<s>[INST] Solve for $x, y \in \mathbb R$: $x^2+y^2=2x^2y^2$ and $(x+y)(1+xy)=4x^2y^2$ 
Solve the following system of equations. $$\large \left\{ \begin{aligned} x^2 + y^2 &= 2x^2y^2\\ (x + y)(1 + xy) &= 4x^2y^2 \end{aligned} \right.$$

From the system of equations, we have that
$$\left\{ \begin{align*} (x + y)^2 \le 2(x^2 + y^2) = 4x^2y^2\\ 4x^2y^2 = (x + y)(1 + xy) \le \frac{[(x + 1)(y + 1)]^2}{4} \le \frac{(x + y + 2)^4}{4^3} \end{align*} \right.$$
$$\implies (x + y)^2 \le \frac{(x + y + 2)^4}{4^3} \implies |x + y| \le \left|\frac{x + y + 2}{2^3}\right|$$
I don't know what to do next. Please help me solve this problem.
 [/INST] \begin{align*}
(x+y)^2(1+xy)^2&=(4x^2y^2)^2\\
(x^2+y^2+2xy)(1+xy)^2&=16x^4y^4\\
(2x^2y^2+2xy)(1+xy)^2&=16x^4y^4\\
2xy(1+xy)^3&=16x^4y^4
\end{align*}
So, $xy=0$ or $1+xy=2xy$.
$xy=0$ or $1$.
If $xy=0$, $x=y=0$.
If $xy=1$, $x=y=1$.
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