IMO 1997 SL 17

Origin: CZE

Problem

Find all pairs of integers x, y \geq1 satisfying the equation xy2 = yx.

Solution

We note first that x and y must be powers of the same positive integer. Indeed, if x = p\alpha1 1 \cdot \cdot \cdot p\alphak k and y = p\beta1 1 \cdot \cdot \cdot p\betak k (some of \alphai and \betai may be 0, but not both for the same index i), then xy2 = yx implies \alphai \betai = x y2 = p q for some p, q > 0 with gcd(p, q) = 1, so for a = p\alpha1/p \cdot \cdot \cdot p\alphak/p k we can take x = ap and y = aq. If a = 1, then (x, y) = (1, 1) is the trivial solution. Let a > 1. The given equation becomes apa2q = aqap, which reduces to pa2q = qap. Hence p ̸= q, so we distinguish two cases: (i) p > q. Then from a2q < ap we deduce p > 2q. We can rewrite the equation as p = ap−2qq, and putting p = 2q + d, d > 0, we obtain d = q(ad −2). By induction, 2d −2 > d for each d > 2, so we must have d \leq2. For d = 1 we get q = 1 and a = p = 3, and therefore (x, y) = (27, 3), which is indeed a solution. For d = 2 we get q = 1, a = 2, and p = 4, so (x, y) = (16, 2), which is another solution. (ii) p < q. As above, we get q/p = a2q−p, and setting d = 2q −p > 0, this is transformed to ad = a(2ad−1)p, or equivalently to d = (2ad −1)p. However, this equality cannot hold, because 2ad−1 > d for each a \geq2, d \geq1. The only solutions are thus (1, 1), (16, 2), and (27, 3).