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Today, I came up with the following problem when trying to solve this.

Are there distinct integers $a,b,m,n>1$ such that the equation $$a^b+b^a=m^n+n^m$$ holds? That is, is there ever an integer that can be written as $a^b+b^a$ in more than one way?

I claim that the answer is No, but I think solving this is beyond my knowledge. For a very preliminary observation, the simplest case is to consider the powers of $1,5,6,0$, since they end in those same digits. For example, $$\begin{cases}a\equiv5\pmod{10}\\b\equiv6\pmod{10}\end{cases}\implies a^b+b^a\equiv1\pmod{10}.$$ However, this brings about an issue, since there is hardly any indication as to what values $x$ and $y$ can take other than them having opposite parity.

PARI/GP code is

intfun(a,b,m,n)={for(i=2,a,for(j=2,b,for(k=2,m,for(l=2,n,if(i<>k && i<>l && j<>k && j<>l && i^j+j^i-k^l-l^k==0,print(i," ",j," ",k," ",l))))));}

No solutions have been found up to $a,b,m,n\le100$.

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    We can accelerate the search by assuming $a\le b, x\le y, a\le x$ – Peter Jul 07 '19 at 19:44
  • Still no solution in the range $a,b,x,y\le 200$ – Peter Jul 07 '19 at 20:03
  • Fermat's little theorem will help. –  Jul 07 '19 at 23:41
  • $a^b-x^y=b^a-y^x$ etc mean they either pump out factors on both sides, or are all coprime. –  Jul 08 '19 at 01:31
  • $x^y-a^b$ rather. –  Jul 08 '19 at 01:42
  • If $a | (b-1)$ and $a |(x-1)$ and $ a |( y-1)$ then there are no solutions (obviously). – Grešnik Jul 08 '19 at 06:52
  • from my reconfiguring it, we get if any pair of them share a factor n, we can get a difference of nth powers on one side. –  Jul 08 '19 at 17:28
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    you can do a much broader search if you simply check if $f(a,b) = a^b+b^a$ is an injection for $(a,b) \in \mathbb{Z}^2, a \geq b \geq 2$. I was able to confirm this up for $a,b \leq 1300$ here https://repl.it/repls/OutstandingEarlyAutomaticvectorization. You would easily be able to search through higher values if you ran this locally on your computer instead of through repl, but I don't have access to that at the moment. – Zach Hunter Sep 28 '19 at 16:52
  • Maybe this could help: $a^b+b^a=b^{\frac{b\ln(a)}{\ln(b)}}+b^a=b^a(b^M+1)$ where $M=\ln\left(\dfrac{\sqrt[\ln(b)]{b\ln(a)}}{e^a}\right)$ – Piquito Aug 19 '20 at 13:37
  • Code could use updates: like parfor, or using the fact that if $a,b$ are same parity, $m,n$ are going to need to be same parity as well. Or that The code as written checks $2^5+5^2$ and $5^2+2^5$ you're doing more than $4$ times as many checks as needed. – Roddy MacPhee Sep 19 '21 at 13:37
  • @ZachHunter I ran a code that checks for collisions as you suggested, up to $a, b \le 10000$, and I also ran it again for $a<70000$ and $b<500$. Did someone calculate asymptotics for this question? Can we say something along the lines "the function $f$ gets so big so fact that intuitively it's very unlikely that we'll find such number considering the bruteforces we ran for small numbers"? – Rei Henigman Jul 30 '23 at 16:25

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