# A series of conjectures on $\sum_{x=0}^{(p-1)/2}(\frac{x^5+cx^3+dx}p)$ (I)

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 $$7 with $$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.