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Assume $p$ is prime and $p\ge 3$.

Through experimentation, I can see that it's probably true. Using Wilson's theorem and Fermat's little theorem, it's equivalent to saying $2^2 4^2 6^2 \cdots (p-1)^2 = (-1)^{\frac{p+1}{2}}$ mod $p$, but I can't figure out any more than that.

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    In Wilson's theorem $(p-1)(p-2)\dots 3\cdot 2\cdot 1 \equiv -1 \bmod p$, combine terms $k$ and $p-k$ together for $k=1,2,\dots,(p-1)/2$. Note $p-k \equiv -k \bmod p$.2012-04-13
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    @KCd: sounds like a very good answer to me, please move it below so that it can be accepted.2012-04-13
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    @KCd: Thanks! That was exactly what I needed. I'll mark it as accepted if you put that in an answer instead of a comment.2012-04-13

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