I want to find all integers $a$ and $b$ that satisfy $a^b=b^a$, where $0<a<b$. By guessing I found that $a=2$ and $b=4$ is a solution. I'm not sure whether it is the only solution, though intuitively I think it is. How can I prove this? Thank you!
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been asked before, though I don't have a link. try thinking about remainder when dividing by 2 for each argument. – Sep 02 '17 at 02:31
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it's (2,4),(1,1),(4,2),x=y – Jacob Claassen Sep 02 '17 at 04:00
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@MANMAID Do you know for a fact that the gcd helps? If you do, unless you explain how, your comment is unhelpful. If you do not, it is misleading. Once again, please stop spamming the site... – Did Sep 02 '17 at 09:07
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@Did Let, $b>a$, $\gcd(a,b)=g$, then $a=gk,b=gl$ and $\gcd(k,l)=1$, then $g^bk^b=g^al^a\Rightarrow g^{b-a}=(\dfrac{l^k}{k^l})^g$. Why do you think this might not help? – MAN-MADE Sep 02 '17 at 10:05
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@MANMAID "Why do you think this might not help?" Oh, so you also need to learn to read comments. – Did Sep 02 '17 at 10:08
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I know for sure in this type of problem, if you use gcd in this way the problem becomes simpler, so @Did "for a fact that the gcd helps". Note that since $\gcd(k,l)=1$, if you break into cases(which is not a hard thing to do here), only solution you can find would be $(1,1),(4,2)$. – MAN-MADE Sep 02 '17 at 10:22