Find all pairs (a, b) of positive integers that satisfy a^{b2} = b^{a}.

**Answer**

(1,1), (16,2), (27,3).

**Solution**

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 b^{2}, 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} ≥ 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.

© John Scholes

jscholes@kalva.demon.co.uk

22 Oct 1998

Last corrected/updated 19 Oct 03