Motivated by my paper Quadratic residues and quartic residues modulo primes [Int. J. Number Theory 16 (2020), 1833-1858], here I pose two new conjectures for primes $p\equiv1\pmod8$ based on my computation.
Conjecture 1. For any prime $p\equiv1\pmod 8$, we have $$\prod_{1\le i,j\le(p-1)/2\atop p\nmid i^2+6ij+j^2}(i^2+6ij+j^2)\equiv -2^{(p-1)/4}\pmod p\tag{1}$$ and $$\prod_{1\le i,j\le(p-1)/2\atop p\nmid i^2-6ij+j^2}(i^2-6ij+j^2)\equiv -2^{(p-1)/4}\pmod p.\tag{2}$$
Conjecture 2. Let $p$ be a prime with $p\equiv1\pmod 8$, and write $p=x^2+2y^2$ with $x,y\in\mathbb Z$ and $x\equiv1\pmod4$. Then $$\prod_{1\le i,j\le(p-1)/2\atop p\nmid i^2+4ij+2j^2}(i^2+4ij+2j^2)\equiv (-1)^{(x+3)/4}2^{(p-1)/4}\pmod p\tag{3}$$ and $$\prod_{1\le i,j\le(p-1)/2\atop p\nmid i^2-4ij+2j^2}(i^2-4ij+2j^2)\equiv (-1)^{(x+3)/4}2^{(p-1)/4}\pmod p.\tag{4}$$
It is easy to see that $(1)$ is equivalent to $(2)$, and $(3)$ is equivalent to $(4)$.
Question 1. Are Conjectures 1 and 2 true? How to prove them?
Actually, I also have some other conjectures similar to Conjectures 1 and 2.
For any prime $p\equiv1\pmod8$, it is easy to see that $(1)$ has the following equivalent form: $$\prod_{1\le i<j\le(p-1)/2\atop p\nmid i^2+6ij+j^2}(i^2+6ij+j^2)\equiv\pm2^{(p-1)/8}\pmod p.\tag5$$
Question 2. How to determine the sign in $(5)$ explicitly?
Your comments are welcome!
