Equality of Integers to the Power of Each Other/Proof 1
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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
We have:
\(\ds n^m\) | \(=\) | \(\ds m^n\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds m \log_n n\) | \(=\) | \(\ds n \log_n m\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds m\) | \(=\) | \(\ds n \log_n m\) |
Without loss of generality, let $m > n$.
We have that $m, n \in \N$.
Hence:
- $\log_n m \in \N \implies m = n^k$
where $k \in \N_{>1}$ by hypothesis.
Hence:
- $n^k = k n$
For $n \ne 0$ the solution reads
- $n = k^\frac 1 {k - 1}$
Define $t = k - 1$.
We have:
\(\ds \lim_{k \mathop \to 1} n\) | \(=\) | \(\ds e\) | ||||||||||||
\(\ds \lim_{k \mathop \to \infty} n\) | \(=\) | \(\ds 1\) |
To check for intermediate maximum consider the first derivative:
- $\dfrac {\d n} {\d k} = \dfrac {k^{\frac 1 {k - 1} } } {k - 1} \paren {\dfrac 1 k - \dfrac {\ln k} {k - 1} }$
Our desired solution constrains the prefactor to be positive.
The term in brackets vanishes only for $k = 1$.
Hence for $k > 1$:
- there is no extremum
- $\map n k$ is monotonically decreasing
- $1 < n < e$.
Hence the only natural number solution is $n = 2$.
The only $k$ that satisfies this is $k = 2$.
Therefore:
- $n = 2$
- $m = 2^2 = 4$
$\blacksquare$