(resp.quadratic nonresidue) modulo p if the congruence equation

x^{2}≡ a mod p
is solvable (resp. unsolvable) in integers.

Definition

Let p be an odd prime. Then theLegendre symbol

· p

is defined by

a p

=

0, if p|a,

1, if a is a quadratic residue modulo p,

−1, if a is a quadratic nonresidue modulo p.

## Properties of Legendre symbols

Definition

If f (x ) ∈ Z[x], then we call

J_{f}(p) :=

p−1

X

x =0

f (x) p

aJacobsthal sum.

Proposition We have

• ab p

= a p

b p

,

• a

≡ a^{(p−1)/2} mod p.

## Properties of Legendre symbols

Definition

If f (x ) ∈ Z[x], then we call

J_{f}(p) :=

p−1

X

x =0

f (x) p

aJacobsthal sum.

Proposition We have

• ab p

= a p

b p

,

• a p

≡ a^{(p−1)/2} mod p.

## Properties of Legendre symbols

Definition

If f (x ) ∈ Z[x], then we call

J_{f}(p) :=

p−1

X

x =0

f (x) p

aJacobsthal sum.

Proposition We have

• ab p

= a p

b p

,

• a

≡ a^{(p−1)/2} mod p.

## Gauss’ proof of the Jacobsthal identity

• Set S(n) =

p−1

X

x =0

x^{3}− nx
p

.

• Pairing the term x = a with the term x = p − a, we find S(n) is always even.

• Replacing x by rx , we find S(r^{2}n) = r
p

S(n).

• Let g be a primitive root modulo p. The above shows
S(g) = −S(g^{3}) =S(g^{5}) = −S(g^{7}) = . . . ,
S(g^{2}) = −S(g^{4}) =S(g^{6}) = −S(g^{8}) = . . . .

## Gauss’ proof of the Jacobsthal identity

• Set S(n) =

p−1

X

x =0

x^{3}− nx
p

.

• Pairing the term x = a with the term x = p − a, we find S(n) is always even.

• Replacing x by rx , we find S(r^{2}n) = r
p

S(n).

• Let g be a primitive root modulo p. The above shows
S(g) = −S(g^{3}) =S(g^{5}) = −S(g^{7}) = . . . ,
S(g^{2}) = −S(g^{4}) =S(g^{6}) = −S(g^{8}) = . . . .

## Gauss’ proof of the Jacobsthal identity

• Set S(n) =

p−1

X

x =0

x^{3}− nx
p

.

• Pairing the term x = a with the term x = p − a, we find S(n) is always even.

• Replacing x by rx , we find S(r^{2}n) = r
p

S(n).

• Let g be a primitive root modulo p. The above shows
S(g) = −S(g^{3}) =S(g^{5}) = −S(g^{7}) = . . . ,
S(g^{2}) = −S(g^{4}) =S(g^{6}) = −S(g^{8}) = . . . .

## Gauss’ proof of the Jacobsthal identity

• Set S(n) =

p−1

X

x =0

x^{3}− nx
p

.

• Pairing the term x = a with the term x = p − a, we find S(n) is always even.

• Replacing x by rx , we find S(r^{2}n) = r
p

S(n).

^{3}) =S(g^{5}) = −S(g^{7}) = . . . ,
S(g^{2}) = −S(g^{4}) =S(g^{6}) = −S(g^{8}) = . . . .

## Gauss’ proof of the Jacobsthal identity, continued

• Let S(g) = 2A and S(g^{2}) =2B. Then
2(p − 1)(A^{2}+B^{2}) =

p−1

X

n,x ,y =0

x^{3}− nx
p

y^{3}− ny
p

=

p−1

X

x ,y =0

xy p

^{p−1}
X

n=0

(x^{2}− n)(y^{2}− n)
p

.

• Using

p−1

X

z=0

z(z + r ) p

=

(p − 1, if r ≡ 0 mod p,

−1, if r 6≡ 0 mod p, we find

2(p − 1)(A^{2}+B^{2}) =p

p−1

X

x ,y =0

δ_{x}2,y^{2} =2(p − 1)p.

## Gauss’ proof of the Jacobsthal identity, continued

• Let S(g) = 2A and S(g^{2}) =2B. Then
2(p − 1)(A^{2}+B^{2}) =

p−1

X

n,x ,y =0

x^{3}− nx
p

y^{3}− ny
p

=

p−1

X

x ,y =0

xy p

^{p−1}
X

n=0

(x^{2}− n)(y^{2}− n)
p

.

