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.
For primes $p≡3\pmod 4$, prove that $[(p−1)/2]!≡±1\pmod p$.
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number-theory
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0I edited your title because it is extremely likely you did not really mean $P-12$. – 2012-10-28
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0For any p in the form 4k+3, ((p-1)/2)! is congruent to -1 mod p or 1 mod p – 2012-10-28
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1@Amanda did you mean $(p-12)!$ or $((p-1)/2)!$? – 2012-10-28
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0((p-1)/2)! Sorry! – 2012-10-28
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0A hint: write $p = 4k + 3$ and prove that $((2k + 1)!)^2 = + 1$ – 2012-10-28
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0@JonahSinick Why do I want to square (2k+1)! – 2012-10-28
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0Well, 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)/2 – 2012-10-28