PUMaC 2009 · 代数（A 组） · 第 2 题

PUMaC 2009 — Algebra (Division A) — Problem 2

Given that P ( x ) is the least degree polynomial with rational coefficients such that √ √ √ P ( 2 + 3) = 2 , find P (10). x x x 1 2 10

解析

Given that P ( x ) is the least degree polynomial with rational coefficients such that √ √ √ P ( 2 + 3) = 2 , find P (10). √ √ Solution. 455. We compute the first few powers of x = 2 + 3 and try to find a relation between them. 0 x = 1 √ √ 1 x = 2 + 3 √ 2 x = 5 + 2 6 √ √ 3 x = 11 2 + 9 3 √ 3 At this point, it seems possible that some expression ax + bx gives us 2. Indeed, requiring √ √ √ 3 the coefficients of both 2 and 3 to be 0 in the expression ax + bx − 2 = 0, where a, b are rationals, we find 11 a + b − 1 = 0 and 9 a + b = 0, with the unique solution ( a, b ) = (1 / 2 , − 9 / 2). 3 Hence P ( x ) = x / 2 − 9 x/ 2,and P (10) = 455. We can justify that P ( x ) needs to be of degree √ 2 at least 3: if it were of degree at most two, then we would have ax + bx + c − 2 = 0 for √ √ √ some rational numbers a, b, c . We need the coefficients of 2, 3, 6 and 1 to be 0 (in more advanced terms: these 4 numbers are linearly independent over Q ), which imposes 4 equations on our 3 coefficients, and will not have a solution. x x x 1 2 10