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

Question

$x^4+y^4+z^4$=$m\over n$

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.

Answer 1

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 $m+n=4$.

Answer 2

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 $m + n = 60706 + 50625 = 111331$.