PUMaC 2009 · 数论（B 组） · 第 6 题

PUMaC 2009 — Number Theory (Division B) — Problem 6

Find the sum of all integers x for which there is an integer y , such that x − y = xy + 61. n n

解析

Find the sum of all integers x for which there is an integer y , such that x − y = xy + 61. Solution. 6. It is easy to see that one or more of x and y cannot be zero, because then the equation cannot hold true because of the constant term. So assume that x and y are positive 2 integers. Also, x 6 = y , because then we would have x + 61 = 0, which is inadmissible. If x < y , then the left hand side becomes negative while the right remains positive. So clearly, 3 3 2 2 x > y ≥ 1. We will use the identity x − y = ( x − y )( x + xy + y ) = xy + 61, from which we have 2 2 61 = ( x − y )( x + y ) + ( x − y − 1) xy 2 . 2 2 Then, if x − y ≥ 3, then we must have x ≥ 3 + y = 4, and 61 ≥ 3( x + y ) + ( x − y − 2 2 2 2