Determine all pairs of positive integers $(m,n)$ that satisfies $m!+n!=m^n$ Determine all pairs of positive integers $(m,n)$ that satisfies $m!+n!=m^n$
I found easily the pairs $(2,2)$ and $(2,3)$ but i can't prove that these are the only pairs possibles.
Any hints?

Here we go, we are looking for pairs $(m,n)$:
$$m!+n!=m^n$$
Case 1: Suppose $m>n>1$. Then  $m=n+k$ for some integer $k>0$. We now have $$(n+k)!+n!=(n+k)^n$$
$$(n!)((n+1)^{(k)}+1)=(n+k)^n$$
Now what properties does this equation have? For starters:


*

*Let $Q_n= \{ p\;|\; p\le n,\; p \text{ prime}\}$ and $P_n=\prod_{p \in Q_n} p$, then   $ P_n|\;(n+k)^n$ and we get $k=jP_n$ for some   $j>0.$

*If  $p\;|\;(n+1)^{(k)}$ then $p\;\not|\;(n+1)^{(k)}+1$.


Note that $Q_n$ is just all the primes up to $n$. Both conditions are true. Condition two is a general statement though; It could be said "It's true just because."
Now, here's where this two conditions have problems: for this equation they are both true, however, if $k=n$ then every prime factor of $(n+1)^{(k)}+1$ is greater than $n$ by condition 2, but $n+k=n+(n)=2n$ has no such factors. Similarly, for any $k>n$ the number $(n+1)^{(k)}$ has every prime factor up to $n+k$ at least once, but no prime in $[\frac{n+k}{2},n+k]$ divides $n+k$. Hence, whatever the value of $k$ it must be less than $n$. 
Condition one tells us that $k\ge P_n$, where $P_n$ is the $n^{th}$ primorial...it is well known that $P_n>n$ for $n>2$. Hence, the only allowed cases by condition two are prohibited by condition one.
The idea for case 2, $m<n$ is similar.
