Problem 180

Rational zeros of a function of three variables

For any integer $n$, consider the three functions

$$f_{1,n}(x,y,z)=x^{n+1}+y^{n+1}-z^{n+1}$$
$$f_{2,n}(x,y,z)=(xy+yz+zx)(x^{n-1}+y^{n-1}-z^{n-1})$$
$$f_{3,n}(x,y,z)=xyz(x^{n-2}+y^{n-2}-z^{n-2})$$

and their combination

$$f_n(x,y,z)=f_{1,n}(x,y,z)+f_{2,n}(x,y,z)-f_{3,n}(x,y,z)$$

We call $(x,y,z)$ a golden triple of order $k$ if $x$, $y$, and $z$ are all rational numbers of the form $a / b$ with $0 <a < b \le k$ and there is (at least) one integer $n$, so that $f_n(x,y,z) = 0$.

Let $s(x,y,z) = x + y + z$.
Let $t = u / v$ be the sum of all distinct $s(x,y,z)$ for all golden triples $(x,y,z)$ of order $35$.
All the $s(x,y,z)$ and $t$ must be in reduced form.

Find $u + v$.

有三个自变量的函数的有理零点

对于任意整数$n$，考虑这三个函数

$$f_{1,n}(x,y,z)=x^{n+1}+y^{n+1}-z^{n+1}$$
$$f_{2,n}(x,y,z)=(xy+yz+zx)(x^{n-1}+y^{n-1}-z^{n-1})$$
$$f_{3,n}(x,y,z)=xyz(x^{n-2}+y^{n-2}-z^{n-2})$$

以及它们的组合

$$f_n(x,y,z)=f_{1,n}(x,y,z)+f_{2,n}(x,y,z)-f_{3,n}(x,y,z)$$

考虑三元组$(x,y,z)$，若$x$、$y$和$z$都可以表示为满足$0 <a < b \le k$的有理数$a/b$，且（至少）存在一个整数$n$使得$f_n(x,y,z)=0$，则称其为$k$阶黄金三元组。

记$s(x,y,z) = x + y + z$。
考虑所有$35$阶黄金三元组$(x,y,z)$所对应的$s(x,y,z)$，记其中所有不同的取值之和为$t = u / v$。
所有的$s(x,y,z)$以及$t$都表示为最简分数。

求$u + v$。