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It should be explained that $\overline{nAm} = m + A\times10^{\mathrm{digit}(m)} + n\times 10^{\mathrm{digit}(m) + \mathrm{digit}(A+m)}$, where $\mathrm{digit}{(X)}$ is the number of digits of $X$.

My friends and I had been talking about a joke before we ran into a problem.

We know that $x^2=25$, so by deleting $2$ on both sides, we get $x = 5$, which is one of the solutions to the equation.

After that, we've been thinking about whether some other groups of solutions satisfy $m^n=\overline{nm}$ where $n,m\in\mathbf{N}_+$. We soon discovered that there are no solutions except $(m,n)=(5,2)$. We tried to make this question more interesting, so we decided to put some more numbers between $n$ and $m$. And therefore we tried to find solutions for $m^n=\overline{nAm}$, where $m, n\in\mathbf{N}_+$ and $A\in\mathbf{N}$.

We've looked around and tried multiple methods of this, but none of them seems to work. I would appreciate it if you give me your thoughts or solutions that point me the way.

David H
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One way of looking at this problem:

You are looking for two numbers $m$ and $n$ such that, for $K$ the number of digits of $m$, we have:

$m^n\equiv m \pmod {10^K}$

$n=$ int$(m^n/10^L)$ for some $L\ge K$.

The first condition can be checked numerically fairly easily and then it seems a matter of chance whether or not the second condition holds.

Simple example

When $m=4$, all odd $n$ give a solution of the first condition. Then a quick use of a spreadsheet finds the solution $$4^{17}=171798...4$$

This looks a suitable approach for a computer search.