Applying Wilson's Theorem to Prove a Congruence

Question

Let $n \in \mathbb{N}$ and suppose that $n = kl$, for some $k, l \in \mathbb{N}$ such that $1<k<l<n$.

Show that

$(n-1)! \equiv 0(\mod n)$.

So far I have show Wilson's Theorem, that is

$(p-1)! \equiv -1(\mod p)$

for some prime $p$.

How would I go about this?

Just think about what $(n-1)!$ means. – lulu May 17 '21 at 16:10 — lulu, May 17 '21 at 16:10
$k$ and $l$ are factors of $(n-1)!$ – Bernard May 17 '21 at 16:10 — Bernard, May 17 '21 at 16:10

score 2 · Accepted Answer · answered May 17 '21 at 18:43

2

Since we know that n is not a prime, Wilson's Theorem is actually not going to help us out with this problem.

Hint: $(n-1)!= (n-1)(n-2) \cdots l \cdots k \cdots 1$, since $1<k<l<n.$

This means that (n-1)! is a multiple of $kl$, and we know that $n=kl$. What does that tell us about $(n-1)!$ mod $n$?

answered May 17 '21 at 18:43

CherryPi

Since $(n-1)$ is divisible by $kl$, it must be congruent to $0(\mod n)$ right?! – mathemagic May 17 '21 at 19:08
@mathemagic Close, in your comment you wrote $(n-1)$, but the factorial is very important! So, its $(n-1)!$ is divisible by $kl$ which then means that it is divisible by $n$. – CherryPi May 17 '21 at 19:47

1 Answers1