<s>[INST] If $x$ and $y$ are rational numbers and $x^5+y^5=2x^2y^2,$ then $1-xy$ is a perfect square.  Prove that if $x, y$ are rational numbers and
$$ x^5 +y^5 = 2x^2y^2$$
then $1-xy$ is a perfect square.
 [/INST] I would be so tempted to divide by $x^2 y^2$, so I would consider the following cases:
Case a:  
if $x=0$ then that implies $y=0$.
Case b:
if $y=0$ then that would imply $x=0$
Case c:
$x \neq 0, y \neq 0$
Thus dividing by $x^2y^2$ will be legal
$$
\begin{align*}
\frac{x^3}{y^2} - 2 + \frac{y^3}{x^2} = 0 \\
x\left(\frac{x}{y}\right)^2 - 2 +y\left(\frac{y}{x}\right)^2 = 0
\end{align*}
$$
Substitute $u=\left(\frac{x}{y}\right)^2$
$$
\begin{align*}
 xu - 2 + \frac{y}{u} = 0\\
xu^2-2u+y=0
\end{align*}
$$
This is a quadratic in $u$ and since $x$ and $y$ are rationals, $u$ is rational.  The discriminant is $4(1-xy)$, which has to be a perfect square for $u$ to be rational. 
Thus $1-xy$ is a perfect square.
</s>