Let $p$ be an odd prime, and let $(\frac{\cdot}p)$ be the Legendre symbol. We define the determinant $D(d,p)$ by $$D(d,p):=\det\left[(i^2+dj^2)\left(\frac{i^2+dj^2}p\right)\right]_{1\le i,j\le(p-1)/2}.$$

I have the following conjecture.

**Conjecture.** Let $p$ be an odd prime, and let $\delta_p$ be $0$ or $1$ according as $p=3$ or not. For any integer $d\not\equiv 0\pmod p$, we have
$$\left(\frac{D(d,p)}p\right)=\begin{cases}(\frac dp)^{(p-1)/4}&\text{if}\ p\equiv1\pmod4,\\(\frac dp)^{\delta_p(p+1)/4}(-1)^{(h(-p)-1)/2}&\text{if}\ p\equiv3\pmod4,\end{cases}$$
where $h(-p)$ denotes the calss number of the imaginary quadratic field $\mathbb Q(\sqrt{-p})$.

*Remark*. We have verified the conjecture for all odd primes $p<600$. It is easy to see that $(\frac{D(d,p)}p)$ only depends on the values of $p$ and $(\frac dp)$. For the determinant
$$S(d,p):=\det\left[\left(\frac{i^2+dj^2}p\right)\right]_{1\le i,j\le(p-1)/2},$$ the value $(\frac{S(d,p)}p)$ was determined in the paper arXiv:1308.2900. However, the method there does not work for the present conjecture.