Are there any positive integral solutions $(m,n)$ to the diophatine equation $n=m^{n}$ besides $(m,n)=(1,1)$? Not sure how to approach this question. I got the (obvious) solution by guessing. It seems clear that $m\leq{n}$.
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4Hint: If $m \neq 1$ then $n \geq 2^{n}$. – Kavi Rama Murthy Jan 24 '22 at 08:23
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@BobbyLaspy The given equation gives this (wrong) inequality so we get a contradiction. – Kavi Rama Murthy Jan 24 '22 at 09:06
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It does not really matter, but you used another equation in the title than in the body. – Peter Jan 24 '22 at 09:11
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1@BobbyLaspy Combine what I wrote with the strict inequality $2^{n} >n$ (which follows from Binomial Theorem, for example). – Kavi Rama Murthy Jan 24 '22 at 09:19
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1@KaviRamaMurthy Why this indirect proof ? We just need $2^n>n$ for every positive integer $n$ (can easily be proven by induction). Then, if $m>1$ , we have $m^n\ge 2^n>n$ , hence $n=m^n$ cannot hold. – Peter Jan 24 '22 at 09:22
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@KaviRamaMurthy: I won't argue further, but I don't see a logic in what you wrote. – Jan 24 '22 at 09:25
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$$m=1^m$$ has the only (and obvious) solution $m=1$.
There are no other solutions to the initial problem, as
$$m<2^m<3^m<\cdots$$
[By induction, $1<2^1$ and $m<2^m\implies m+1<2m<2^{m+1}$.]
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No,there are no other positive integral solutions except the one you have already guessed. For this solution ,I would be replacing (n,m) with (x,y).(as I comfortable in solving with those variables)
$\implies$$x=y^{x}$
$\implies y=x^{1/x}$
By differentiating the expression,you will realise that the maximum value of the expression is at $e$ which is $e^{1/e}$. You can prove that $e^{1/e}<2$ since $\frac{1}{e} (≈0.37)< ln(2) {≈0.69}$.
As,you can observe that the only integral value that y can be is 1.Thus the only positive integral solutions is (1,1).