• Using

p−1

X

z=0

z(z + r ) p

=

(p − 1, if r ≡ 0 mod p,

−1, if r 6≡ 0 mod p, we find

2(p − 1)(A^{2}+B^{2}) =p

p−1

X δ_{x}2,y^{2} =2(p − 1)p.

## Arithmetic-geometric approach

Idea.

Consider the elliptic curve En:y^{2}=x^{3}− nx. We have

#En(Fp) =1 +

p−1

X

x =0

1 + x^{3}− nx
p

=p + 1 +

p−1

X

x =0

x^{3}− nx
p

.

Thus,

L(E_{n}/Q, s)^{−1}=Y

p

1 +

p−1

X

x =0

x^{3}− nx
p

p^{−s}+p^{1−2s}

.

Since E_{1}and E_{n} are isomorphic over Q(√^{4}

n), the two L-functions
L(E_{1}/Q, s) and L(En/Q, s) must be related in some way, which give
information about the Jacobsthal sums.

## Arithmetic-geometric approach

Idea.

Consider the elliptic curve En:y^{2}=x^{3}− nx. We have

#En(Fp) =1 +

p−1

X

x =0

1 + x^{3}− nx
p

=p + 1 +

p−1

X

x =0

x^{3}− nx
p

.

Thus,

L(E_{n}/Q, s)^{−1}=Y

p

1 +

p−1

X

x =0

x^{3}− nx
p

p^{−s}+p^{1−2s}

.
Since E_{1}and E_{n} are isomorphic over Q(√^{4}

n), the two L-functions
L(E_{1}/Q, s) and L(En/Q, s) must be related in some way, which give

## Arithmetic-geometric approach

Idea.

Consider the elliptic curve En:y^{2}=x^{3}− nx. We have

#En(Fp) =1 +

p−1

X

x =0

1 + x^{3}− nx
p

=p + 1 +

p−1

X

x =0

x^{3}− nx
p

.

Thus,

L(E_{n}/Q, s)^{−1}=Y

p

1 +

p−1

X

x =0

x^{3}− nx
p

p^{−s}+p^{1−2s}

.

Since E_{1}and E_{n} are isomorphic over Q(√^{4}

n), the two L-functions
L(E_{1}/Q, s) and L(En/Q, s) must be related in some way, which give
information about the Jacobsthal sums.

## Tate modules and Galois representations

Let ` be a prime. Let E be an elliptic curve over a number field K and
E [`^{n}]be the subgroup of `^{n}-torsion points.

Consider the Tate module

T_{`}(E ) = lim

←−E [`^{n}].

The absolute Galois group G_{K} = Gal(Q/K ) acts on T`(E ), yielding a
Galois representation

ρE ,`:G_{K} → GL(2, Q`).

Then L(ρ_{E ,`},s) = L(E /K , s).

## Tate modules and Galois representations

Let ` be a prime. Let E be an elliptic curve over a number field K and
E [`^{n}]be the subgroup of `^{n}-torsion points.

Consider the Tate module

T_{`}(E ) = lim

←−E [`^{n}].

The absolute Galois group G_{K} = Gal(Q/K ) acts on T`(E ), yielding a
Galois representation

ρE ,`:G_{K} → GL(2, Q`).

Then L(ρ_{E ,`},s) = L(E /K , s).

## Tate modules and Galois representations

Let ` be a prime. Let E be an elliptic curve over a number field K and
E [`^{n}]be the subgroup of `^{n}-torsion points.

Consider the Tate module

T_{`}(E ) = lim

←−E [`^{n}].

The absolute Galois group G_{K} = Gal(Q/K ) acts on T`(E ), yielding a
Galois representation

ρE ,`:G_{K} → GL(2, Q`).

Then L(ρ_{E ,`},s) = L(E /K , s).

## Tate modules and Galois representations

^{n}]be the subgroup of `^{n}-torsion points.

Consider the Tate module

T_{`}(E ) = lim

←−E [`^{n}].

The absolute Galois group G_{K} = Gal(Q/K ) acts on T`(E ), yielding a
Galois representation

ρE ,`:G_{K} → GL(2, Q`).

Then L(ρ_{E ,`},s) = L(E /K , s).

## A lemma

Lemma (Clifford)

(Under suitable conditions on G and ρ) Assume that H G and G/H is cyclic of finite order.

