[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)

Question

[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)

anonymous · Answer

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.

kinggeorge · Answer

Perfect solution.