IMO 1986 SL 4

Origin: CZS

Problem

Let n be a positive integer and let p be a prime number, p > 3. Find at least 3(n+1) [easier version: 2(n+1)] sequences of positive integers x, y, z satisfying xyz = pn(x + y + z) that do not differ only by permutation.

Solution

Let x = p\alphax′, y = p\betay′, z = p\gammaz′ with p ∤x′y′z′ and \alpha \geq\beta \geq\gamma. From the given equation it follows that pn(x + y) = z(xy −pn) and consequently z′ | x + y. Since also p\gamma | x + y, we have z | x + y, i.e., x + y = qz. The given equation together with the last condition gives us xy = pn(q + 1) and x + y = qz. (1) Conversely, every solution of (1) gives a solution of the given equation.

For q = 1 and q = 2 we obtain the following classes of n + 1 solutions each: q = 1 : (x, y, z) = (2pi, pn−i, 2pi + pn−i) for i = 0, 1, 2, . . ., n; q = 2 : (x, y, z) =  3pj, pn−j, 3pj+pn−j

for j = 0, 1, 2, . . ., n. For n = 2k these two classes have a common solution (2pk, pk, 3pk); oth- erwise, all these solutions are distinct. One further solution is given by (x, y, z) =
1, pn(pn + 3)/2, p2 + 2  , not included in the above classes for p > 3. Thus we have found 2(n + 1) solutions. Another type of solution is obtained if we put q = pk + pn−k. This yields the solutions (x, y, z) = (pk, pn + pn−k + p2n−2k, pn−k + 1) for k = 0, 1, . . . , n. For k < n these are indeed new solutions. So far, we have found 3(n+1)−1 or 3(n + 1) solutions. One more solution is given by (x, y, z) = (p, pn + pn−1, pn−1 + pn−2 + 1).