Solve the following system: $(x^2+1)(y^2+1) = 12xy$ and $(x+1)^2(y+1)^2=30xy$ I have been given the following system to solve:
$$(x^2+1)(y^2+1) = 12xy$$
$$(x+1)^2(y+1)^2=30xy$$
I noticed that the system is symmetric, but any of the methods I know to solve symmetric systems doesn't seem right. I have also tried writing the second equation as $$((x^2+1)+2x)((y^2+1)+2y) = (x^2+1)(y^2+1)+2y(x^2+1)+2x(y^2+1)+4xy = 30xy$$,
but this didn't get me anywhere. I would appreciate any hints towards the right solution.
 A: Divide both sides of both equations by $xy$ and let $u = x+\frac1x + 1$, $y = y+\frac1y + 1$. We have
$$\begin{cases}(u-1)(v-1) &= 12\\ (u+1)(v+1) &= 30\end{cases}  
\implies \left\{\begin{align}
uv &= \frac12(30+12)-1 = 20\\
u + v &= \frac12(30-12) = 9
\end{align}\right.
$$
This implies $u,v$ are roots of $\lambda^2 - 9\lambda + 20 = (\lambda - 4)(\lambda - 5)$.
Up to permutation of $x, y$, we have
$$
\begin{cases}
x + \frac1x = u - 1 = 3\\
y + \frac1y = v - 1 = 4
\end{cases}
\iff
\begin{cases}
x^2 - 3x + 1 = 0\\
y^2 - 4y + 1 = 0
\end{cases}
$$
This leads to 8 solutions of the problem (see illustrations below):
$$(x, y) = \left(\frac{3 \pm \sqrt{5}}{2},2\pm \sqrt{3}\right)
\text{ or }
\left(2\pm \sqrt{3},\frac{3 \pm \sqrt{5}}{2}\right)
$$

A: $$
\left\{ 
\begin{array}{c}
\left(x^2+1 \right)\left(y^2+1 \right)=12xy \\ 
\left(x+1 \right)^2\left(y+1 \right)^2=30xy
\end{array}
\right. 
$$
$$
\iff \left\{ 
\begin{array}{c}
\left(x^2+1 \right)\left(y^2+1 \right)=12xy \\ 
\left(x^2+1+2x \right)\left(y^2+1+2y \right)=30xy
\end{array}
\right. 
$$
$$
\iff \left\{ 
\begin{array}{c}
\left(x+\frac{1}{x} \right)\left(y+\frac{1}{y} \right)=12 \\ 
\left(x+\frac{1}{x}+2 \right)\left(y+\frac{1}{y}+2 \right)=30
\end{array}
\right. 
$$
Put $A=x+\frac{1}{x}$ and $B=y+\frac{1}{y}$, we have
$$
\left\{ 
\begin{array}{c}
AB=12 \\ 
\left(A+2 \right)\left(B+2 \right)=30
\end{array}
\right. 
$$
Solving the equations, we have $A=3, B=4$ or $A=4, B=3$.
When $A=3, B=4$ , we have
$$
\left\{ 
\begin{array}{c}
x+\frac{1}{x}=3 \\ 
y+\frac{1}{y}=4
\end{array}
\right. 
$$
$$
\left\{ 
\begin{array}{c}
x=\frac{3\pm\sqrt{5}}{2}\\ 
y=2\pm\sqrt{3}
\end{array}
\right. 
$$
When $A=4, B=3$ , we have
$$
\left\{ 
\begin{array}{c}
x+\frac{1}{x}=4 \\ 
y+\frac{1}{y}=3
\end{array}
\right. 
$$
$$
\left\{ 
\begin{array}{c}
x=2\pm\sqrt{3}\\ 
y=\frac{3\pm\sqrt{5}}{2}
\end{array}
\right. 
$$
A: Hint
From a formal point of view, you face an octic polynomial.
Use the first equation and solve for $y$
$$y_\pm=\frac{6x\pm\sqrt{-x^4+34 x^2-1}}{x^2+1}$$ Using $y_+$, plug it in the second, group terms and factor; you should have
$$12 x \left(x^2-4 x+1\right) \left(x^2-3 x+1\right)+$$ $$2 \left(x^2-4 x+1\right) \left(x^2-3 x+1\right)\sqrt{-x^4+34 x^2-1}=0$$ that is to say
$$\left(x^2-4 x+1\right) \left(x^2-3 x+1\right)(6x+\sqrt{-x^4+34 x^2-1})=0$$
Now, it is simple.
Do the same using $y_-$.
A: 
You can find the answer by solving the last equation

