There are real numbers $x, y$ such that $x \ne 0$, $y \ne 0$, $xy + 1 \ne 0$ and $x + y \ne 0$. Suppose the numbers $x + \frac{1}{x} + y + \frac{1}{y}$ and $x^3+\frac{1}{x^3} + y^3 + \frac{1}{y^3}$ are rational. Prove that then the number $x^2+\frac{1}{x^2} + y^2 + \frac{1}{y^2}$ is also rational.
Problem
Source: Polish Math Olympiad 2021 2nd round p1 day 2
Tags: algebra, retional
WallyWalrus
01.06.2021 20:59
Denote $a=x+\dfrac{1}{x};b=y+\dfrac{1}{y}$. $a+b=\dfrac{(x+y)(xy+1)}{xy}\in\mathbb{Q}\setminus\{0\}\Longrightarrow (a+b)^2\in\mathbb{Q}\Longrightarrow$ $\Longrightarrow a^2+b^2+2ab\in\mathbb{Q}$. $x^3+\dfrac{1}{x^3}+y^3+\dfrac{1}{y^3}=a^3-3a+b^3-3b=(a+b)(a^2-ab+b^2-3)\in\mathbb{Q}\Longrightarrow$ $\Longrightarrow a^2-ab+b^2-3=\dfrac{x^3+\dfrac{1}{x^3}+y^3+\dfrac{1}{y^3}}{a+b}\in\mathbb{Q}\Longrightarrow a^2+b^2-ab\in\mathbb{Q}$. $\begin{cases}a^2+b^2+2ab\in\mathbb{Q}\\a^2+b^2-ab\in\mathbb{Q}\end{cases} \Longrightarrow 3ab\in\mathbb{Q}\Longrightarrow ab\in\mathbb{Q}$. Results: $x^2+\dfrac{1}{x^2}+y^2+\dfrac{1}{y^2}=a^2-2+b^2-2=(a+b)^2-2ab-4\in\mathbb{Q}$.
teomihai
28.12.2021 08:00
WallyWalrus wrote: Denote $a=x+\dfrac{1}{x};b=y+\dfrac{1}{y}$. $a+b=\dfrac{(x+y)(xy+1)}{xy}\in\mathbb{Q}\setminus\{0\}\Longrightarrow (a+b)^2\in\mathbb{Q}\Longrightarrow$ $\Longrightarrow a^2+b^2+2ab\in\mathbb{Q}$. $x^3+\dfrac{1}{x^3}+y^3+\dfrac{1}{y^3}=a^3-3a+b^3-3b=(a+b)(a^2-ab+b^2-3)\in\mathbb{Q}\Longrightarrow$ $\Longrightarrow a^2-ab+b^2-3=\dfrac{x^3+\dfrac{1}{x^3}+y^3+\dfrac{1}{y^3}}{a+b}\in\mathbb{Q}\Longrightarrow a^2+b^2-ab\in\mathbb{Q}$. $\begin{cases}a^2+b^2+2ab\in\mathbb{Q}\\a^2+b^2-ab\in\mathbb{Q}\end{cases} \Longrightarrow 3ab\in\mathbb{Q}\Longrightarrow ab\in\mathbb{Q}$. Results: $x^2+\dfrac{1}{x^2}+y^2+\dfrac{1}{y^2}=a^2-2+b^2-2=(a+b)^2-2ab-4\in\mathbb{Q}$. beautiful solution!