Let $x$ and $y$ be rational numbers, such that $x^{5}+y^{5}=2x^{2}y^{2}$. Prove that $1-xy$ is the square of a rational number.
Problem
Source: IMOTC 2014 Practice Test 2 Problem1
Tags: quadratics, algebra, quadratic formula, algebra unsolved
xxp2000
12.07.2014 22:48
$(x^3-y^2)^2=y^4(1-xy)$ $1-xy=1$ when $y=0$; $1-xy=\left (\dfrac{x^3}{y^2}-1\right )^2$ when $y\ne 0$.
professordad
13.07.2014 07:42
Set $xy=a$, $a \in \mathbb{Q}$. We know
\[x^5+\frac{a^5}{x^5} = 2a^2 \Longrightarrow x^{10}-2a^2x^5+a^5=0\] $y^{10}-2a^2y^5+a^5$ also equals 0. It follows from Quadratic Formula that $x^5$ and $y^5$ are $a^2(1 \pm \sqrt{1-a})$. Since they're both rational, $\sqrt{1-a}$ must be rational, so $1-xy$ is the square of a rational number.
sayantanchakraborty
13.07.2014 19:08
I think we can proceed more easily in the following way: Let $x=\frac{x_1}{k},y=\frac{y_1}{k}$,where $x_1,y_1,k \in \mathbb{Z}$ and $k \neq 0$.This follows from the fact that we can make the denominators of the two fractions same by computing the lcm.Then by the given condition we have ${x_1}^5+{y_1}^5=2{x_1}^2{y_1}^2k \Rightarrow k=\frac{{x_1}^5+{y_1}^5}{2{x_1}^2{y_1}^2}$.So $1-xy$ $=1-\frac{x_1y_1}{k^2}$ $=1-\frac{4{x_1}^5{y_1}^5}{({x_1}^5+{y_1}^5)^2}$ $=(\frac{{x_1}^5-{y_1}^5}{{x_1}^5+{y_1}^5})^2$
Sardor
14.07.2014 12:17
We have $ x^{10}+y^{10}+2x^5y^5=4x^4y^4 $ then $ (x^5-y^5)^2=4x^4y^4(1-xy) $ or $ 1-xy=\frac{(x^5-y^5)^2}{(2x^2y^2)^2} $
Smita
21.06.2018 17:30
professordad wrote:
Set $xy=a$, $a \in \mathbb{Q}$. We know
\[x^5+\frac{a^5}{x^5} = 2a^2 \Longrightarrow x^{10}-2a^2x^5+a^5=0\]$y^{10}-2a^2y^5+a^5$ also equals 0. It follows from Quadratic Formula that $x^5$ and $y^5$ are $a^2(1 \pm \sqrt{1-a})$. Since they're both rational, $\sqrt{1-a}$ must be rational, so $1-xy$ is the square of a rational number.
Beautiful solution