<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] These are symmetric equations, so denoting $s=x+y,p=xy$ we have
$$s^2-2p=2p^2$$
$$s(1+p)=4p^2$$
Now eliminate $s$:
$$s^2=2p^2+2p=\left(\frac{4p^2}{1+p}\right)^2$$
$$2p(1+p)^3=16p^4$$
If $p=0$ then $s=0$ and obviously $x=y=0$. Otherwise, divide by $p$:
$$(1+p)^3=(2p)^3$$
Assuming we are only solving in the reals:
$$1+p=2p\qquad p=1,s=2$$
By Viète's formulas, $x$ and $y$ are the roots of $t^2-2t+1$, whereby we get $x=y=1$.
</s>