$x^4+y^4+z^4=\frac{m}{n}$, find $m+n$. $x^4+y^4+z^4$=$m\over n$
x, y, z are all real numbers,
satisfying $xy+yz+zx=1$ and $5\left(x+\frac{1}{x}\right)=12\left(y+\frac{1}{y}\right)=13\left(z+\frac{1}{z}\right)$
m, n are positive integers and their greatest common divisor is 1. Calculate m+n.
My thinkings so far are as follows:

*

*to operate such that we can arrange 5,12,13 into some kind of pythagorean triplets.

*to do whole square of x+1/x or x^2 + 1/x^2 such that
the third term is a constant and can be moved to other side.


Taking resultants of the polynomials given by
$$
f=5\left(x+\frac{1}{x}\right)-12\left(y+\frac{1}{y}\right),g=5\left(x+\frac{1}{x}\right)-13\left(z+\frac{1}{z}\right), h=xy+yz+zx-1,
$$
set to zero, i.e., with
\begin{align*}
0 & = 5x^2y - 12xy^2 - 12x + 5y,\\
0 & = 5x^2z - 13xz^2 - 13x + 5z,\\
0 & = xy + xz + yz - 1
\end{align*}
we obtain the linear equation $13z=12y+25x$, and then by substituting $z$ and taking resultants again,
$$
(x,y,z)=\left(\frac{1}{5},\frac{2}{3},1\right),\; \left(-\frac{1}{5},-\frac{2}{3},-1\right).
$$
Hence we have
$$
x^4+y^4+z^4=\frac{60706}{50625}.
$$
So we have $(m,n)=(60706,50625)$, so that
$$
m+n=111331.
$$
Note: Over the complex numbers we obtain two additional solutions, namely
$$
(x,y,z)=(i,-i,i),(-i,i,-i).
$$
But then $x^4+y^4+z^4=3$, so that $m+n=4$.
