As in Question 319254, for an odd prime $p$ and integers $c,d$ we define $$S_p(c,d):=\sum_{x=0}^{(p-1)/2}\left(\frac{x^5+cx^3+dx}p\right),$$ where $(\frac{\cdot}p)$ is the Legendre symbol.

In my paper arXiv:1308.2900, I conjectured that the determiannt $$[8,18]_p:=\det\left[\left(\frac{i^2+8ij+18j^2}p\right)\right]_{0\le i,j\le p-1}$$ vanishes for any prime $p\equiv 13,17\pmod{24}$. This is still open, and it is implied by part (i) of my following conjecture.

Conjecture. (i) We have $S_p(8d,18d^2)=0$ for any prime $p\equiv13,17\pmod{24}$ and integer $d$.

(ii) Suppose that $p\equiv1\pmod{24}$ is a prime and $p=a^2+6b^2$ with $a,b\in\mathbb Z$ and $a\equiv1\pmod 3$. Then, for any $d\in\mathbb Z$ we have $$S_p(8d,18d^2)=\begin{cases}-2a&\text{if}\ 2^{(p-1)/8}\equiv (3d)^{(p-1)/4}\pmod p,\\2a&\text{if}\ 2^{(p-1)/8}\equiv-(3d)^{(p-1)/4}\pmod p \\0&\text{if}\ 2^{(p-1)/4}\not\equiv(\frac dp)\pmod p.\end{cases}$$

(iii) If $p\equiv5\pmod{24}$ and $p=2a^2+3b^2$ with $a,b\in\mathbb Z$, then $S_p(8d,18d^2)=\pm2a$ for any integer $d\not\equiv0\pmod p$.

Remark. Conjecture 1 in Question 319254 implies that $S_p(8d,18d^2)\not=0$ for each integer $d$ and any prime $p\equiv7\pmod 8$.


1 Answer 1


Here is a proof of (i):

Since the relevant primes $p$ are $\equiv 1 \bmod 4$, we have $S_p(c,d) = \frac{1}{2} T_p(c,d)$, where $$ T_p(c,d) = \sum_{x=0}^{p-1} \left(\frac{x^5+cx^3+dx}{p}\right) \,. $$ Note that $T_p(c,d)$ is, up to a sign, the trace of Frobenius at $p$ for the Jacobian variety of the curve $$ C_{c,d} \colon y^2 = x^5 + cx^3 + dx \,. $$

The curves $C_{8d,18d^2}$ are over $\bar{\mathbb Q}$ all isomorphic to $$ C \colon y^2 = x^5 + \frac{4}{3}\sqrt{2} x^3 + x \,,$$ which has 2-to-1 maps to two elliptic curves $E$ and $E'$ with complex multiplication by ${\mathbb Z}[\sqrt{-6}]$. Let $J_d$ be the Jacobian of $C_{8d,18d^2}$; then $J_d$ has a $(2,2)$-isogeny $\phi$ to $E \times E'$. Now consider the endomorphism $\alpha$ of $J_d$ given as the composition $$ J_d \stackrel{\phi}{\longrightarrow} E \times E' \stackrel{(\sqrt{-6},\sqrt{-6})}{\longrightarrow} E \times E' \stackrel{\hat{\phi}}{\longrightarrow} J_d \,, $$ where $\hat{\phi}$ is the isogeny dual to $\phi$. Then $\alpha$ acts as scalar multiplication by $2\sqrt{-6}$ on the cotangent space of $J_d$ at the origin (because $\hat{\phi} \circ \phi$ is multiplication by 2). Since the cotangent space has a basis defined over $\mathbb Q$ (coming from the differentials $\frac{dx}{y}, \frac{x\,dx}{y}$ on the curve), the absolute Galois group of $\mathbb Q$ acts on $\alpha$ via the quadratic character of $-6$.

If $p \equiv 13$ or $17 \bmod 24$, then $\left(\frac{-6}{p}\right) = -1$, hence the Frobenius at $p$ conjugates $\alpha$ to its negative when both are acting on the $\ell$-adic Tate module $T_\ell J_d$. Since $\alpha$ is non-degenerate (because $\alpha^2$ is multiplication by $-24$), this implies that the trace of Frobenius, which is $-T_p(8d, 18d^2) = -2 S_p(8d, 18d^2)$, vanishes.

A similar argument applies to other pairs $(c,d)$ with the property that the elliptic curves that are covered by $C_{c,d}$ have CM. For example, $$ \sum_{x=0}^{p-1} \left(\frac{x^5 + 5365 d x^3 + 7195797 d^2 x}{p}\right) = 0 $$ whenever $$ p \equiv 5, 13, 17, 29, 45, 57, 61, 69, 89, 93, 97, 105, 109, 113, 117, 125, 129, 133 \bmod 148. $$

Regarding (ii), when $2$ is a square mod $p$, then $x \leftarrow 3 \sqrt{2} d/x$ induces an isomorphism of $C_{8d,18d^2}$ with its quadratic twist by $3d\sqrt{2}$. If $3d\sqrt{2}$ is not a square mod $p$, then the trace must be zero, since the trace for the nontrivial quadratic twist over $\mathbb F_p$ is the negative of the original trace. This gives the last case in (ii).

If $3d\sqrt{2}$ is a square mod $p$, then the two maps to elliptic curves are defined over $\mathbb F_p$, and the trace will be the sum of the traces of the two elliptic curves. For CM elliptic curves, the traces can be given in terms of Hecke characters; this results in expressions like in the first two cases in (ii). It should not be too hard to work out what the elliptic curves are in this case and thus verify Conjecture (ii).

  • $\begingroup$ Great! Thank you for proving part (i) of my conjecture! $\endgroup$ Dec 24, 2018 at 11:27

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