Determine the number of solutions of the equation $n^m = m^n$ 
Possible Duplicate:
$x^y = y^x$ for integers $x$ and $y$ 

Determine the number of solutions of the equation 
$n^m = m^n$ 
where both m and n are integers.

This is a nice problem for a calculus class, to describe all pairs $(x,\xi)$ of positive real numbers with $x\ne\xi$ and $x^\xi=\xi^x$. From $\xi\log x=x\log\xi$ you get $(\log x)/x=(\log\xi)/\xi$, in other words, you’re looking for horizontal lines that intersect the graph of $f(x)=(\log x)/x$ twice. Since the function is defined and differentiable on $\langle0,\infty\rangle$ with a single maximum, all you need to do is spot where that maximum happens, and by differentiating, you see that it’s at $x=e$. Using the fact that $f(1)=0$, you see that for any $x$ in $\langle1,e\rangle$, there’s a unique $\xi>e$ for which $x^\xi=\xi^x$. And of course you see that there’s only one integer in the open interval $\langle1,e\rangle$.
