IMO 1995 SL N4

Origin: BUL | Category: Number Theory

Problem

Find all positive integers x and y such that x+y2+z3 = xyz, where z is the greatest common divisor of x and y.

Solution

Let x = za and y = zb, where a and b are relatively prime. The given Diophantine equation becomes a + zb2 + z2 = z2ab, so a = zc for some c \inZ. We obtain c + b2 + z = z2cb, or c = b2+z z2b−1. (i) If z = 1, then c = b2+1 b−1 = b + 1 + b−1, so b = 2 or b = 3. These values yield two solutions: (x, y) = (5, 2) and (x, y) = (5, 3). (ii) If z = 2, then 16c = 16b2+32 4b−1 = 4b + 1 + 4b−1, so b = 1 or b = 3. In this case (x, y) = (4, 2) or (x, y) = (4, 6). (iii) Let z \geq3. First, we see that z2c = z2b2+z3 z2b−1 = b + b+z3 z2b−1. Thus b+z3 z2b−1 must be a positive integer, so b + z3 \geqz2b −1, which implies b \leq z2−z+1 z−1 . It follows that b \leqz. But then b2 +z \leqz2 +b < z2b−1, with the last inequality because (z2−1)(b−1) > 2. Therefore c = b2+z z2b−1 < 1, a contradiction. The only solutions for (x, y) are (4, 2), (4, 6), (5, 2), (5, 3).