QUESTION. Is my following conjecture true? If true, how to prove it?

Conjecture. Let $p$ be a prime with $p\equiv5\pmod 6$, and define the matrices $A_p$ and $B_p$ by $$A_p:=\left[\frac1{i^2-ij+j^2}\right]_{1\le i,j\le p-1} \ \ \text{and}\ \ B_p=\left[\frac1{i^2-ij+j^2}\right]_{1\le i,j\le (p-1)/2}.$$ Then $$\left(\frac{\det A_p}p\right)=\left(\frac 2p\right)\ \ \text{and}\ \ \text{ord}_p(\det B_p)=\frac{p+1}6,$$ where $(\frac{\cdot}p)$ is the Legendre symbol and $\text{ord}_p(x)$ is the $p$-adic order (or valuation) of $x$.

I formulated the conjecture in 2013 and checked it via Mathematica. It appeared in Remark 1.3 of my recent paper [Finite Fields Appl. 56(2019), 285-307].

I think some of you might be able to prove the conjecture. Your comments are welcome!

  • $\begingroup$ How can $\dfrac{p+1}{6}$ be an order modulo $p$? Or do you mean the $p$-valuation of a rational number and the whole matrices are over $\mathbb{Z}$ rather than $\mathbb{Z}/p$ ? $\endgroup$ – darij grinberg Mar 8 at 20:55
  • $\begingroup$ As $p\equiv2\pmod3$, both $\det A_p$ and $\det B_p$ are $p$-adic integers. As usual, $\text{ord}_p(x)$ is the largest $n\in\mathbb N$ such that $x\equiv 0\pmod {p^n}$. $\endgroup$ – Zhi-Wei Sun Mar 8 at 21:39
  • 1
    $\begingroup$ Ah, so you do mean the $p$-valuation. I am used to calling it $v_p\left(x\right)$, while $\operatorname{ord}_p\left(x\right)$ stands for the order of an invertible element $x$ in the multiplicative group $\mathbb{F}_p^\times$. $\endgroup$ – darij grinberg Mar 8 at 21:45

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