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?

  • $\begingroup$ Have you calculated the kernels for small values of $p\equiv 3\pmod 4$? Maybe there is some pattern. $\endgroup$ – Fedor Petrov Jun 9 '18 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 $\alpha^{n-1}=\alpha^{2k-1}=\pm \alpha\ne 1$.

  • $\begingroup$ 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 $p\equiv 3\pmod4$. $\endgroup$ – Zhi-Wei Sun Jun 8 '18 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 $$\det\left[\left(\frac{i^2-j^2}p\right)\right]_{1\le i,j\le (p-1)/2}=0.$$

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    $\begingroup$ @Amdeberhan Have you really proved the equivalence? $\endgroup$ – Zhi-Wei Sun Jun 8 '18 at 4:57
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    $\begingroup$ By (1.16) of my paper arxiv.org/abs/1308.2900, for any prime $p\equiv 3\pmod 4$ we have $\det[(\frac{i^2-j^2}p)]_{1\le i,j\le(p-1)/2}=0$. $\endgroup$ – Zhi-Wei Sun Jun 8 '18 at 4:58
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    $\begingroup$ 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) $\endgroup$ – Fedor Petrov Jun 8 '18 at 8:59

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