Congruence mod p involving a product

Question

Let $p$ be a prime, $p\equiv 3$ mod $4$. Numerically it appears that $$ \prod_{n=1}^{p-1}\left(1+n^2\right)\equiv 4\mod p. $$ How can one prove this? For $p\equiv 1$ mod $4$, the product is $0$ mod $p$ because $-1$ is a quadratic residue.

Answer 1

Let's work over the finite field $\Bbb F_p$ with $p\equiv3\pmod 4$. One can adjoin $i$ to this with $i^2=-1$ to get the finite field $k=\Bbb F_{p^2}$. Your product is $$\prod_{n=1}^{p-1}(n^2+1)=\prod_{n=1}^{p-1}(n-i)(n+i)=f(i)f(-i)$$ where $$f(X)=\prod_{n=1}^{p-1}(X+n)=X^{p-1}-1$$ (we are working in characteristic $p$). Then $$f(i)=-1-1=-2.$$ Likewise, $f(-i)=-2$ and so the original product is $4$.