Assume that ρ_{1}:G → GL(V_{1})and ρ_{2}:G → GL(V_{2})are irreducible
representations over an algebraically closed of characteristic not
dividing |G/H| such that ρ_{1}

Hand ρ_{2}

Hhave a common isomorphic irreducible subrepresentations of H.

Then

ρ_{1}' ρ_{2}⊗ χ

for some representation of G of degree 1 that is lifted from that of G/H.

## A lemma

Lemma (Clifford)

(Under suitable conditions on G and ρ) Assume that H G and G/H is cyclic of finite order.

Assume that ρ_{1}:G → GL(V_{1})and ρ_{2}:G → GL(V_{2})are irreducible
representations over an algebraically closed of characteristic not
dividing |G/H| such that ρ_{1}

Hand ρ_{2}

Hhave a common isomorphic irreducible subrepresentations of H.

Then

ρ_{1}' ρ_{2}⊗ χ

for some representation of G of degree 1 that is lifted from that of G/H.

## A lemma

Lemma (Clifford)

(Under suitable conditions on G and ρ) Assume that H G and G/H is cyclic of finite order.

Assume that ρ_{1}:G → GL(V_{1})and ρ_{2}:G → GL(V_{2})are irreducible
representations over an algebraically closed of characteristic not
dividing |G/H| such that ρ_{1}

Hand ρ_{2}

Hhave a common isomorphic irreducible subrepresentations of H.

Then

ρ_{1}' ρ_{2}⊗ χ

for some representation of G of degree 1 that is lifted from that of G/H.

## Arithmetic-geometric approach

Let E_{n}:y^{2}=x^{3}− nx. It is isomorphic to E1over Q(√^{4}

n), which is not abelian over Q.

Extend the base field to K = Q(i). Then L = Q(√^{4}

n, i) is cyclic over Q.

Let G_{K} = Gal(Q/K ) and GL= Gal(Q/L).

The elliptic curves Enhave CM by Z[i], so ρEn,`

GK = πn⊕ π_{n},
where π_{n}are representations of G_{K} of degree 1.

## Arithmetic-geometric approach

Let E_{n}:y^{2}=x^{3}− nx. It is isomorphic to E1over Q(√^{4}

n), which is not abelian over Q.

Extend the base field to K = Q(i). Then L = Q(√^{4}

n, i) is cyclic over Q.

Let G_{K} = Gal(Q/K ) and GL= Gal(Q/L).

The elliptic curves Enhave CM by Z[i], so ρEn,`

GK = πn⊕ π_{n},
where π_{n}are representations of G_{K} of degree 1.

## Arithmetic-geometric approach

Let E_{n}:y^{2}=x^{3}− nx. It is isomorphic to E1over Q(√^{4}

n), which is not abelian over Q.

Extend the base field to K = Q(i). Then L = Q(√^{4}

n, i) is cyclic over Q.

Let G_{K} = Gal(Q/K ) and GL= Gal(Q/L).

The elliptic curves Enhave CM by Z[i], so ρEn,`

GK = πn⊕ π_{n},
where π_{n}are representations of G_{K} of degree 1.

## Arithmetic-geometric approach

E_{1}/K and En/K are isomorphic over L, so
π_{1}

G_{L} ' π_{n}
G_{L}.
By the lemma above,

π_{n}= π_{1}⊗ χ

for some linear character χ on G_{K} with G_{L}⊂ kerχ, i.e., a character on
G_{K}/G_{L}' Gal(L/K ).

## Arithmetic-geometric approach

E_{1}/K and En/K are isomorphic over L, so
π_{1}

G_{L} ' π_{n}
G_{L}.
By the lemma above,

π_{n}= π_{1}⊗ χ

for some linear character χ on G_{K} with G_{L}⊂ kerχ, i.e., a character on
G_{K}/G_{L}' Gal(L/K ).

## Arithmetic-geometric approach

A character on G_{K} with G_{L}⊂ kerχ has the following description. The
Galois group Gal(L/K ) is generated by

σ :√^{4}

n 7−→ i√^{4}
n.

For each prime p of K not dividing 2n, the Frobenius Frob_{p} is the
element σ^{j} ∈ Gal(L/K ) such that

σ^{j}(√^{4}

n) ≡ (√^{4}

n)^{Np} mod p,
where Np denotes the norm of p.

Then there exists k ∈ {1, 3} such that χ satisfies
χ(Frob_{p}) =i^{jk}

## Arithmetic-geometric approach

A character on G_{K} with G_{L}⊂ kerχ has the following description. The
Galois group Gal(L/K ) is generated by

σ :√^{4}

n 7−→ i√^{4}
n.

