I want to show that for every odd prime $p$ the following congruence holds:
$$ 2^2\cdot4^2\cdot\dots\cdot(p-3)^2\cdot(p-1)^2 \equiv (-1)^{\frac{1}{2}(p+1)}\mod{p} $$
How would I approach this problem?
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If $k$ is even, then $p-k$ is odd. You can split each square into the product of two complementary terms to cover every nonzero residue class:
$$\begin{align} \prod_{k=1}^{(p-1)/2} (2k)^2 &\equiv \prod_{k=1}^{(p-1)/2} (-1) \cdot (2k) \cdot (p-2k) & \pmod p \\&= (-1)^{(p-1)/2} \prod_{n=1}^{p-1} n \\&= (-1)^{(p-1)/2} (p-1)! \end{align}$$
and Wilson's theorem implies
$$ (p-1)! \equiv -1 \pmod p $$