-4
$\begingroup$

Let $p$ be an odd prime, and let $(\frac{\cdot}p)$ be the Legendre symbol. We define the determinant $D(d,p)$ by $$D(d,p):=\det\left[(i^2+dj^2)\left(\frac{i^2+dj^2}p\right)\right]_{1\le i,j\le(p-1)/2}.$$

I have the following conjecture.

Conjecture. Let $p$ be an odd prime, and let $\delta_p$ be $0$ or $1$ according as $p=3$ or not. For any integer $d\not\equiv 0\pmod p$, we have $$\left(\frac{D(d,p)}p\right)=\begin{cases}(\frac dp)^{(p-1)/4}&\text{if}\ p\equiv1\pmod4,\\(\frac dp)^{\delta_p(p+1)/4}(-1)^{(h(-p)-1)/2}&\text{if}\ p\equiv3\pmod4,\end{cases}$$ where $h(-p)$ denotes the calss number of the imaginary quadratic field $\mathbb Q(\sqrt{-p})$.

Remark. We have verified the conjecture for all odd primes $p<600$. It is easy to see that $(\frac{D(d,p)}p)$ only depends on the values of $p$ and $(\frac dp)$. For the determinant $$S(d,p):=\det\left[\left(\frac{i^2+dj^2}p\right)\right]_{1\le i,j\le(p-1)/2},$$ the value $(\frac{S(d,p)}p)$ was determined in the paper arXiv:1308.2900. However, the method there does not work for the present conjecture.

$\endgroup$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.