how to prove 437 divides 18!+1?

Question

I was solving some problems and i came across this problem. I didn't understand how to approach this problem. Can we solve this with out actually calculating 18!?

Answer 1

Note that $437=(19)(23)$. We prove that $19$ and $23$ divide $18!+1$. That is enough, since $19$ and $23$ are relatively prime.

The fact that $19$ divides $18!+1$ is immediate from Wilson's Theorem, which says that if $p$ is prime then $(p-1)!\equiv -1\pmod{p}$.

For $23$ we need to calculate a bit. We have $22!\equiv -1\pmod{23}$ by Wilson's Theorem.

Now $(18!)(19)(20)(21)(22)=22!$.

But $19\equiv -4\pmod{23}$, $20\equiv -3\pmod{23}$, and so on. So $(19)(20)(21)(22)\equiv 24\equiv 1\pmod{23}$. It follows that $18!\equiv 22!\pmod{23}$, and we are finished.

Answer 2

That's not so big numbers. This is a demonstration: 18! + 1 = 6402373705728001 = 14650740745373 * 437

Of course that's not as interesting as André's one.