# Does $\det\left[\left(\frac{i^2-\frac{p-1}2!\,j}p\right)\right]_{1\le i,j\le(p-1)/2}$ vanish for every prime $p\equiv3\pmod 4$?

For any odd prime $p$, let $D_p$ denote the determinant $$\det\left[\left(\frac{i^2-\frac{p-1}2!\times j}p\right)\right]_{1\le i,j\le (p-1)/2},$$ where $(\frac{\cdot}p)$ is the Legendre symbol. Then \begin{gather*}D_3=0,\ D_5=-1,\ D_7=D_{11}=0,\ D_{13}=-8, \\ D_{17}=-72,\ D_{19}=D_{23}=0,\ D_{29}=-2061248.\end{gather*}

QUESTION: Let $p$ be an odd prime. Is it true that $D_p=0$ if and only if $p\equiv 3\pmod4$?

In 2013 I formulated this problem and conjectured that the answer is yes. I have verified this for all primes $p<2300$. By a result of L. Mordell [Amer. Math. Monthly 68(1965), 145-146], for any prime $p>3$ with $p\equiv3\pmod4$ we have $$\frac{p-1}2!\equiv(-1)^{(h(-p)+1)/2}\pmod p,$$ where $h(-p)$ is the class number of the imaginary quadratic field $\mathbb Q(\sqrt{-p})$.

Any ideas towards the solution?

• Have you calculated the kernels for small values of $p\equiv 3\pmod 4$? Maybe there is some pattern. Jun 9, 2018 at 10:03

Warning: this is mod $p$ answer only. Thus it answers the original question for $p=4k+1$ but $p=4k+3$ remains open.

Denote $\alpha=-(\frac{p-1}2)!$ and apply the formula given by Darij Grinberg in the comment here https://mathoverflow.net/a/302143/4312 where $n=(p-1)/2$, $x_j=\alpha\cdot j$, $y_j=j^2$ for $j=1,\dots,n$. Note that all elementary symmetric polynomials $e_k$, $1\leqslant k\leqslant n-1$, for $y$'s are equal to 0 (mod $p$ of course), since they are coefficeints of a polynomial $(t^2-1^2)(t^2-2^2)\dots (t^2-n^2)=t^{p-1}-1$. Thus only two summands remain: $e_0(x)e_n(y)$ and $e_n(x)e_0(y)$. They go with the same (and non-zero) coefficient, thus $p$ divides your determinant if and only if $p$ divides $e_n(x)+e_n(y)=\prod x_i+\prod y_i=-\alpha^{n+1}+\alpha^2=\alpha^2(\alpha^{n-1}-1)$. We have $\alpha^2=(-1)^{n+1}$ (from above polynomial or from Wilson theorem). If $p=4k+3$, $n=2k+1$, this gives $\alpha^2=1$ and indeed $\alpha^{n-1}-1=0$. If $p=4k+1$, $n=2k$, this gives $\alpha^2=-1$ and $\alpha^{n-1}=\alpha^{2k-1}=\pm \alpha\ne 1$.

• I'm glad to see the progress. You have proved that $p\mid D_p$ for any prime $p\equiv 3\pmod4$, but my conjecture says that $D_p$ is actually zero for any prime $p\equiv 3\pmod4$. Jun 8, 2018 at 5:03

This is not an answer but what seems to be an interesting variant of the same behavior.

$$\det\left[\left(\frac{i^2-\frac{p-1}2!\times j}p\right)\right]_{1\le i,j\le (p-1)/2}=0$$ if and only if $$\det\left[\left(\frac{i^2-j^2}p\right)\right]_{1\le i,j\le (p-1)/2}=0.$$

• @Amdeberhan Have you really proved the equivalence? Jun 8, 2018 at 4:57
• By (1.16) of my paper arxiv.org/abs/1308.2900, for any prime $p\equiv 3\pmod 4$ we have $\det[(\frac{i^2-j^2}p)]_{1\le i,j\le(p-1)/2}=0$. Jun 8, 2018 at 4:58
• The second matrix is simply antisymmetric of odd order, but the first is not even if we permute the rows and columns (there are not enough many zero entries) Jun 8, 2018 at 8:59