Peter wrote: Find all pairs $ (a,b)$ of positive integers that satisfy the equation \[ a^{b^{2}} = b^{a}. \] Solution at http://www.kalva.demon.co.uk/imo/isoln/isoln975.html Notice first that if we have $ a^m = b^n$, then we must have $ a = c^e, b = c^f$, for some c, where m=fd, n=ed and d is the greatest common divisor of m and n. [Proof: express a and b as products of primes in the usual way.] In this case let d be the greatest common divisor of a and b2, and put $ a = de, b^2 = df$. Then for some c, $ a = c^e, b = c^f$. Hence f $ c^e = e c^{2f}$. We cannot have e = 2f, for then the c's cancel to give e = f. Contradiction. Suppose 2f > e, then $ f = e c^{2f - e}$. Hence e = 1 and $ f = c^{2f - 1}$. If c = 1, then f = 1 and we have the solution a = b = 1. If c ≥ 2, then $ c^{2f - 1} \geq 2^f > f,$ so there are no solutions. Finally, suppose 2f < e. Then $ e = f c^{e - 2f}$. Hence f = 1 and $ e = c^{e - 2}$. $ c^{e - 2}$ ≥ $ 2^{e - 2}$ ≥ e for e ≥ 5, so we must have e = 3 or 4 (e > 2f = 2). e = 3 gives the solution a = 27, b = 3. e = 4 gives the solution a = 16, b = 2.

$(1,1)$ and $(16,2)$ are the only solutions. I would post the complete solution later as I am busy.

My calculator says that $\left(27,3\right)$ is also a solution.

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=3845&sid=8bd9c89cbce5a3485b8f120f8c8a3b5a#p3845