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