Talk:Equality of Integers to the Power of Each Other
Possible solution
$n^m = m^n \implies m \log_n n = n \log_n m \implies m = n \log_n m$
Assume $m > n$.
Since $m, n \in \N$, $\log_n m \in \N$ if $m = n^k$, where $k \in \N$ and $k > 1$ by assumption, but otherwise a free parameter.
Hence, $n^k = k n$.
For $n \ne 0$ the solution reads
- $n = k^\frac 1 {k - 1}$
Define $t = k - 1$.
$\ds \lim_{k \to 1} n = e$
$\ds \lim_{k \to \infty} n = 1$
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 constraints 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, and $1 < n < e$.
The only natural solution is $n = 2$
The only $k$ that satisfies this is $k = 2$.
Therefore, $n = 2$, $m = 2^2 = 4$. Now assume $m < n$ and repeat the calculation. --Julius (talk) 09:03, 18 May 2017 (EDT)
Another Possible Solution
Inspired by the final lines in 1 plus Perfect Power is not Prime Power except for 9.
Without loss of generality suppose $m > n$.
Write $m = n + x$, where $x$ is an integer.
Then:
| \(\ds m^n\) | \(=\) | \(\ds n^m\) | ||||||||||||
| \(\ds \paren {n + x}^n\) | \(=\) | \(\ds n^{n + x}\) | ||||||||||||
| \(\ds \paren {1 + \frac x n}^n\) | \(=\) | \(\ds n^x\) | Divide both sides by $n^n$ |
From Real Sequence (1 + x over n)^n is Convergent:
- $\paren {1 + \dfrac x n}^n$ is increasing and has limit $e^x$.
Hence we must have $n^x < e^x$.
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 $k = 2^{k - 1}$.
$2^{k - 1} \ge 1 + k - 1 = k$ is guaranteed by Bernoulli's Inequality, where equality holds if and only if $k - 1 = 0$ or $1$.
(The above equality condition is curiously not documented in the linked article)
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. --RandomUndergrad (talk) 10:45, 7 June 2020 (EDT)
- Both of these have been implemented. --prime mover (talk) 20:58, 13 September 2021 (UTC)