HMMT 二月 2021 · 团队赛 · 第 1 题

HMMT February 2021 — Team Round — Problem 1

[40] Let a and b be positive integers with a > b . Suppose that √ √ √ √ √ √ a + b + a − b is a integer. √ (a) Must a be an integer? √ (b) Must b be an integer? ◦

解析

[40] Let a and b be positive integers with a > b . Suppose that √ √ √ √ √ √ a + b + a − b is a integer. √ (a) Must a be an integer? √ (b) Must b be an integer? Proposed by: Daniel Zhu Answer: (a) Yes (b) No √ √ √ √ √ √ √ √ 2 2 2 2 Solution 1: Let r = a + b and s = a − b . We know r + s = 2 a and r − s = 2 b . If r + s is an integer k , then 2 2 2 2 2 2 √ r + s ( r + s ) + ( r − s ) k + 4 b/k a = = = , 2 4 4 √ which is rational. Recall that since a is a positive integer, a is either an integer or an irrational √ 2 2 number. (Proof: if a = p/q for relatively prime positive integers p, q , then p /q is an integer, which implies q = 1.) Thus the answer to part (a) is yes. The answer to part (b) is no because √ √ 2 (2 ± 2) = 6 ± 32 , meaning that setting a = 36 and b = 32 is a counterexample. √ √ √ √ √ √ √ √ Solution 2: Second solution for part (a): Squaring a + b + a − b , we see that 2 a +2 a − b √ √ is an integer. Hence a − b = m − a for some rational number m . Squaring both sides of this, we √ √ 2 m + b 2 see that a − b = m − 2 m a + a, so a = , a rational number. As in the first solution, it follows 2 m √ that a is an integer. ◦