For each prime p of K not dividing 2n, the Frobenius Frob_{p} is the
element σ^{j} ∈ Gal(L/K ) such that

σ^{j}(√^{4}

n) ≡ (√^{4}

n)^{Np} mod p,
where Np denotes the norm of p.

Then there exists k ∈ {1, 3} such that χ satisfies
χ(Frob_{p}) =i^{jk}
for all p.

## Arithmetic-geometric approach

A character on G_{K} with G_{L}⊂ kerχ has the following description. The
Galois group Gal(L/K ) is generated by

σ :√^{4}

n 7−→ i√^{4}
n.

For each prime p of K not dividing 2n, the Frobenius Frob_{p} is the
element σ^{j} ∈ Gal(L/K ) such that

σ^{j}(√^{4}

n) ≡ (√^{4}

n)^{Np} mod p,
where Np denotes the norm of p.

Then there exists k ∈ {1, 3} such that χ satisfies
χ(Frob_{p}) =i^{jk}

## Proof of Jacobsthal’s identity

Now for a prime p ≡ 1 mod 4, a prime of K lying over p has norm p.

If n is a quadratic nonresidue modulo p, then
n^{(p−1)/2} ≡ −1 mod p,
which implies that

(√^{4}

n)^{Np}≡ ±i√^{4}

n mod p.

That is,

χ(Frob_{p}) = ±i.

## Proof of Jacobsthal’s identity

Now for a prime p ≡ 1 mod 4, a prime of K lying over p has norm p.

If n is a quadratic nonresidue modulo p, then
n^{(p−1)/2} ≡ −1 mod p,
which implies that

(√^{4}

n)^{Np}≡ ±i√^{4}

n mod p.

That is,

χ(Frob_{p}) = ±i.

## Proof of Jacobsthal’s identity

Now for a prime p ≡ 1 mod 4, a prime of K lying over p has norm p.

If n is a quadratic nonresidue modulo p, then
n^{(p−1)/2} ≡ −1 mod p,
which implies that

(√^{4}

n)^{Np}≡ ±i√^{4}

n mod p.

That is,

χ(Frob_{p}) = ±i.

## Proof of Jacobsthal’s identity

It is well-known that
L(E_{1}/Q, s) = Y

p≡1 mod 4

1

1 − 2_{p}a_{p}p^{−s}+p^{1−2s}

Y

p≡3 mod 4

1
1 + p^{1−2s},
where for p ≡ 1 mod 4, apand bpare positive integers with ap odd
and b_{p} even such that p = a^{2}_{p}+b^{2}_{p}, and

p= −1 ap

(−1)^{b}^{p}^{/2}.

Thus, for a prime p of K = Q(i) lying over p ≡ 1 mod 4,
π_{1}(p) = ±ap± bpi

Then

π_{n}(p) = π_{1}(p)χ(p) = ±b_{p}± api

## Proof of Jacobsthal’s identity

It is well-known that
L(E_{1}/Q, s) = Y

p≡1 mod 4

1

1 − 2_{p}a_{p}p^{−s}+p^{1−2s}

Y

p≡3 mod 4

1
1 + p^{1−2s},
where for p ≡ 1 mod 4, apand bpare positive integers with ap odd
and b_{p} even such that p = a^{2}_{p}+b^{2}_{p}, and

p= −1 ap

(−1)^{b}^{p}^{/2}.

Thus, for a prime p of K = Q(i) lying over p ≡ 1 mod 4,
π_{1}(p) = ±ap± bpi

Then

π_{n}(p) = π_{1}(p)χ(p) = ±b_{p}± api

## Proof of Jacobsthal’s identity

It is well-known that
L(E_{1}/Q, s) = Y

p≡1 mod 4

1

1 − 2_{p}a_{p}p^{−s}+p^{1−2s}

Y

p≡3 mod 4

1
1 + p^{1−2s},
where for p ≡ 1 mod 4, apand bpare positive integers with ap odd
and b_{p} even such that p = a^{2}_{p}+b^{2}_{p}, and

p= −1 ap

(−1)^{b}^{p}^{/2}.

Thus, for a prime p of K = Q(i) lying over p ≡ 1 mod 4,
π_{1}(p) = ±ap± bpi

Then

π_{n}(p) = π_{1}(p)χ(p) = ±b_{p}± api

## Proof of Jacobsthal’s identity

Therefore, the p-factor of L(En,s) is

(1 ± 2b_{p}p^{−s}+p^{1−2s})^{−1}.
That is,

p−1

X

x =0

x^{3}− nx
p

= ±2bp, which gives us the Jacobsthal identity.

