Let $x,y,$ and $z$ be real numbers satisfying the system of equations
\begin{align*} xy + 4z &= 60 \\ yz + 4x &= 60 \\ zx + 4y &= 60. \end{align*}
Let $S$ be the set of possible values of $x.$ Find the sum of the squares of the elements of $S.$
We first subtract the second equation from the first, noting that they both equal $60$. 
\begin{align*} xy+4z-yz-4x&=0 \\ 4(z-x)-y(z-x)&=0 \\ (z-x)(4-y)&=0 \end{align*}
Case 1: Let $y=4$.
The first and third equations simplify to:
\begin{align*} x+z&=15 \\ xz&=44 \end{align*}
from which it is apparent that $x=4$ and $x=11$ are solutions.
Case 2: Let $x=z$.
The first and third equations simplify to:
\begin{align*} xy+4x&=60 \\ x^2+4y&=60 \end{align*}
We subtract the following equations, yielding:
\begin{align*} x^2+4y-xy-4x&=0 \\ x(x-4)-y(x-4)&=0 \\ (x-4)(x-y)&=0 \end{align*}
We thus have $x=4$ and $x=y$, substituting in $x=y=z$ and solving yields $x=6$ and $x=-10$.
Then, we just add the squares of the solutions (make sure not to double count the $4$), and get \[4^2+11^2+(-6)^2+10^2=16+121+36+100=\boxed{273}.\]
~SAHANWIJETUNGA