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I know to use Wilson's Theorem and that each element in the second half is congruent to the negative of the first half, but I'm not sure how to construct a proof for it.

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    I edited your title because it is extremely likely you did not really mean $P-12$.2012-10-28
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    For any p in the form 4k+3, ((p-1)/2)! is congruent to -1 mod p or 1 mod p2012-10-28
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    @Amanda did you mean $(p-12)!$ or $((p-1)/2)!$?2012-10-28
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    ((p-1)/2)! Sorry!2012-10-28
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    A hint: write $p = 4k + 3$ and prove that $((2k + 1)!)^2 = + 1$2012-10-28
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    @JonahSinick Why do I want to square (2k+1)!2012-10-28
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    Well, somebody already answered, but the point is that the combination of the two facts that you mentioned gives that the square of (2k+1)! Is 1, and then you can conclude that the square root is plus or minus 1, then use that 2k+1 is (p-1)/22012-10-28

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