## Proof of Jacobsthal’s identity

Therefore, the p-factor of L(En,s) is

(1 ± 2b_{p}p^{−s}+p^{1−2s})^{−1}.
That is,

p−1

X

x =0

x^{3}− nx
p

= ±2bp, which gives us the Jacobsthal identity.

## Cubic analogue of the Jacobsthal identity

Theorem (Chan-Long-Y)

Let p ≡ 1 mod 6. Assume that n is an integer such that x^{3}≡ n
mod p is not solvable in integers. Set

A =

p−1

X

x =0

x^{3}− 1
p

, B =

p−1

X

x =0

x^{3}− n
p

.

Then

A^{2}+AB + B^{2}=3p.

## Cubic analogue of the Jacobsthal identity

Theorem (Chan-Long-Y)

Let p ≡ 1 mod 6. Assume that n is an integer such that x^{3}≡ n
mod p is not solvable in integers. Set

A =

p−1

X

x =0

x^{3}− 1
p

, B =

p−1

X

x =0

x^{3}− n
p

.

Then

A^{2}+AB + B^{2}=3p.

## Question

Let −d be the discriminant of an imaginary quadratic number field such that Q(√

−d ) has class number 1.

Let

f (x , y ) =

(x^{2}+ (d /4)y^{2}, if d ≡ 0 mod 4,
x^{2}+xy + ((1 + d )/4)y^{2}, if d ≡ 3 mod 4.

Then whether p = f (x , y ) is solvable depends only on

−d p

. Question. When

−d p

=1, can we express the integers A and B in p = f (A, B) in terms of Jacobsthal sums in a uniform way?

## Question

Let −d be the discriminant of an imaginary quadratic number field such that Q(√

−d ) has class number 1.

Let

f (x , y ) =

(x^{2}+ (d /4)y^{2}, if d ≡ 0 mod 4,
x^{2}+xy + ((1 + d )/4)y^{2}, if d ≡ 3 mod 4.

Then whether p = f (x , y ) is solvable depends only on

−d p

. Question. When

−d p

=1, can we express the integers A and B in p = f (A, B) in terms of Jacobsthal sums in a uniform way?

## Question

Let −d be the discriminant of an imaginary quadratic number field such that Q(√

−d ) has class number 1.

Let

f (x , y ) =

(x^{2}+ (d /4)y^{2}, if d ≡ 0 mod 4,
x^{2}+xy + ((1 + d )/4)y^{2}, if d ≡ 3 mod 4.

Then whether p = f (x , y ) is solvable depends only on

−d p

. Question. When

−d p

=1, can we express the integers A and B in p = f (A, B) in terms of Jacobsthal sums in a uniform way?

## Jacobsthal identity for Q( √

## −2), Part I

Theorem (Hashimoto-Long-Y)

Assume that p ≡ 1 mod 8 and n is a quadratic nonresidue modulo p.

Set

A = 1 2

p−1

X

x =0

x^{3}+4x^{2}+2x
p

, B = 1

4

p−1

X

x =0

x^{5}+nx
p

.

Then A and B are integers satisfying p = A^{2}+2B^{2}.

## Jacobsthal identity for Q( √

## −d ), Part II

Theorem (Hashimoto-Long-Y) Assume that p ≡ 3 mod 8. Set

A = 1 2

p−1

X

x =0

x^{3}+4x^{2}+2x
p

,

B = 1 4

1 +

p−1

X

x =0

x^{6}+4x^{5}+10x^{4}− 20x^{2}− 16x − 8
p

.
Then A and B are integers satisfying p = A^{2}+2B^{2}.

## The elliptic curve y

^{2}

## = x

^{3}

## + 4x

^{2}

## + 2x

Lemma

The elliptic curve y^{2}=x^{3}+4x^{2}+2x has CM by Z[√

−2] and its L-function is

Y

p≡1,3 mod 8

1

1 − 2papp^{−s}+p^{1−2s}

Y

p≡5,7 mod 8

1
1 + p^{1−2s},
where a_{p}and b_{p}are positive integers such that p = a^{2}_{p}+2b^{2}_{p}and

_{p} =

2(−1)^{b}^{p}^{/2}

−2 ap

, if p ≡ 1 mod 8,

−2

−2 ap

, if p ≡ 3 mod 8.

