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 $a,b,m,n\le100$.