<s>[INST] 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.
 [/INST] 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):
$$(x, y) = \left(\frac{3 \pm \sqrt{5}}{2},2\pm \sqrt{3}\right)
\text{ or }
\left(2\pm \sqrt{3},\frac{3 \pm \sqrt{5}}{2}\right)
$$

</s>