Let $p$ be an odd prime and define $$D_p^+:=\det\left[(i+j)\left(\frac{i+j}p\right)\right]_{1\le i,j\le(p-1)/2}$$ and $$D_p^{-}:=\det\left[(i-j)\left(\frac{i-j}p\right)\right]_{1\le i,j\le(p-1)/2},$$ where $(\frac{\cdot}p)$ is the Legendre symbol.

QUESTION. Is my following conjecture true? How to solve it?

Conjecture. Let $p>5$ be a prime. If $p\equiv1\pmod4$, then $$\left(\frac{D_p^+}p\right)=1=\left(\frac{D_p^-}p\right).$$ If $p\equiv3\pmod4$, then $p\nmid D_p^+D_p^-$. Moreover, when $p\equiv7\pmod8$ we have $$\left(\frac{D_p^-}p\right)=(-1)^{(h(-p)-1)/2},$$ where $h(-p)$ denotes the class number of the imaginary quadrtic field $\mathbb Q(\sqrt{-p})$.

Remark. (i) I have verified the conjecture for all primes $5<p<2000$.

(ii) For any prime $p\equiv1\pmod4$, clearly $D_p^-$ is a skew-symmetric determinant and hence it is an integer square by a result of Cayley. But I'm unable to show that $p\nmid D_p^-$ for all primes $p>5$ with $p\equiv 1\pmod4$.

In my opinion, the conjecture looks not so difficult. Your comments towards its solution are welcome!


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