$\prod_{i=1}^{p-1} (i^2+1) \equiv 4 \pmod p$ if $p$ is a prime $\equiv3\pmod 4$

Question

I find that $\prod_{i=1}^{p-1} (i^2+1) \equiv 4 \pmod p$ if $p$ is a prime congruent to $3$ mod $4$, which is verified for small primes. I'd like to ask how to prove it.

Answer 1

Let $u$ be a square root of $-1$ in the finite field $\Bbb F_{p^2}$. Then your product equals $$\prod_{a=0}^{p-1}(a+u)(a-u)=f(u)f(-u)$$ where $$f(X)=\prod_{a=0}^{p-1}(a+X)=X^p-X$$ over the field $\Bbb F_{p^2}$. Thus $f(u)=u^p-u=-2u$, as $p\equiv3\pmod 4$, and similarly $f(-u)=2u$. Your product is then $-4u^2=4$ in $\Bbb F_{p^2}$.