On the determinants $\det\left[(i\pm j)\left(\frac{i\pm j}p\right)\right]_{1\le i,j\le(p-1)/2}$

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.

(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!