On $\det\left[\frac1{i^2-ij+j^2}\right]_{1\le i,j\le p-1}$ and $\det\left[\frac1{i^2-ij+j^2}\right]_{1\le i,j\le (p-1)/2}$

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!

• 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 $\mathbb{Z}/p$ ? – darij grinberg Mar 8 at 20:55
• 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 $x\equiv 0\pmod {p^n}$. – Zhi-Wei Sun Mar 8 at 21:39
• 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 $\mathbb{F}_p^\times$. – darij grinberg Mar 8 at 21:45