2
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Find all positive integers $m$ and $n$ with $m\ne n$ that satisfy the equation

$~~~~~~~\displaystyle \sqrt[m]{m} ~=~ \sqrt[n]{n}$

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  • $\begingroup$ Did you want to add the requirement that $m$ and $n$ are distinct? $\endgroup$ – Ian MacDonald Jul 14 '17 at 14:09
  • $\begingroup$ Good catch @Ian MacDonald, yes 'm' and 'n' are distinct positive numbers! $\endgroup$ – Mea Culpa Nay Jul 14 '17 at 14:18
  • $\begingroup$ @Gamow ... thanks for editing the question and making it pretty ! $\endgroup$ – Mea Culpa Nay Jul 17 '17 at 8:08
5
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$2^{1/2} = 4^{1/4}$ is the only solution.

If you analyse the function $f(x)=x^{1/x}$, you'll find it has a maximum between $x=2$ and $x=3$ (actually at $x=e$) and is strictly decreasing afterwards with asymptote at $1$. Therefore any such solution pairs will consist of one number less than $3$ and that is $3$ or larger. The smallest cannot be $1$, since $f(1)=1$ and the function never reaches 1 after. This leaves only the pair $2$ and $4$.

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5
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First of all, there are the trivial solutions where $m=n$. Then we have $\{m,n\}=\{2,4\}$. I claim these are all. Note first of all that we're looking for multiple $x$ where $x^{1/x}$ has the same value; equivalently, where $\frac{\log x}x$ does. It's easy to show that this function is increasing from 0 to $e$, and decreasing thereafter; in particular, given any pair $\{m,n\}$ one of the numbers must be less than $e$ and one greater. But the only positive integers below $e$ are 1 and 2. We have already seen what to do with 2; and $1^{1/1}=1$ is smaller than $n^{1/n}$ for any other positive integer $n$.

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  • $\begingroup$ (In case anyone's wondering, my answer was posted slightly before Jaap's.) $\endgroup$ – Gareth McCaughan Jul 14 '17 at 16:31
  • $\begingroup$ Yes, I think you beat me by about 30 seconds or so. I gave you a +1. $\endgroup$ – Jaap Scherphuis Jul 14 '17 at 19:42

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