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So the question is :

Determine all pairs $(a, b)$ of positive integers satisfying $a^{b^2} = b^a$.

I tried it for 2 hours by different methods like taking it even, odd and by modulus method but cannot find any solution. Please help me in this question.

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  • $\begingroup$ Running a quick Mathematica program, I found that if $b\le1000$ then $(b,a)\in\{(1,1),(2,16),(3,27)\}$ are the only possible solutions. $\endgroup$
    – Vepir
    Jan 10, 2020 at 16:10

1 Answer 1

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Just follow your nose; let $d:=\gcd(a,b)$ so that $a=du$ and $b=dv$ with $u$ and $v$ coprime. Then $$b^a=(dv)^{du}=((dv)^u)^d \qquad\text{ and }\qquad a^{b^2}=(du)^{d^2v^2}=((du)^{dv^2})^d,$$ from which it follows that $(dv)^u=(du)^{dv^2}$. Because $u$ and $v$ are coprime we either have $u=1$ or $v=1$.

If $u=1$ then $dv=d^{dv^2}$, and so $v=d^{dv^2-1}$ from which it quickly follows that also $v=d=1$ and hence $a=b=1$.

If $v=1$ then $d^u=(du)^d$ from which it follows that $u^d=d^{u-d}$, and in particular $u\geq d$. Let $c:=\gcd(d,u)$ so that $d=ce$ and $u=cw$ with $e$ and $w$ coprime and $w\geq e$. Then $$u^d=(cw)^{ce}=((cw)^e)^c \qquad\text{ and }\qquad d^{u-d}=(ce)^{cw-ce}=((ce)^{w-e})^c,$$ from which it follows that $(cw)^e=(ce)^{w-e}$. As $e$ and $w$ are coprime and $w\geq e$ it follows that $e=1$, so $$cw=c^{w-1},$$ and hence $w=c^{w-2}$, from which it quickly follows that $w\leq4$. We check these few cases:

  1. If $w=1$ then $c=d=1$ which leads to $a=b=1$.
  2. If $w=2$ then $u=2d$ and hence $a=2b^2$, and plugging this in shows that $$b^{2b^2}=(2b^2)^{b^2}=2^{b^2}b^{2b^2},$$ contradicting the fact that $b$ is a positive integer.
  3. If $w=3$ then $c=d=3$ which leads to $a=27$ and $b=3$.
  4. If $w=4$ then $c=d=2$ which leads to $a=16$ and $b=2$.
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  • $\begingroup$ Hi, I'm looking for a solution using $v_p$, mind if I ask you to help? $\endgroup$ Feb 18, 2022 at 7:54

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