HMMT 十一月 2024 · 团队赛 · 第 2 题
HMMT November 2024 — Team Round — Problem 2
题目详情
- [20] Compute the sum of all positive integers x such that ( x − 17) x − 1+( x − 1) x + 15 is an integer.
解析
- [20] Compute the sum of all positive integers x such that ( x − 17) x − 1+( x − 1) x + 15 is an integer. Proposed by: Edward Yu Answer: 11 Solution: First, we prove the following claim. √ √ Claim 1. If integers a , b , c , d , n satisfy a and c are nonzero, b and d are nonnegative, and a b + c d = n , then either n = 0 or both b and d are perfect squares. √ √ Proof. We know a b = n − c d . Squaring both sides, we get √ √ 2 2 2 2 a b = ( n − c d ) = n + c d − 2 nc d, √ √ √ √ so 2 nc d is an integer. If d is not an integer, then it is not rational, so 2 nc d = 0. Since c and d are nonzero, we must have n = 0. √ Similarly, if b is not an integer, n = 0. Thus either n = 0 or both b and d are perfect squares. Applying our claim to the given expression, we get three cases. • Case 1: Either x − 17 or x − 1 is zero. It is easy to verify x = 1 is a solution, while x = 17 is not. √ √ • Case 2: ( x − 17) x − 1 + ( x − 1) x + 15 = 0. Then 2 2 ( x − 17) ( x − 1) = ( x − 1) ( x + 15) = ⇒ ( x − 1)( − 48 x + 304) = 0 . 304 Thus x = 1 or x = , and the latter isn’t an integer. 48 2 2 • Case 3: x − 1 and x + 15 are both perfect squares. Let x + 15 = y and x − 1 = z for nonnegative integers y and z . Then, 2 2 ( y + z )( y − z ) = y − z = ( x + 15) − ( x − 1) = 16 . Observe that y + z is nonnegative and y + z ≥ y − z , so we have the following cases for y + z and y − z : 17 15 – y + z = 16, y − z = 1 = ⇒ y = , z = 2 2 – y + z = 8, y − z = 2 = ⇒ y = 5, z = 3 – y + z = 4, y − z = 4 = ⇒ y = 4, z = 0. The only nonnegative integer solutions are ( y, z ) = (5 , 3) or ( y, z ) = (4 , 0), which correspond to x = 10 and x = 1. Both of these are indeed solutions. Hence, the only x that work are x = 1 and x = 10, for a total of 11 .