## The hyperelliptic curve y

^{2}

## = x

^{5}

## + x

Lemma

For C : y^{2}=x^{5}+x , we have

L(C/Q, s) = L(E1/Q, s)L(E2/Q, s), (1)
where E_{1}:y^{2}=x^{3}+4x^{2}+2x , E_{2}:y^{2}=x^{3}− 4x^{2}+2x .

Proof.

There are 2-to-1 coverings

(x , y ) 7−→ (X , Y ) = (x ± 1)^{2}

x ,y (x ± 1)
x^{2}

from C to E_{1}and E_{2}. Considering the associated Galois
representations, we get (1).

## The hyperelliptic curve y

^{2}

## = x

^{5}

## + x

Lemma

For C : y^{2}=x^{5}+x , we have

L(C/Q, s) = L(E1/Q, s)L(E2/Q, s), (1)
where E_{1}:y^{2}=x^{3}+4x^{2}+2x , E_{2}:y^{2}=x^{3}− 4x^{2}+2x .

Proof.

There are 2-to-1 coverings

(x , y ) 7−→ (X , Y ) = (x ± 1)^{2}

x ,y (x ± 1)
x^{2}

from C to E_{1}and E_{2}. Considering the associated Galois
representations, we get (1).

## L-function of y

^{2}

## = x

^{5}

## + x

Corollary

For C : y^{2}=x^{5}+x , let

1

(1 − α_{p,1}p^{−s}) . . . (1 − α_{p,4}p^{−s})
be the p-factor of L(C/Q, s).

• If p ≡ 1 mod 8, then
α_{p,j} = −2

a

(−1)^{b/2}(a ± b√

−2),

each with multiplicity 2, where a and b are the positive integers
such that p = a^{2}+2b^{2}.

• If p ≡ 3 mod 8, then α_{p,j} = ±a ± b√

−2, where a and b are
integers such that p = a^{2}+2b^{2}.

√ √

## L-function of y

^{2}

## = x

^{5}

## + x

Corollary

For C : y^{2}=x^{5}+x , let

1

(1 − α_{p,1}p^{−s}) . . . (1 − α_{p,4}p^{−s})
be the p-factor of L(C/Q, s).

• If p ≡ 1 mod 8, then
α_{p,j} = −2

a

(−1)^{b/2}(a ± b√

−2),

each with multiplicity 2, where a and b are the positive integers
such that p = a^{2}+2b^{2}.

• If p ≡ 3 mod 8, then α_{p,j} = ±a ± b√

−2, where a and b are

## The curve y

^{2}

## = x

^{6}

## + 4x

^{5}

## + 10x

^{4}

## − 20x

^{2}

## − 16x − 8

Lemma

The hyperelliptic curve X_{1}:y^{2}=x^{6}+4x^{5}+10x^{4}− 20x^{2}− 16x − 8 is
isomorphic to X_{2}:y^{2}=x^{5}+x over a field of degree 16 over Q, which
is cyclic of degree 4 over Q(ζ8).

Proof.

Setting

x =

√

2(x_{1}+1)

x_{1}− 1 , y = y_{1}
(x_{1}− 1)^{3},
we get y_{1}^{2}=128(2 +√

2)x_{1}(x_{1}^{4}+3 − 2√
2).

The proof of the theorem follows the argument in the case of the classical Jacobsthal identity (although more complicated).

## The curve y

^{2}

## = x

^{6}

## + 4x

^{5}

## + 10x

^{4}

## − 20x

^{2}

## − 16x − 8

Lemma

The hyperelliptic curve X_{1}:y^{2}=x^{6}+4x^{5}+10x^{4}− 20x^{2}− 16x − 8 is
isomorphic to X_{2}:y^{2}=x^{5}+x over a field of degree 16 over Q, which
is cyclic of degree 4 over Q(ζ8).

Proof.

Setting

x =

√

2(x_{1}+1)

x_{1}− 1 , y = y_{1}
(x_{1}− 1)^{3},
we get y_{1}^{2}=128(2 +√

2)x_{1}(x_{1}^{4}+3 − 2√
2).

The proof of the theorem follows the argument in the case of the

## The curve y

^{2}

## = x

^{6}

## + 4x

^{5}

## + 10x

^{4}

## − 20x

^{2}

## − 16x − 8

Lemma

The hyperelliptic curve X_{1}:y^{2}=x^{6}+4x^{5}+10x^{4}− 20x^{2}− 16x − 8 is
isomorphic to X_{2}:y^{2}=x^{5}+x over a field of degree 16 over Q, which
is cyclic of degree 4 over Q(ζ8).

