For any odd integer $n>1$ and integers $c$ and $d$, define
$$(c,d)\_n:=\det\left[\left(\frac{i^2+cij+dj^2}n\right)\right]\_{1\le i,j\le n-1},$$
where $(\frac{\cdot}n)$ is the Jacobi symbol. It is easy to show that $(c,d)\_n=0$ if $(\frac dn)=-1.$

QUESTION: Does $n^2$ divide $(2,3)\_n$ for each odd integer $n>3$?

My computation led me to conjecture that $n^2\mid (2,3)\_n$ for any odd integer $n>3$ and also $n^2\mid (6,15)\_n$ for any odd integer $n>5$.

For any positive integer $n\equiv\pm5\pmod{12}$, we have $(\frac 3n)=-1$ and hence $(2,3)\_n=0$. Also, I observe that $(2,3)\_n=0$ if $n>3$ and $n\equiv 3\pmod 6$. But I don't see why $(2,3)\equiv0\pmod{n^2}$ for each positive integer $n\equiv\pm1\pmod{12}$. Note that
$$\frac{(2,3)\_{11}}{11^2}=-2^3\ \ \text{and}\ \ \frac{(2,3)\_{37}}{37^2}=2^{44}\times467^2.$$

I also have some other questions concerning $(c,d)\_n$. For example, I conjecture that $(6,1)\_n=0$ for any positive integer $n\equiv 3\pmod4$. For more such questions, one may consult my preprint [On some determinants with Legendre symbol entries](http://arxiv.org/abs/1308.2900).

Any ideas helpful to the question ?