Krdytkyu

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 $1leq i<p$, $i^{p-1}equiv 1 pmod p$ 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.