# Problem from a Theorem

If $$p$$ is an odd prime number, then prove that $$\left( \left( \dfrac{p-1}{2} \right)! \right)^2 + (-1)^\frac{p-1}{2}$$ is divisible by $$p$$.

Note by Fahim Shahriar Shakkhor
3 years, 11 months ago

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From Wilson's Theorem, $$(p-1)! \equiv -1 (\mod p)$$

$$\Rightarrow 1\times2\times........\times \frac{p-1}{2}\times\frac{p+1}{2} \times ..........\times (p-2)\times (p-1) \equiv -1 (\mod p)$$

$$\Rightarrow 1\times2\times........\times \frac{p-1}{2}\times(-\frac{p-1}{2}) \times ..........\times (-2)\times (-1) \equiv -1 (\mod p)$$

$$\Rightarrow ((\frac{p-1}{2})!)^2 \times (-1)^{\frac{p-1}{2}} \equiv -1 (\mod p)$$

$$\Rightarrow ((\frac{p-1}{2})!)^2 \equiv (-1)\times(-1)^{\frac{p-1}{2}} (\mod p)$$

$$\Rightarrow ((\frac{p-1}{2})!)^2 + (-1)^{\frac{p-1}{2}} \equiv 0 (\mod p)$$

- 3 years, 11 months ago

Nice and tricky!

For convenience, we let $$p=2n+1$$ and so we wanna prove that $$(n!)^2=n!\cdot n!\equiv (-1)^{n+1}\pmod p$$. Now note that $$n\equiv n-p=-(p-n)\pmod p$$. So

$n!=\prod_{k=1}^n k \equiv \prod_{k=1}^n -(p-k)=(-1)^n \prod_{k=1}^n (p-k)=(-1)^n \prod_{k=1}^n (2n+1-k)=(-1)^n \prod_{k=n+1}^{2n} k\pmod p.$

And hence we have

$n!\cdot n! \equiv n!\cdot (-1)^n \prod_{k=n+1}^{2n} k=(-1)^n\cdot (2n)!=(-1)^n\cdot (p-1)!\equiv (-1)^n\cdot (-1)=(-1)^{n+1}\pmod p$

and the last congruence step follows by Wilson's theorem.

- 3 years, 11 months ago