Proof.

Setting

x =

√

2(x_{1}+1)

x_{1}− 1 , y = y_{1}
(x_{1}− 1)^{3},
we get y_{1}^{2}=128(2 +√

2)x_{1}(x_{1}^{4}+3 − 2√
2).

The proof of the theorem follows the argument in the case of the classical Jacobsthal identity (although more complicated).

## How do we find the curves?

Assume p ≡ 1, 3 mod 8 and p = a^{2}+2b^{2}= (a + b√

−2)(a − b√

−2).

If C : y^{2}=f (x ) is a curve such that the p-factor of L(C/Q, s) is
1

(1 ± (a + b√

−2)p^{−s})(1 ± (a − b√

−2)p^{−s}),
then

p−1

X

x =0

f (x) p

= ±2a.

Thus, we are looking at elliptic curves with CM by Z[√

−2].

## How do we find the curves?

Assume p ≡ 1, 3 mod 8 and p = a^{2}+2b^{2}= (a + b√

−2)(a − b√

−2).

If C : y^{2}=f (x ) is a curve such that the p-factor of L(C/Q, s) is
1

(1 ± (a + b√

−2)p^{−s})(1 ± (a − b√

−2)p^{−s}),
then

p−1

X

x =0

f (x) p

= ±2a.

Thus, we are looking at elliptic curves with CM by Z[√

−2].

## How do we find the curves?

Assume p ≡ 1, 3 mod 8 and p = a^{2}+2b^{2}= (a + b√

−2)(a − b√

−2).

If C : y^{2}=f (x ) is a curve such that the p-factor of L(C/Q, s) is
1

(1 ± (a + b√

−2)p^{−s})(1 ± (a − b√

−2)p^{−s}),
then

p−1

X

x =0

f (x) p

= ±2a.

Thus, we are looking at elliptic curves with CM by Z[√

−2].

## How do we find the curves?

To get b, we observe that (a + b√

−2)(ζ8+ ζ_{8}^{3}) + (a − b√

−2)(ζ_{8}^{5}+ ζ_{8}^{7}) = −4b.

Thus, we are looking for curves y^{2}=f (x ) whose L-function has
p-factor

1
(1 ± ζ_{8}(a + b√

−2)p^{−s}) . . . (1 ± ζ_{8}^{7}(a − b√

−2)p^{−s}),
i.e., a hyperelliptic curve of genus 2.

To find such a curve, we shall find “its L-function” first.

## How do we find the curves?

To get b, we observe that (a + b√

−2)(ζ8+ ζ_{8}^{3}) + (a − b√

−2)(ζ_{8}^{5}+ ζ_{8}^{7}) = −4b.

Thus, we are looking for curves y^{2}=f (x ) whose L-function has
p-factor

1
(1 ± ζ_{8}(a + b√

−2)p^{−s}) . . . (1 ± ζ_{8}^{7}(a − b√

−2)p^{−s}),
i.e., a hyperelliptic curve of genus 2.

To find such a curve, we shall find “its L-function” first.

## How do we find the curves?

To get b, we observe that (a + b√

−2)(ζ8+ ζ_{8}^{3}) + (a − b√

−2)(ζ_{8}^{5}+ ζ_{8}^{7}) = −4b.

Thus, we are looking for curves y^{2}=f (x ) whose L-function has
p-factor

1
(1 ± ζ_{8}(a + b√

−2)p^{−s}) . . . (1 ± ζ_{8}^{7}(a − b√

−2)p^{−s}),
i.e., a hyperelliptic curve of genus 2.

To find such a curve, we shall find “its L-function” first.

## Hecke characters

Let K be a number field. For each place v , let Kv be the completion of K with respect to | · |v and Ov be the valuation ring of Kv when v is a finite place.

Let

IK = (

(x_{v}) ∈Y

v

K_{v}^{∗} :x_{v} ∈ O^{∗}_{v} for all but finitely many v
)

be the idele group of K , equipped with the product topology.

Definition

AHecke character (Grössencharakter)χis a continuous group
homomorphism from the idele class group IK/K^{∗} to C^{∗}.

## Hecke characters

Let K be a number field. For each place v , let Kv be the completion of K with respect to | · |v and Ov be the valuation ring of Kv when v is a finite place.

