# On the determinant $\det[(i^2+dj^2)(\frac{i^2+dj^2}p)]_{1\le i,j\le(p-1)/2}$ with $p$ an odd prime

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.