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

As in Question 319254, for an odd prime $$p$$ and integers $$c,d$$ we let $$S_p(c,d):=\sum_{x=0}^{(p-1)/2}\left(\frac{x^5+cx^3+dx}p\right).$$ If $$p\equiv1\pmod4$$, then we obviously have \begin{align}&\sum_{x=0}^{p-1}\left(\frac{x^5+cx^3+dx}p\right) \\=&\sum_{x=0}^{(p-1)/2}\left(\left(\frac xp\right)+\left(\frac{p-x}p\right)\right)\left(\frac{x^5+cx^3+dx}p\right)=2S_p(c,d). \end{align}

Conjecture 1. Let $$p\equiv1\pmod{12}$$ be a prime and write $$p=a^3+3b^2$$ with $$a,b\in\mathbb Z$$ and $$a\equiv1\pmod3$$. Suppose that $$d\in\mathbb Z$$ is a quadratic residue mod $$p$$, then $$S_p(10d,9d^2)=\begin{cases}-2a&\text{if}\ 3d\ \text{is a quartic residue mod}\ p,\\2a&\text{otherwise}.\end{cases}$$

Remark 1. In my preperint arXiv:1812.08080 joint with F. Petrov, D. Krachun vand Vsemirnov, we proved that $$\sum_{x=0}^{p-1}\left(\frac{x^5+10x^3y+9xy^2}p\right)=0\tag{*}$$ for any prime $$p\equiv5\pmod{12}$$ and integer $$y$$. By Conjecture 2 in Question 319254, for any prime $$p\equiv1\pmod{12}$$ and integer $$y$$ with $$(\frac yp)=-1$$, we should also have $$(*)$$.

Conjecture 2. Let $$p\equiv1,9\pmod{20}$$ be a prime and write $$p=a^2+5b^2$$ with $$a,b\in\mathbb Z$$. If $$d$$ is an integer with $$(\frac d p)\equiv 5^{(p-1)/4}\pmod p$$, then $$S_p(5d,5d^2)=\pm2a$$.

Remark 2. In the preperint arXiv:1812.08080, we proved that $$\sum_{x=0}^{p-1}\left(\frac{x^5+5x^3y+5xy^2}p\right)=0\tag{**}$$ for any prime $$p\equiv13,17\pmod{20}$$ and integer $$y$$. By Conjecture 2 in Question 319254, for any prime $$p\equiv1,9\pmod{20}$$ and integer $$y$$ with $$(\frac yp)\equiv-5^{(p-1)/4}\pmod p$$, we should also have $$(**)$$.

• By Conjecture 1 in Question 319254, for each integer $d$ and any prime $p>3$ with $p\equiv3\pmod{4}$ we should have $S_p(10d,9d^2)=0=S_p(5d,5d^2)$. – Zhi-Wei Sun Dec 22 '18 at 13:12