Let $$f(X)=a_{n}X^{n}+a_{n-1}X^{n-1}+\cdots +a_{1}X+a_{0}$$be a polynomial with real coefficients which satisfies $$a_{n}\geq a_{n-1}\geq \cdots \geq a_{1}\geq a_{0}>0.$$Prove that for every complex root $z$ of this polynomial, we have $|z|\leq 1$.
Problem
Source: Moldova TST 2017, Day 1, Problem 2
Tags: algebra, polynomial
rmtf1111
06.03.2017 22:57
If someone finds it useful, this problem is present here as problem number 19: http://yufeizhao.com/imo2008/zhao-polynomials.pdf
siddigss
07.03.2017 00:04
Assume that $P$ has a root $|z|>1$. We have $$0=|(z-1)P(z)|=|a_nz^{n+1}+(a_{n-1}-a_n)z^n+(a_{n-2}-a_{n-1})z^{n-1}+...+(a_0-a_1)z-a_0|$$$$\geq |a_nz^{n+1}|-|(a_{n-1}-a_n)z^n|-|(a_{n-2}-a_{n-1})z^{n-1}|...-|a_0|$$$$=a_n|z|^{n+1}-(a_{n}-a_{n-1})|z|^n-(a_{n-1}-a_{n-2})|z|^{n-1}-...-|a_0|$$$$>a_n|z|^{n+1}-(a_{n}-a_{n-1})|z|^{n+1}-(a_{n-1}-a_{n-2})|z|^{n+1}-...-|a_0||z|^{n+1}=0$$Contradiction!
sqing
07.03.2017 10:28
rmtf1111 wrote: If someone finds it useful, this problem is present here as problem number 19: http://yufeizhao.com/imo2008/zhao-polynomials.pdf Nice. Thanks. Let $a, b, c$ be nonzero integers such that both $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}$ and $\frac{b}{a}+\frac{c}{b}+\frac{a}{c}$ are integers. Prove that $|a| = |b| = |c|.$