HMMT 十一月 2025 · GEN 赛 · 第 4 题
HMMT November 2025 — GEN Round — Problem 4
题目详情
- Given that a , b , and c are integers with c ≤ 2025 such that | x + ax + b | = c has exactly 3 distinct integer solutions for x , compute the number of possible values of c . ◦
解析
- Given that a , b , and c are integers with c ≤ 2025 such that | x + ax + b | = c has exactly 3 distinct integer solutions for x , compute the number of possible values of c . Proposed by: Srinivas Arun Answer: 31 2 Solution: The answer is 31. The possible values of c are c = 2 k for integers 1 ≤ k ≤ 31. These work 2 2 2 2 2 2 by taking a = 0 and b = − 2 k ; then | x − 2 k | = 2 k is equivalent to x = 0 or 4 k , so it has three integer solutions: x = 0 , − 2 k, 2 k . We will now prove no other values of c work. The problem condition is equivalent to the two quadratic equations 2 2 x + ax + b = c and x + ax + b = − c having three distinct integer solutions in total. If c = 0, then these two quadratics are the same, so they have at most two roots in total, contradiction. Hence c ̸ = 0, so the two quadratics share no roots. If one of these quadratics has an integer root, then both its roots are integers. This follows from Vieta’s formulas: for each quadratic, the sum of the roots is − a , an integer. If only one quadratic has integer roots, then the two quadratics have at most two integer roots in total, contradiction. So both quadratics have all integer roots. Because the two quadratics have three distinct integer roots in total, and all roots of both quadratics are integers, the two quadratics have three distinct roots in total. Since the quadratics share no roots, one quadratic has two roots and the other has one root. Graphically, this means that the vertex of 2 x + ax + b lies on either the line y = c or y = − c . Since this parabola faces upward, the case y = c is 2 impossible (since then x + ax + b never intersects y = − c ), so the vertex lies on y = − c . Refer to the diagram below. 2 f ( x ) = x + ax + b y = c y = − c a ( − , − c ) 2 © 2025 HMMT a 2 2 2 Because x + ax + b = − c has one root, x + ax + ( b + c ) is a perfect square; it must be ( x + ) , and 2 a a 2 so x + ax + ( b + c ) = 0 has the single solution x = − . This implies is an integer. 2 2 2 Next we use the condition that x + ax + b = c has integer roots. Complete the square to obtain that 2 a 2 2 x + ax + b = c = ⇒ x + ax + ( b + c ) = x + = 2 c. 2 a Since and x are both integers, 2 c must be a (nonzero) perfect square. Note that 2 c is even, so let 2 2 2 2 2 c = (2 k ) for some integer k . This rewrites as c = 2 k . Since 1 ≤ c ≤ 2025, 1 ≤ k ≤ 1012, so 1 ≤ k ≤ 31. Hence these are the only values of c that work, as desired. ◦