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KingGeorge
[SOLVED] Before I go to bed, have an interesting little number theory problem because I'm bored. Suppose \(p\) is an odd prime. Show that \[\large\left(\left(\frac{p-1}{2}\right)!\right)^2\equiv(-1)^{\frac{p+1}{2}}\pmod{p}\] (The whole factorial is squared)
First note that:\[\left(\left(\frac{p-1}{2}\right)!\right)^2 = \left(1\cdot2\cdot 3\cdots \frac{p-1}{2}\right)\left(1\cdot2\cdot 3\cdots \frac{p-1}{2}\right)\]Since we are looking at things mod p, we see that:\[1 = -(p-1),2 = -(p-2),\ldots , \frac{p-1}{2}=-\left(\frac{p-1}{2}+1\right)\]This gives us:\[\left(1\cdot2\cdot 3\cdots \frac{p-1}{2}\right)\left(1\cdot2\cdot 3\cdots \frac{p-1}{2}\right)\]\[=\left(1\cdot2\cdot 3\cdots \frac{p-1}{2}\right)\left(-\left(\frac{p-1}{2}+1\right)\cdots -(p-1)\right)\]\[=(-1)^{\frac{p-1}{2}}(p-1)!\]After this, an application of Wilson's Theorem will do the trick.