PUMaC 2014 · 个人决赛（B 组） · 第 3 题

PUMaC 2014 — Individual Finals (Division B) — Problem 3

解析

Solve the following equation for x, y 2 N 3 3 x + y = 101 xy + 101 Solution: If 101 | x , we must also have 101 | y and thus x, y 101. Hence with out loss of generality, we 3 3 can assume x y and thus we have x 101 xy and y > 101 which means eqality cannot hold. Hence 101 - x and 101 - y . 3 3 2 2 We see that 101 | x + y and hence 101 | ( x + y )( x xy + y ). We wish to show that 2 2 2 2 101 - ( x xy + y ). Assuming the contrary, let 101 | ( x xy + y ). Hence we must have 2 2 2 2 2 2 101 | (4 x 4 xy + y + 3 y ) = (2 x y ) + 3 y . Hence we see that 3 y is a quadratic ✓ ◆ 2 3 y residue of 101 and hence, making use of the Legendre Symbol, we have 1 = = 101 ✓ ◆ ✓ ◆ ✓ ◆ 2 1 3 y . 101 101 101 ✓ ◆ ✓ ◆ 2 1 y (101 1) / 2 We see that = ( 1) = 1 and since 101 - y , = 1, we have 1 = 101 101 ✓ ◆ ✓ ◆ ✓ ◆ 3 101 101 1 3 1 2 2 2 = ( 1) = = 1which is contradictory and thus we must have 101 3 3 2 2 101 - ( x xy + y ). 2 2 2 Hence 101 | ( x + y ) and we have x + y = 101 k . Thus, we have k ( x + y ) = ( k + 1) xy + 1 Since 2 2 k ( x + y ) 2 kxy , we have ( k + 1) xy + 1 2 kxy and hence xy + 1 kxy . Therefore k = 1. 2 2 2 This leaves us with x xy + y = xy + 1. Hence ( x y ) = 1 and we see that the only solutions are ( x, y ) = (51 , 50) or (50 , 51). 3