I'd like to ask how to calculate
$$\prod_{i=1}^{p-1} (i^{2st}+1) \pmod p$$ where $s<t$ are primes congruent to $1$ modulo $4$, for any odd prime $p$.
Here're some partial results. For any $1\leq i<p$, $i^{p-1}\equiv 1 \pmod p$ from Euler's theorem. Let $2st=\text{quotient}\cdot (p-1)+2x$ where $2x$ is the remainder. Therefore, $$\prod_{i=1}^{p-1} (i^{2st}+1) \equiv \prod_{i=1}^{p-1}[(i^x)^{2}+1].$$ For primes $p$ congruent to $1$ modulo $4$, there is a quadratic residue $a$ satisfying $a^2+1 \equiv 0 \pmod p$. Since there's a discrete logarithm $x$ for $i^x=a$, the product is congruent to $0$.
I'm not sure how to extend the results to primes $p$ congruent to $3$ modulo $4$. But for most primes, experimental results show that the product is congruent to $4=\prod_{i=1}^{p-1}(i^{2}+1)$. Especially, for primes with $x$ being multiple of $s$ or $t$, the product may deviate from $4$. I'm still finding the pattern.
WLOG, let $x=sy$. I find that for small primes, the products $[i^{2x}+1]$ can be grouped into $\frac{t-y}{\text{quotient}}$ groups according to their equivalence. I'm not sure whether this is general.