HMMT 二月 2003 · 冲刺赛 · 第 3 题

HMMT February 2003 — Guts Round — Problem 3

[5] If a and b are positive integers that can each be written as a sum of two squares, then ab is also a sum of two squares. Find the smallest positive integer c such that 3 3 3 3 c = ab , where a = x + y and b = x + y each have solutions in integers ( x, y ), but 3 3 c = x + y does not. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HARVARD-MIT MATHEMATICS TOURNAMENT, MARCH 15, 2003 — GUTS ROUND 1 2 3

解析

If a and b are positive integers that can each be written as a sum of two squares, then ab is also a sum of two squares. Find the smallest positive integer c such that c = ab , 3 3 3 3 3 3 where a = x + y and b = x + y each have solutions in integers ( x, y ), but c = x + y does not. Solution: 4 3 3 3 3 We can’t have c = 1 = 1 + 0 or c = 2 = 1 + 1 , and if c = 3, then a or b = ± 3 which is not a sum of two cubes (otherwise, flipping signs of x and y if necessary, we would get either a sum of two nonnegative cubes to equal 3, which clearly does not happen, or a difference of two nonnegative cubes to equal 3, but the smallest difference between 3 3 two successive cubes ≥ 1 is 2 − 1 = 7). However, c = 4 does meet the conditions, 3 3 with a = b = 2 = 1 + 1 (an argument similar to the above shows that there are no 3 3 x, y with 4 = x + y ), so 4 is the answer. 1 2 3