What is $\prod_{k=1}^{(p-1)/2} (2k-1)^2 \bmod p$ where $p$ is an odd prime?
Note: Someone just asked this and it was deleted while I was working on it, so I am posting it with my answer.
I get $(-1)^{(p+1)/2}$.
Here is how.
Working mod $p$,
$\begin{array}\\ \prod_{k=1}^{(p-1)/2} (2k-1)^2 &=\prod_{k=1}^{(p-1)/2} (2k-1)^2\dfrac{\prod_{k=1}^{(p-1)/2} (2k)^2}{\prod_{k=1}^{(p-1)/2} (2k)^2}\\ &=\dfrac{\prod_{k=1}^{p-1} (k)^2}{2^{p-1}\prod_{k=1}^{(p-1)/2} (k)^2}\\ &=\dfrac{(p-1)!^2}{2^{p-1}((p-1)/2)!^2}\\ &\equiv \dfrac{1}{((p-1)/2)!^2} \qquad\text{since }(p-1)! \equiv -1\text{ and }2^{p-1}\equiv 1\\ \end{array} $
Also,
$\begin{array}\\ (p-1)! &=\prod_{k=1}^{p-1} k\\ &=\prod_{k=1}^{(p-1)/2} k\prod_{k=(p+1)/2}^{p-1} k\\ &=((p-1)/2)!(-1)^{(p-1)/2}\prod_{k=(p+1)/2}^{p-1} (-k)\\ &\equiv((p-1)/2)!(-1)^{(p-1)/2}\prod_{k=(p+1)/2}^{p-1} (p-k)\\ &\equiv((p-1)/2)!(-1)^{(p-1)/2}\prod_{k=1}^{(p-1)/2} k\\ &\equiv((p-1)/2)!^2(-1)^{(p-1)/2}\\ \text{so}\\ -1 &\equiv ((p-1)/2)!^2(-1)^{(p-1)/2}\\ \text{or}\\ ((p-1)/2)!^2 &\equiv -(-1)^{(p-1)/2} \qquad\text{since } (-1)^{p-1} \equiv 1\\ &\equiv (-1)^{(p+1)/2}\\ \text{so that}\\ \prod_{k=1}^{(p-1)/2} (2k-1)^2 &\equiv (-1)^{(p+1)/2}\\ \end{array} $
