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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 $(**)$.

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

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