$2^4=16=4^2$. In fact, $\{2,4\}$ is the only pair of natural numbers with that property, i.e. if $m 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 My question: Is there an elementary proof? By elementary I mean most of all no irrational numbers, no calculus.
Elementary proof of $m^n\neq n^m$ for almost all natural numbers $m\neq n$
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1If 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 :( – 2011-06-27
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0use unique factorization – 2011-06-27
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0I *think* this is a duplicate, but I can't find the previous one... – 2011-06-27
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1@Arturo: Did you mean this: http://math.stackexchange.com/questions/9505/xy-yx-x-and-y-are-integers? – 2011-06-27
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2@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... – 2011-06-27
4 Answers
The relation $m^n=n^m$ implies that $n,m$ have the same prime factors $p_1
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 $m=4$.
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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0Thanks for the hint. The second $\Rightarrow$ needs unique factorization, right? – 2011-06-28
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0@Stefan Unique factorization is a bit overkill. It suffices to apply the [Rational Root Test](http://en.wikipedia.org/wiki/Rational_root_theorem) 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). – 2011-06-28
Have a look here (in particular from "Try this: Substitute").
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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2It is the (nearly) same method suggested by the OP at the second paragraph. – 2011-06-28
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0It does look that way. – 2011-06-28