# Prove gcd(nn!, n!+1)=1

1. Apr 10, 2012

### youvecaughtme

1. The problem statement, all variables and given/known data
For any $n \in \mathbb{N}$, find $\mathrm{gcd}(n!+1,(n+1)!+1)$. First come up with a conjecture, then prove it.

2. The attempt at a solution
By testing some values, it seems like $\mathrm{gcd}(n!+1,(n+1)!+1) = 1$

I'm trying to prove this by induction. I'll leave out the inductive assumption and base case verification because I can do those.

I have $\mathrm{gcd}(n!+1,(n+1)!+1) = 1$ and I'm trying to show that $\mathrm{gcd}((n+1)!+1,(n+2)!+1) = 1$.

I can simplify what's given to me to $\mathrm{gcd}(nn!, n!+1)=1$ but I can't find out how to get it into the form I want it. Can anybody look at what I'm doing and give me any guidance?

$\mathrm{gcd}(n!+1,(n+1)!+1) = 1 \implies \mathrm{gcd}(n!+1,(n+1)n!+1) = 1 \implies \mathrm{gcd}(n!+1,nn!+n!+1) = 1 \implies \mathrm{gcd}((n)n!, n!+1) = 1$

2. Apr 11, 2012

### morphism

I wouldn't bother with induction here.

Try to argue as follows. If a prime divides both n!+1 and (n+1)!+1, it will divide their difference. What does this imply?