Determine all pairs of positive integers $(m,n)$ that satisfies $m!+n!=m^n$

Question

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?

Answer 1

This is not quite a full answer but it has the main ideas you'll need.

First, let us get the ability to assert a lower bound on $m$ after checking only finitely many pairs. This can be done by noting that $m^n>n!$, so by Stirling's approximation, $n<em$. Thus we can divide the problem between $m<m^*$ and $m \geq m^*$ by directly checking the cases $m=1,2,\dots,m^*-1$ and $n=1,2,\dots,\lfloor em \rfloor$.

Second, let us see what happens if $m$ is large. Similar to the above, if we assume $m>e$, then Stirling's approximation applied to $m^n>m!$ implies

$$m \leq \frac{n}{1-1/\ln(m)}.$$

So it suffices to provide an upper bound on $n$ in order to finish this case.

Third, let $r=\min \{ m,n \}$. Then $r!$ divides $m^n$ and so $r\#$ divides $m$, where $\#$ is the primorial function. Thus in particular $r\# \leq m$.

So there are two cases: $r=m$ and $r=n$. Consider $r=m$. By Bertand's postulate, if $m>2$ then there is a prime $p$ between $m/2$ and $m$, and so $m\# \geq 2p>m$. So $r=m$ can only occur for $m \leq 2$.

Now consider $r=n$. In this case $n\# \leq m$. If we assume $m \geq 8$, then $1-1/\ln(m) \leq 2$, so for $m \geq 8$ the inequality above gives

$$n\# \leq 2n.$$

Applying Bertrand's postulate twice in the same manner as the previous case implies that $n\# \geq n^2/8$. Thus $n \leq 16$.

Putting the pieces together, we divide the problem between $m<8$ and $m \geq 8$. In the case $m<8$ we have $n \leq \lfloor em \rfloor$, which is on the order of 100 cases to check. In the case $m \geq 8$ we have $n \leq 16$ and $m \leq 2n$, which is again on the order of 100 cases to check.

Answer 2

You can using mathematical induction to prove it, but its not verified for n=m=1 Or make table , row contains n! and column m! Then see any numbers satisfying this equation.

Answer 3

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:

1. 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 $j>0.$
2. 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.