## Exponential Symmetry

#### Problem

Given that $m$ and $n$ are positive integers, solve $m^n = n^m$.

#### Solution

For the purpose of this problem we shall assume that $m = n$ is a trivial solution, and it can be seen that if $m = 1$ then $1^n = n^1 \Rightarrow m = n = 1$.

So without loss of generality we can assume that $2 \le m \lt n$.

Begin by dividing both sides by $m^m$:

$$\begin{align}\dfrac{m^n}{m^m} &= \dfrac{n^m}{m^m}\\\therefore m^{n-m} &= \left(\dfrac{n}{m}\right)^m\end{align}$$As the left hand side is integer it follows that the right hand side is integer.

Let the integer, $k = \dfrac{n}{m} \Rightarrow n = km$.

As we have ruled out the trivial case $m = n$ and $n = km$, it is clear that $k \ge 2$.

For $k = 2, m^1 = 2 \Rightarrow n = 4$

Let us consider $k \ge 3$:

When $k = 3, m^2 = 3$, and as $m \ge 2$, it follows that $m^2 \gt 3$.

We shall now prove inductively that $m^{k-1} \gt k$ and thus $m^{k-1} \ne k$ for $k \ge 3$.

If true then $m^{k-2} \gt k-1$.

$\therefore m^{k-1} = m \cdot m^{k-2} \gt m(k - 1)$

But as $m \ge 2, m^{k-1} \gt 2k - 2 \gt k$, which is the expected result.

Hence $m^{k-1} \gt k$ for all values of $k \ge 3$.

Therefore we prove that $2^4 = 4^2$ is the only solution in positive integers.

What if we relax the condition of $m$ and $n$ being positive?

What if we allow $m$ and $n$ to be rational numbers?