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Let $p$ be an odd prime. Here I introduce the sum $$S_p(c,d):=\sum_{x=0}^{(p-1)/2}\left(\frac{x^5+cx^3+dx}p\right)$$ with $c,d\in\mathbb Z$, where $(\frac{\cdot}p)$ is the Legendre symbol.

I have a series of conjectures on such sums. Here I state two general ones.

Conjecture 1. Let $p>7$ be a prime with $p\equiv 3\pmod4$, and let $c,d\in\mathbb Z$. If $S_p(c,d)=0$, then $$\left(\frac{c^2-4d}p\right)=1\not=\left(\frac dp\right).$$

Remark 1. I have verified this for all primes $7<p<1000$ with $p\equiv3\pmod4$. Note that $S_3(1,1)=0$ but $(\frac{1^2-4\cdot1}3)=0$ and $(\frac 1p)=1$, and that $S_7(5,1)=S_7(6,2)=0$ but $(\frac{5^2-4\cdot1}7)=(\frac{6^2-4\cdot2}7)=0$ and $(\frac 17)=(\frac 27)=1$.

Conjecture 2. Let $p\equiv 1\pmod4$ be a prime, and let $c,d\in\mathbb Z$.

(i) If $d^{(p-1)/4}\equiv-1\pmod p$ (i.e., $d$ is a quadratic residue and a quartic nonresidue mod $p$), then $S_p(c,d)=0$.

(ii) If $S_p(c,d)=0$ but $d^{(p-1)/4}\not\equiv-1\pmod p$, then $$\left(\frac{c^2-4d}p\right)=\left(\frac dp\right).$$

Remark 2. I have verified Conjecture 2 for all primes $p<500$ with $p\equiv1\pmod4$.

I'll pose more conjectures later.

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