Find all pairs (a, b) of positive integers that satisfy ab2 = ba.
Answer
(1,1), (16,2), (27,3).
Solution
Notice first that if we have am = bn, then we must have a = ce, b = cf, 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, b2 = df. Then for some c, a = ce, b = cf. Hence f ce = e c2f. We cannot have e = 2f, for then the c's cancel to give e = f. Contradiction. Suppose 2f > e, then f = e c2f-e. Hence e = 1 and f = c2f-1. If c = 1, then f = 1 and we have the solution a = b = 1. If c ≥ 2, then c2f-1 ≥ 2f > f, so there are no solutions.
Finally, suppose 2f < e. Then e = f ce-2f. Hence f = 1 and e = ce-2. ce-2 ≥ 2e-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