For which $0\leq a<p^2$, where $p$ is an odd prime, we have that $(2p-1)!\equiv a\mod{p^2}$

Question

Let $p$ be an odd prime. I need to find for which $0\leq a < p^2$, $(2p-1)!\equiv a\mod{p^2}$.

If $a\equiv (2p-1)!\mod{p^2}$, then we have that $a = kp^2 + (2p-1)!$, and therefore $p\mid a$, which means $a$ is an integer multiple of $p$, and necessarily not zero, since we have only one instance of $p$ in $(2p-1)!$

How can I proceed from here?

Answer 1

Your have $(2p-1)! = \prod_{i=1}^{2p-1} j =p \prod_{j=1}^{p-1}j(j+p)$. Thus modulo $p^2$ this is $p \prod_{j=1}^{p-1}j^2= p ((p-1)!)^2$.

Now one knows that $(p-1)!$ is congruent $-1$ modulo $p$; this is not hard to to show, and you can find it as part of a proof of Wilson's theorem.

Thus in total we get that the result is $p$.

Answer 2

We have $$ (p-1)! \equiv \frac{(2p-1)!}{p!} \equiv -1 \pmod p $$ by Wilson's theorem (edit: the division in the above expression is regular integer division, not modular division). So $\frac{(2p-1)!}{p} \equiv 1\pmod p$, or in other words, there is a $k$ such that $$ \frac{(2p -1)!}{p} = kp + 1. $$Multiply both sides by $p$ to get $a = p$.