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$2^4=16=4^2$. In fact, $\{2,4\}$ is the only pair of natural numbers with that property, i.e. if $m<n$ are natural numbers and $m^n=n^m$, then $m=2$ and $n=4$.

This is easily seen with some analysis: For $m,n\in\mathbf{N}\backslash\{0\}$, the equation $m^n=n^m$ is equivalent to $\sqrt[m]{m}=\sqrt[n]{n}$. By calculus, we can show that the real function $t\mapsto \sqrt[t]{t}$ is strictly increasing for $t<e$ and strictly decreasing for $t>e$. So the smaller of the two numbers has to be $<e$ and the proposition follows.

My question: Is there an elementary proof? By elementary I mean most of all no irrational numbers, no calculus.

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    $\begingroup$ If only I had my time machine (or a better memory) I could give you an answer. I once gave a presentation related to this (as well as whether $e^{\pi}$ was greater than $\pi^e$), but the details escape me right now. I remember an analytical approach like what you described, and I can't remember what else I did :( $\endgroup$ Jun 27, 2011 at 20:58
  • $\begingroup$ use unique factorization $\endgroup$
    – yoyo
    Jun 27, 2011 at 21:00
  • $\begingroup$ I think this is a duplicate, but I can't find the previous one... $\endgroup$ Jun 27, 2011 at 21:03
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    $\begingroup$ @Arturo: Did you mean this: math.stackexchange.com/questions/9505/…? $\endgroup$
    – Aryabhata
    Jun 27, 2011 at 21:04
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    $\begingroup$ @Aryabhata: Hmmm... yes, but I somehow have the impression of having seen the problem solved by considering the prime factorization (e.g., for every prime $p$ we must have $n\mathrm{ord}_p(m) = m\mathrm{ord}_p(n)$), and that question doesn't do that... $\endgroup$ Jun 27, 2011 at 21:08

4 Answers 4

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The relation $m^n=n^m$ implies that $n,m$ have the same prime factors $p_1<p_2<...<p_k$. Suppose $m=\prod p_i^{\alpha_i},\ n=\prod p_i^{\beta_i}$. By the unique factorization theorem and the relation $m^n=n^m$ we get that $\alpha_i n=\beta_i m$. Suppose $m>n$. This implies $\alpha_i>\beta_i$ since $\alpha_i/\beta_i=m/n$. Therefore $n|m$.

Denote $m=dn$ and $(dn)^n=n^{(dn)}$ i.e. $dn=n^d$ or $d=n^{d-1}$. For $n \geq 2$ we have $n^{d-1}\geq d$ with equality for $d=1$ (which is not good since $m>n$) or $d=2,n=2$ and therefore $m=4$.

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HINT $\ $ wlog $\rm\displaystyle\ m > n\ \Rightarrow\ \bigg(\frac{m}n\bigg)^n =\: n^{m-n}\in \mathbb Z\ \Rightarrow\ k := \frac{m}n \in\mathbb Z\ \Rightarrow\ k = n^{k-1}\ \Rightarrow\:\cdots$

NOTE $\ $ Above I have implicitly invoked the Rational Root Test to infer that for $\rm\:j\in \mathbb Z,\: n\in \mathbb N\:,\:$ rational roots of $\rm\ x^n -j\ $ must be integers. $\:$ Above is the special case $\rm\ x = m/n,\ j = n^{m-n}\:.$

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  • $\begingroup$ Thanks for the hint. The second $\Rightarrow$ needs unique factorization, right? $\endgroup$
    – Stefan
    Jun 28, 2011 at 21:32
  • $\begingroup$ @Stefan Unique factorization is a bit overkill. It suffices to apply the Rational Root Test to $\rm\:x^n - j\:,\:$ to infer that any rational root is integral - just as in irrationality proofs. Said more technically, it uses only that $\mathbb Z$ is integrally closed (as is any UFD). $\endgroup$ Jun 28, 2011 at 22:24
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Have a look here (in particular from "Try this: Substitute").

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There is a rather pretty picture: given any real numbers $a,b > 1,$ the relation $$ a^b = b^a $$ is equivalent to $$ \frac{\log a}{a} = \frac{\log b}{b}.$$ So what you do is draw the graph $$ y = \log x $$ and then draw any line through the origin with positive slope (but smaller than $1/e$). The line intersects the curve in two points, with $x$-values $a,b$ satisfying $ a^b = b^a.$ The smaller of the two numbers $a,b$ lies between $1$ and $e,$ so the only possible integer is $2.$ In that case, the other $x$-value is $4.$

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    $\begingroup$ It is the (nearly) same method suggested by the OP at the second paragraph. $\endgroup$
    – user10676
    Jun 28, 2011 at 11:15
  • $\begingroup$ It does look that way. $\endgroup$
    – Will Jagy
    Jun 28, 2011 at 18:38

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