$x^4+y^4+z^4=\frac{m}{n}$, 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.

 A: 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 $m+n=4$.
A: Notice if $(x,y,z)$ satisfies the two conditions, so does $(-x,-y,-z)$. Furthermore, it is clear $x,y,z$ have the same sign. Since $(x,y,z)$ and $(-x,-y,-z)$ gives the same $\frac{m}{n}$, we only need to consider the case $x,y,z > 0$.
Take three numbers $\alpha,\beta,\gamma \in (0,\frac{\pi}{2})$ such that
$$x = \tan\alpha, y = \tan\beta, z = \tan\gamma$$
Recall the three angle summation formula for tangent, we have
$$\begin{align}\tan(\alpha+\beta+\gamma) 
&= \frac{\tan\alpha + \tan\beta + \tan\gamma - \tan\alpha\tan\beta\tan\gamma}{1 - \tan\alpha\tan\beta - \tan\beta\tan\gamma - \tan\gamma\tan\alpha}\\
&= \frac{x+y+z - xyz}{1 -xy - yz -zx}\\&= \infty\end{align}$$
This implies $\alpha + \beta + \gamma = (N + \frac12)\pi$ for some integer $N$.
Since $\alpha + \beta + \gamma \in (0,\frac32\pi)$, this forces $N = 0$ and
hence $\alpha + \beta + \gamma = \frac{\pi}{2}$.
Notice
$$\frac{2}{x+\frac1x} = \sin(2\alpha),\quad
\frac{2}{y+\frac1y} = \sin(2\beta)\quad\text{ and }\quad
\frac{2}{z+\frac1z} = \sin(2\gamma)$$
the condition
$$5\left(x+\frac1x\right) = 12\left(y  + \frac1y\right) = 13\left(z + \frac1z\right)$$
is equivalent to
$$\sin(2\alpha) : \sin(2\beta) : \sin(2\gamma) = 5 : 12 : 13 \tag{*1}$$
Since $5^2 + 12^2 = 13^2$, we find
$$\sin^2(2\alpha) + \sin^2(2\beta) = \sin^2(2\gamma)\tag{*2}$$
With a little bit of algebra, one can verify when $\alpha + \beta + \gamma = \frac{\pi}{2}$, one has the "cosine rule":
$$\cos(2\gamma) = \frac{\sin^2(2\alpha) + \sin^2(2\beta) - \sin^2(2\gamma)}{2\sin(2\alpha)\sin(2\beta)}$$
$(*2)$ tell us  $\cos(2\gamma) = 0 \implies \gamma = \frac{\pi}{4} \implies z = \tan\gamma = 1$. Plug this into $(*1)$, we get
$$
\begin{cases}
\sin(2\alpha) = \frac{5}{13}\\
\sin(2\beta)  = \frac{12}{13}
\end{cases}
\quad\implies\quad
\begin{cases}
\cos(2\alpha) = \frac{12}{13}\\
\cos(2\beta)  = \frac{5}{13}
\end{cases}
$$
This leads to
$$\begin{align}
x &= \tan \alpha = \sqrt{\frac{1-\cos(2\alpha)}{1 + \cos(2\alpha)}}
= \sqrt{\frac{1-\frac{12}{13}}{1+\frac{12}{13}}} = \frac15\\
y &= \tan \beta = \sqrt{\frac{1-\cos(2\beta)}{1 + \cos(2\beta)}}
= \sqrt{\frac{1-\frac{5}{13}}{1+\frac{5}{13}}} = \frac23\\
\end{align}
$$
As a result
$$\frac{m}{n} = x^4 + y^4 + z^4 = \frac1{5^4} + \frac{2^4}{3^4} + 1^4 = \frac{60706}{50625}$$
This leads to $m + n = 60706 + 50625 = 111331$.
