Let $p$ be an odd prime, and let $(\frac{\cdot}p)$ denote the Legendre symbol. Motivated by the evaluation of the determinants $$\det\left[\left(\frac{j+k}p\right)\right]_{1\le j,k\le(p-1)/2}\ \ \text{and}\ \ \det\left[\left(\frac{j-k}p\right)\right]_{1\le j,k\le(p-1)/2}$$ (cf. Question 470324), I have formulated the following conjecture based on my computation via Mathematica.
Conjecture. Let $p>3$ be a prime.
(i) We have $$\begin{aligned}&\ \left|\left(\frac{j+k}p\right)+\left(\frac{j-k}p\right)\right|_{1\le j,k\le (p-1)/2} \\=&\ \begin{cases}(\frac 2p)p^{(p-5)/4}&\text{if}\ p\equiv1\pmod4, \\(-1)^{(h(-p)-1)/2}p^{(p-3)/4}&\text{if}\ p\equiv3\pmod4, \end{cases} \end{aligned}\tag{1}$$ where $h(-p)$ denotes the class number of the imaginary quadratic field $\mathbb Q(\sqrt{-p})$. When $p\equiv1\pmod4$, we also have $$\left|\left(\frac{j+k}p\right)-\left(\frac{j-k}p\right)\right|_{1\le j,k\le (p-1)/2} =(-p)^{(p-1)/4}.\tag{2} $$
(ii) We have $$\begin{aligned}&\ \det\left[x+\left(\frac{j+k}p\right)+\left(\frac{j-k}p\right)\right]_{0\le j,k\le(p-1)/2} \\ =&\ \begin{cases}(\frac 2p)p^{(p+3)/4}x&\text{if}\ p\equiv1\pmod4, \\(-1)^{(h(-p)-1)/2}p^{(p+1)/4}x&\text{if}\ p\equiv3\pmod4. \end{cases}\end{aligned}\tag{3}$$
QUESTION. What method or tool can be used to study the above problem? Are the quadratic Gauss sums helpful?
Your comments are welcome!
