Exponential Symmetry


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


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

$$\begin{align}\therefore m^{km-m} &= k^m\\m^{m(k-1)} &= k^m\\\left(m^{m(k-1)} \right)^{1/m} &= \left(k^m \right)^{1/m}\\\therefore m^{k-1} &= k\end{align}$$

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?

Problem ID: 344 (09 Jul 2008)     Difficulty: 4 Star

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