Equality of Integers to the Power of Each Other/Proof 2

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Theorem

$2$ and $4$ are the only pair of positive integers $m, n$ such that $m \ne n$ such that:

$m^n = n^m$


Thus: $2^4 = 4^2 = 16$


Proof

Without loss of generality suppose $m > n$.

Write $m = n + x$, where $x$ is an integer.

Then:

\(\ds m^n\) \(=\) \(\ds n^m\)
\(\ds \paren {n + x}^n\) \(=\) \(\ds n^{n + x}\)
\(\ds \paren {1 + \frac x n}^n\) \(=\) \(\ds n^x\) dividing both sides by $n^n$

From Real Sequence $\paren {1 + \dfrac x n}^n$ is Convergent:

$\paren {1 + \dfrac x n}^n$ is increasing and has limit $e^x$.

Hence:

$n^x < e^x$

This forces $n = 2$.

We have:

$m^2 = 2^m$

showing that $m$ is a power of $2$.

Write $m = 2^k$.

Then:

$2^{2 k} = 2^{2^k}$

giving:

$k = 2^{k - 1}$

By Bernoulli's Inequality:

$2^{k - 1} \ge 1 + k - 1 = k$

where equality holds if and only if $k - 1 = 0$ or $k - 1 = 1$.

We can skip Bernoulli's Inequality by induction on $k$ for $k > 2$.

Either way, this gives:

$k = 1$ or $2$
$m = 2$ or $4$.

We reject $m = 2$ since $n = 2$.

Hence $2^4 = 4^2$ is the only solution.

$\blacksquare$