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KingGeorge
 2 years ago
[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 \[\large\left(\left(\frac{p1}{2}\right)!\right)^2\equiv(1)^{\frac{p+1}{2}}\pmod{p}\]
(The whole factorial is squared)
KingGeorge
 2 years ago
[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 \[\large\left(\left(\frac{p1}{2}\right)!\right)^2\equiv(1)^{\frac{p+1}{2}}\pmod{p}\] (The whole factorial is squared)

This Question is Closed

joemath314159
 2 years ago
Best ResponseYou've already chosen the best response.3First note that:\[\left(\left(\frac{p1}{2}\right)!\right)^2 = \left(1\cdot2\cdot 3\cdots \frac{p1}{2}\right)\left(1\cdot2\cdot 3\cdots \frac{p1}{2}\right)\]Since we are looking at things mod p, we see that:\[1 = (p1),2 = (p2),\ldots , \frac{p1}{2}=\left(\frac{p1}{2}+1\right)\]This gives us:\[\left(1\cdot2\cdot 3\cdots \frac{p1}{2}\right)\left(1\cdot2\cdot 3\cdots \frac{p1}{2}\right)\]\[=\left(1\cdot2\cdot 3\cdots \frac{p1}{2}\right)\left(\left(\frac{p1}{2}+1\right)\cdots (p1)\right)\]\[=(1)^{\frac{p1}{2}}(p1)!\]After this, an application of Wilson's Theorem will do the trick.
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