Let

IK = (

(x_{v}) ∈Y

v

K_{v}^{∗} :x_{v} ∈ O^{∗}_{v} for all but finitely many v
)

be the idele group of K , equipped with the product topology.

Definition

AHecke character (Grössencharakter)χis a continuous group
homomorphism from the idele class group IK/K^{∗} to C^{∗}.

## Hecke characters

Let K be a number field. For each place v , let Kv be the completion of K with respect to | · |v and Ov be the valuation ring of Kv when v is a finite place.

Let

IK = (

(x_{v}) ∈Y

v

K_{v}^{∗} :x_{v} ∈ O^{∗}_{v} for all but finitely many v
)

be the idele group of K , equipped with the product topology.

Definition

AHecke character (Grössencharakter)χis a continuous group
homomorphism from the idele class group IK/K^{∗} to C^{∗}.

## Hecke L-functions and their functional equations

Definition

Let χ be a Hecke character. Write χ =Q

vχ_{v}. TheHecke L-functionis
defined by

L(s, χ) = Y

v finite, χv(O^{∗}v)=1

1

1 − χv(πv)Nv^{−s},

where πv is any uniformizer of Kv and Nv is the norm of the prime ideal corresponding to v .

## Hecke L-functions and their functional equations

Proposition

Let K be an imaginary quadratic number field. Suppose that k is the
positive integer such that |χ(x )| = |x |^{k −1}for all x ∈ IK/K^{∗}. Setting

Λ(s, χ) = 2π
pd_{K}d_{χ}

!−s

Γ(s)L(s, χ),

we have

Λ(s, χ) = Λ(k − s, χ)

for some root of unity , where d_{K} is the discriminant of K and d_{χ}is the
norm of the modulus of χ.

Remark

We getCM modular formsfrom Hecke characters on imaginary

## Hecke L-functions and their functional equations

Proposition

Let K be an imaginary quadratic number field. Suppose that k is the
positive integer such that |χ(x )| = |x |^{k −1}for all x ∈ IK/K^{∗}. Setting

Λ(s, χ) = 2π
pd_{K}d_{χ}

!−s

Γ(s)L(s, χ),

we have

Λ(s, χ) = Λ(k − s, χ)

for some root of unity , where d_{K} is the discriminant of K and d_{χ}is the
norm of the modulus of χ.

Remark

We getCM modular formsfrom Hecke characters on imaginary quadratic number field.

## Finding curves

• Let K = Q(√

−2). We construct Hecke characters χ1and χ_{2}of
modulus 8 so that χ takes value ζ_{8}^{j}(a + b√

−2).

• We then look for a hyperelliptic curve whose L-function coincide
with L(s, χ_{1})L(s, χ_{2}). Specifically, we look for such a curve among
hyperelliptic curves with an automorphism defined over Q(√

−2).

• In practice, we consider curves

y^{2}=x^{6}+mx^{5}+nx^{4}− 2nx^{2}− 4mx − 8,
which has an automorphism

(x , y ) 7−→ 2 x,

√−8y
x^{3}

,

## Finding curves

• Let K = Q(√

−2). We construct Hecke characters χ1and χ_{2}of
modulus 8 so that χ takes value ζ_{8}^{j}(a + b√

−2).

• We then look for a hyperelliptic curve whose L-function coincide
with L(s, χ_{1})L(s, χ_{2}). Specifically, we look for such a curve among
hyperelliptic curves with an automorphism defined over Q(√

−2).

• In practice, we consider curves

y^{2}=x^{6}+mx^{5}+nx^{4}− 2nx^{2}− 4mx − 8,
which has an automorphism

(x , y ) 7−→ 2 x,

√−8y
x^{3}

,

and search for m and n such that the L-function is L(s, χ_{1})L(s, χ_{2}).

## Finding curves

• Let K = Q(√

−2). We construct Hecke characters χ1and χ_{2}of
modulus 8 so that χ takes value ζ_{8}^{j}(a + b√

−2).

• We then look for a hyperelliptic curve whose L-function coincide
with L(s, χ_{1})L(s, χ_{2}). Specifically, we look for such a curve among
hyperelliptic curves with an automorphism defined over Q(√

−2).

• In practice, we consider curves

y^{2}=x^{6}+mx^{5}+nx^{4}− 2nx^{2}− 4mx − 8,
which has an automorphism

(x , y ) 7−→ 2 x,

√−8y
x^{3}

,

## Problem

Problem. For each imaginary quadratic number field K with class number 1, find an analogous identity.