PUMaC 2014 · 数论（A 组） · 第 7 题

PUMaC 2014 — Number Theory (Division A) — Problem 7

[ 7 ] Find the number of positive integers n ≤ 2014 such that there exists integer x that satisfies x + n the condition that is an odd perfect square. x − n 2

解析

[ 7 ] Find the number of positive integers n ≤ 2014 such that there exists integer x that satisfies x + n the condition that is an odd perfect square. x − n Solution: x + n 2 n 2 2 2 = k ⇐⇒ 1 + = k = ⇒ 2 n = ( x − n )( k − 1) x − n x − n 2 2 By the problem condition, k − 1 is even. Then, n is also even, since k − 1 6 ≡ 2 (mod 4). So, 2 letting k = 2 a − 1 and n = 2 b , we get b = ( x − 2 b )( a − a ). 2 2 It is sufficient to find all b ≤ 1007 such that b is divisible by a − a . Note that 2 − 2 = 2, so 2 any even b works. However, since a − a is always even, no odd b works. So, there are 503 integers n that satisfy the problem statement. 2