Does there exist a positive integer $n$ and a) complex numbers $a_0, a_1, \ldots, a_n;$ b) reals $a_0, a_1, \ldots, a_n, $ such that $P(x) Q(x)=x^{2024}+1$ where $P(x)=a_nx^n+\ldots +a_1x+a_0$ and $Q(x)=a_0x^n+a_1x^{n-1}+\ldots+a_n?$
Problem
Source: Serbia Additional IMO TST 2024, P1 (out of 4)
Tags: algebra
30.05.2024 13:49
30.05.2024 14:07
30.05.2024 14:11
30.05.2024 18:57
Cute polynomial problem ! Part a):The answer is yes ! Example: Let $P(X)=aX^{1012}+b$ and $Q(X)=bX^{1012}+a$ and we need that $ab=1$ and $a^2+b^2=0$ or $a^4+1=0$ so take for example $a=cos\frac{\pi}{4}+isin \frac{\pi}{4}$ and $b=\frac{1}{a}$ and it works Part b): take $\omega=cos \frac{2\pi}{3}+isin \frac{2\pi}{3}$ Denote $a=a_0+a_3+\dots+ a_{1011}$ $b=a_1+a_4+\dots +a_{1012}$ $c=a_2+a_5+\dots+ a_{1010}$ Plugging $X=\omega$ we have : $$(a+b \omega+c \omega^2)(b+a \omega+c \omega^2)=-\omega$$and denote $x=a-c$ and $y=b-c$ when we open the paranthesis we have : $x^2+y^2-xy=-1$ which is obviously a contradiction($x^2+y^2-xy\ge 0$)