Let $P(x)$ be a polynomial with the coefficients being $0$ or $1$ and degree $2023$. If $P(0)=1$, then prove that every real root of this polynomial is less than $\frac{1-\sqrt{5}}{2}$.
Problem
Source: Azerbaijan NMO 2024. Senior P4
Tags: algebra, polynomial, AZE SENIOR NATIONAL MO
Tintarn
08.07.2024 19:20
Nice problem!
Clearly $P$ has only negative roots. Suppose that $-x$ is a real root so that $0<x \le \varphi=\frac{\sqrt{5}-1}{2}$.
Then $1+x^{g_1}+\dots+x^{g_k}=x^{u_1}+\dots+x^{u_m}$ for even integers $g_1<\dots<g_k \le 2022$ and odd integers $u_1<\dots<u_m=2023$.
Proving that this is impossible is clearly equivalent to showing that $RHS<LHS$ in this equation. But this is equivalent to
\[x+x^3+x^5+\dots+x^{2023}<1\]which is true because the LHS is less than the geometric series $\frac{x}{1-x^2}$ and by asumption we have $x \le 1-x^2$. Done!
wizixez
16.12.2024 12:18
The problem itself is not hard but you need to find two key ideas::: Let $\frac{1-\sqrt{5}}{2}=\zeta $ It is obvious that $P$ only has positive roots. The polynomial $Q(x)=x^{2023}+x^{2021}+...+x^3+x+1$ Let us evaluate $Q(\zeta )$ We know that $\zeta^2-\zeta -1$ so $\zeta =\zeta^2-1$ $Q(x)=x(x^{2022}+x^{2020}+...+x^2+1)+1$ Since $\zeta = \zeta^2-1$ $Q(\zeta )=(\zeta^2-1)(\zeta^{2022}+...+\zeta^2+1)+1=\zeta^{2024}>0$ we get a contradictin since $\zeta \in R$ We know that by quadratic formula for $u \in (\zeta :\zeta +\sqrt{5})$ the expression $x^2-x-1<0$ Lets assume there exists a root $z\in (\zeta :0)$ By the bound of $z$ we say that $z>z^2-1$ So $0=Q(z)=z(z^{2022}+...+z^2+1)+1>(z^2-1)(z^{2022}+...+z^2+1)+1=z^{2024}>0$ we get a contradiction