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> If $p$ is congruent to $3 \pmod 4$, and $n$ is the number of quadratic residues less than $p/2$, how can I show that $2\cdot 4\cdot 6\cdots (p-1)$ is congruent to $(-1)^{n+k}$ modulo $p$?
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So $p=4k+3$ and I know $(-1/p)$ is congruent to $(-1)^{(p-1)/2}$ etc but can't think of a way to show the product of the even numbers is congruent to $(-1)^{n+k}$.
Any hints/help would be much appreciated.