So I was trying to prove that the only positive integers m and n for which: $$m \ne n$$ $$m^n=n^m$$
Are $m=2$ and $n=4$, and vice versa.
And in doing this I converted m to x and n to y, and graphed $x^y = y^x$ on desmos. The link is https://www.desmos.com/calculator/dcsuehcgvt. I clearly saw there was a line where $x=y$ in the middle, and a curve which fulfills the requirement $m \ne n$.
The curve has a vertical asymptote at $x=1$ and a horizontal one at $y=1$. This leads me to believe there are no solutions past $[4, 2]$ on the curve because $y$ will be less than 2 and because of the asymptote, larger than 1, and therefore not an integer.
Similarly, there are no solutions past $[2,4]$ for the same reason.
Therefore the only feasible regions are $x = [2,4]$ and $y = [2,4]$. The only integers that fit the constraints in this area are the pairs $[2,4]$ and $[4,2]$.
However, there are two steps which I feel are not properly done:
- I didn't prove that the curve $x^y = y^x$ has the two asymptotes referenced above - this could be helped by being able to solve for x.
- I didn't prove there aren't individual points not on the curve that work - this could be proven by a proof of continuity.
Take the logarithm.
$$y\log x=x\log y,$$ or
$$\frac{\log x}x=\frac{\log y}y.$$
For $x>1$, there are two solutions in $y$, one of course being $y=x$. The other can be expressed by means of the Lambert $W$ function.
The asymptotes are justified by the fact that for $x\to1$, the other solution is $y\to\infty$.
In the intervals $(0,e)$ and $(e,\infty)$, the function $\log x/x$ is continuous and strictly monotonous, hence continuously invertible, with codomains $(-\infty,1/e)$ and $(0,1/e)$.