# A determinant problem for primes $p\equiv 1\pmod4$

Let $p$ be an odd prime, and let $A_p$ denote the matrix $$[a_{ij}]_{1\le i,j\le (p-1)/2},$$ where $$a_{1j}=\left(\frac jp\right),\ \ \text{and}\ \ a_{ij}=\left(\frac{i^2+j^2}p\right)\ \text{for}\ i>1,$$ with $(\frac{\cdot}p)$ the Legendre symbol.

QUESTION: Is $-\det A_p$ always an odd square for each prime $p\equiv1\pmod4$?

Define $a_p=\sqrt{-\det A_p}$ for any prime $p\equiv1\pmod4$. Then \begin{gather*}a_5=1,\ a_{13}=3,\ a_{17}=21,\ a_{29}=83,\ a_{37}=9095,\ a_{41}=98835, \\a_{53}=4689023,\ a_{61}=700882875,\ a_{73}=267187918095. \end{gather*} I have computed the values of $a_p$ for all primes $p\equiv1\pmod4$ with $p<2000$. Based on the numerical data, I conjecture that the above question has an affirmative answer.

This question is my supplement to the previous question http://mathoverflow.net/questions/310192.