Equality of Integers to the Power of Each Other

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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 1

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:

$n = 2$
$m = 2^2 = 4$

$\blacksquare$


Proof 2

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.

$\blacksquare$


Sources

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