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.
 A: 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 $x=y=1$.
A: The first equation is equivalent to $(x+y)^2=2xy(1+xy)$.
Now multiplying the second equation by the previous relation yields: $$(x+y)^3=(2xy)^3$$ thus $x+y=2xy$ and then: $$(x+y)^2=2\times 2x^2y^2=2(x^2+y^2)\iff (x-y)^2=0$$
A: \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 $xy=1$, $x=y=1$.
A: 
My simple solution is $(\!x^{2}+ y^{2}- 2x^{2}y^{2}\!)+ \left (\!(\!x+ y\!)(\!1+ x^{2}y^{2}\!)\!\right )= (\!x+ y- 2xy\!)(\!3xy+ x+ y+ 1\!)$

