Let $p$ be an odd prime and let $S_p$ denote the determinant $$\det\left[\left(\frac{i^2+j^2}p\right)\right]_{1\le i,j\le (p-1)/2}$$ with $(\frac{\cdot}p)$ the Legendre symbol. By Theorem 1.2 of my paper arXiv:1308.2900 available from http://arxiv.org/abs/1308.2900, $-S_p$ is a quadratic residue modulo $p$. Here I ask a further question.

QUESTION. Is it true that for each prime $p\equiv3\pmod4$ the number $-S_p$ is always a positive square divisible by $2^{(p-3)/2}$?

Define $a_p=\sqrt{-S_p}/2^{(p-3)/4}$ for any prime $p\equiv3\pmod4$. Then \begin{gather*}a_3=a_7=a_{11}=1,\ a_{19}=2,\ a_{23}=1,\ a_{31}=29,\ a_{43}=254, \\a_{47}=367,\ a_{59}=9743,\ a_{67}=305092,\ a_{71}=29,\ a_{79}=1916927. \end{gather*} I have computed the values of $a_p$ for all primes $p\equiv3\pmod4$ with $p<2000$. Based on the numerical data, I conjecture that the above question has an affirmative answer but I'm unable to prove this.

Any ideas towards the solution? Your comments are welcome!