For $n=1,2,3,\ldots$ let $a_n$ denote the determinant $\det[(i^2+j^2)^n]_{0\le i,j\le n-1}$. Then $$a_1=0,\ a_2=-1,\ a_3=-17280,\ a_4= 1168415539200.$$

QUESTION: Is it true that $(2n)!\mid a_n$ for all $n=3,4,\ldots$?

I even conjecture that $$a_n'=\frac{(-1)^{n(n-1)/2}a_n}{2\prod_{k=1}^n(k!(2k-1)!)}$$ is a positive integer for every integer $n>2$. Note that \begin{gather*}a_3'=1,\ a_4'=559,\ a_5'=10767500,\ a_6'=9372614611500. \end{gather*}

The question is similar to my previous question http://mathoverflow.net/questions/302130. But it seems that darij grinberg's method there does not work for the present question.

Any comments are welcome!