PUMaC 2012 · 加试 · 第 1 题

PUMaC 2012 — Power Round — Problem 1

On any problem, you may use without proof any result or remark from earlier in the test, even if it’s a problem your team has not solved. You may cite results from conjectures or subsequent problems only if your team solved them independently of the problem where you wish to cite them. You may not cite parts of your proof of other problems: if you wish to use a lemma in multiple problems, please reproduce it in each one.

解析

If f = 0 , then we pick q = r = 0 . Suppose that f 6 = 0 . Consider the set of polynomials S = { p = f − qg : q ∈ Q [ X ] } If 0 ∈ S , then we are done, so suppose S contains only nonzero elements. We know S is nonempty because it contains p = f . Let r = f − qg be of minimal degree in S . If s = deg r − deg g ≥ 0 , then we can subtract off s a constant multiple of X g ( X ) from r to produce another element of S of degree strictly lower than r , contradicting minimality of r . Therefore, deg r < deg g as needed. s