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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 $$S_p(8d,18d^2)=\begin{cases}-2a&\text{if}\ 2^{(p-1)/8}3^{(p-1)/4}\equiv d^{(p-1)/4}\pmod p,\\2a&\text{if}\ 2^{(p-1)/8}3^{(p-1)/4}\equiv-d^{(p-1)/4}\pmod p \\0&\text{if}\ 2^{(p-1)/4}\not\equiv(\frac dp)\pmod p.\end{cases}$$

(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 $p\equiv7\pmod 8$.

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 $p\equiv7\pmod 8$.

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 $p\equiv7\pmod 8$.

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 $$S_p(8d,18d^2)=\begin{cases}-2a&\text{if}\ 2^{(p-1)/8}3^{(p-1)/4}\equiv d^{(p-1)/4}\pmod p,\\2a&\text{if}\ 2^{(p-1)/8}3^{(p-1)/4}\equiv-d^{(p-1)/4}\pmod p \\0&\text{if}\ 2^{(p-1)/4}\not\equiv(\frac dp)\pmod p.\end{cases}$$

(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 $p\equiv7\pmod 8$.

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 $$S_p(8d,18d^2)=\begin{cases}\pm2a&\text{if}\ (\frac d p)\equiv 2^{(p-1)/4}\pmod p, \\0&\text{otherwise}.\end{cases}$$ Moreover, when $2^{(p-1)/4}\equiv1\pmod p$ we have $$S_p(8,18)=\begin{cases}-2a&\text{if}\ 2^{(p-1)/8}\equiv 3^{(p-1)/4}\pmod p,\\2a&\text{if}\ 2^{(p-1)/8}\equiv-3^{(p-1)/4}\pmod p.\end{cases}$$ (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 $p\equiv7\pmod 8$.

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 $p\equiv7\pmod 8$.