Let $p \in \mathbb{N}$ be an odd prime number. Show that $(p − 1)! \equiv p − 1 \pmod{2p}$.

Question

The furthest I've gotten on this one is I know I need to apply Wilson's Theorem but I'm confused because I don't understand how the $2p$ affects the Theorem itself.

Answer 1

So you know Wilson's theorem, which states that $(p-1)! \equiv -1 \mod p$ for any prime number $p$. What does this mean? Let's be really silly and just say the obvious; it means that $(p-1)!=kp-1$ for some integer $k$.

Lets consider the integer $k$. Can $k$ be even? If it were, then $kp-1$ would be odd, which is impossible since $(p-1)!$ contains a factor of $2$, so long as $p$ itself is greater than 2.

If you require more of a hint, let me know in a comment.