I'm curious as to how many positive integers $m\neq n$ exist such that $n^m = m^n$.
Is $n = 2$, $m=4$ the only case?
I've plotted two surfaces (one surface represents $n^m$ where the other represents $m^n$) on a log-scale. Where they intersect is where $n^m = m^n$.
On
Hint:
Translate the equation as $$m\log n=n\log m \iff\frac{\log n}n=\frac{\log m}m.$$ Then determine the variations of the function $\dfrac{\log x}x$ on $(1,+\infty)$ and show that for any $x\in (1,\mathrm e)$ there exists exactly one $x'\in(1,+\infty)$ such that $$\frac{\ln x'}{x'}=\frac{\ln x}{x}.$$
We have $$ m^n=n^m\\ n\ln m=m\ln n\\ \frac{\ln m}m=\frac{\ln n}n $$ so we have $f(m)=f(n)$ for the function $f(x)=\frac{\ln x}x$. Differentiating tells us that $f$ is strictly increasing on $[1,e]$ and strictly decreasing on $[e,\infty)$, so for any pair $(m,n)$, assuming $m<n$ we have $m<e<n$. The only positive integers $m<e$ are $1$ and $2$, and $1$ clearly doesn't work.