For any odd prime $p$, let $D_p$ denote the determinant $$\det\left[\left(\frac{i^2-\frac{p-1}2!\times j}p\right)\right]_{1\le i,j\le (p-1)/2},$$ where $(\frac{\cdot}p)$ is the Legendre symbol. Then \begin{gather*}D_3=0,\ D_5=-1,\ D_7=D_{11}=0,\ D_{13}=-8, \\ D_{17}=-72,\ D_{19}=D_{23}=0,\ D_{29}=-2061248.\end{gather*}

QUESTION: Let $p$ be an odd prime. Is it true that $D_p=0$ if and only if $p\equiv 3\pmod4$?

In 2013 I formulated this problem and conjectured that the answer is yes. I have verified this for all primes $p<2300$. By a result of L. Mordell [Amer. Math. Monthly 68(1965), 145-146], for any prime $p>3$ with $p\equiv3\pmod4$ we have $$\frac{p-1}2!\equiv(-1)^{(h(-p)+1)/2}\pmod p,$$ where $h(-p)$ is the class number of the imaginary quadratic field $\mathbb Q(\sqrt{-p})$.

Any ideas towards the solution?