Odd-residue squared product modulo a prime (${\prod }_{j=1}^{(p-3)/2}(2j+1{)}^2\equiv (-1{)}^{(p+1)/2}(\mathrm{mod}p)$)

For an odd prime $p$, the even residues are the negatives of the odd residues. If $O=1\cdot 3\cdots (p-2)$, then $(p-1)!\equiv (-1{)}^{(p-1)/2}{O}^2$. Apply Wilson theorem to obtain the displayed congruence. The empty product for $p=3$ is one.