Let $p$ be an odd prime. For $d\in\mathbb Z$ we define $$T(d,p):=\det\left[\left(\frac{i^2+dj^2}p\right)\right]_{0\le i,j\le(p-1)/2},$$ where $(\frac{\cdot}p)$ is the Legendre symbol. By (1.17) of my paper arXiv:1308.2900, if$(\frac dp)=-1$ then $(\frac{T(d,p)}p)=1$.

Suppose that $p\equiv3\pmod4$. Then, by (1.14) of arXiv:1308.2900, $T(d,p)=T(-1,p)$ for any $d\in\mathbb Z$ with $(\frac dp)=-1$. As $T(-1,p)$ is a skew-symmetric determinant of even order, it is an integer square.

In the case $p\equiv1\pmod4$, if $d$ and $d'$ are both quadratic nonresidues modulo $p$, then we clearly have $T(d,p)=\pm T(d',p)$.

I have the following conjecture which seems quite challenging.

**Conjecture**. Let $p\equiv1\pmod4$ be a prime and write $p=x^2+4y^2$ with $x$ and $y$ positive integers. Then, for any integer $d\in\mathbb Z$ with $(\frac dp)=-1$, there is a positive integer $t(p)$ (not depending on $d$) such that $$|T(d,p)|=2^{(p-1)/2}t(p)^2y.$$

Via Mathematica, I find that \begin{gather}t(5)=1,\ t(13)=3,\ t(17)=4,\ t(29)=91,\ t(37)=81,\ t(41)=180, \\t(53)=1703,\ t(61)=87120,\ t(73)=16104096,\ t(89)=3947892146, \\ t(97)=19299520512,\ t(101)=885623936875,\ t(109)=36548185365.\end{gather}

Your comments are welcome!

PS: I have verified the conjecture for all primes $p<5000$ with $p\equiv1\pmod4$.