PUMaC 2009 · 数论（B 组） · 第 1 题

PUMaC 2009 — Number Theory (Division B) — Problem 1

解析

Find the number of pairs of integers x and y such that x + xy + y = 28. 2 2 Solution. 12. We multiply both sides by 4, and rewrite the equation as (2 x + y ) + 3 y = 112. From there, we can simply take possible values for y sequentially to get equations that work, 2 i.e. if we are considering some y , we check that 112 − 3 y is a perfect square, and if it is, it is 2 2 immediately an admissible answer. We end up with three possible equations: 10 +3 · 2 = 112, 2 2 2 2 8 + 3 · 4 = 112 and 2 + 3 · 6 = 112. For each of these, we can find a possible x and y , all nonzero, satisfying the equation. But note that, given that x and y , ± x and ± y serve the purpose just as well. So it follows that for each of those three equations, there are four possible values of ( x, y ). Hence the total number of pairs of integers is 12.