## For any integer $n$, consider the three functions

\begin{align}

f_{1, n}(x, y, z) &= x^{n + 1} + y^{n + 1} – z^{n + 1}\\

f_{2, n}(x, y, z) &= (xy + yz + zx) \cdot (x^{n – 1} + y^{n – 1} – z^{n – 1})\\

f_{3, n}(x, y, z) &= xyz \cdot (x^{n – 2} + y^{n – 2} – z^{n – 2})

\end{align}

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 $0 \lt a \lt b \le k$ 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 $u + v$.

### This problem involves some higher level mathematical analysis and knowledge of number theory. Here is the detailed solution:

We start by observing the form of our function $f_n(x, y, z)$. We note $f_n(x, y, z)$ to be a homogeneous polynomial of degree $n + 1$ – each term has total degree $n + 1$ when summing the powers of $x$, $y$ and $z$. Because of this homogeneity, provided that $f_n(x, y, z) = 0$ for some $n$, then scaling $x, y, z$ by the same factor will also give a root, ie: $f_n(ax, ay, az) = 0$.

Following this, for a given $(x, y, z)$ we can thus scale these values so that $x,y,z$ are all integers. Since they were originally rational numbers of the form $a / b$ with $a < b$, we know that our integer scaling factors will at most need to be up to $35$. Hence the integers fall within the range: $-35 \le x,y,z \le 35$. A brute force search within this range will then yield integer solutions to $f_n(x, y, z)$ for some $n$. Note that for a given $(x, y, z)$, we have that $f_{1,n}(x,y,z), f_{2,n}(x,y,z), f_{3,n}(x,y,z)$ are in arithmetic progression with common difference $f_{1,n-1}(x,y,z)$. Hence if we have a triple $(x, y, z)$ such that $f_n(x, y, z) = 0$, then we also have $f_{n+1}(x, y, z) = 0$. The problem statement implies that $f_{1,n}(x,y,z) \neq 0$ (Otherwise, we would have infinitely many solutions). So we first check each integer triple in the range for being a solution to $f_{1,n}(x,y,z) = 0$, which happens only for $(x,y,z) = (2,3,5)$ with $n=2$. This triple generates other solutions to $f_n(x, y, z) = 0 $ for higher $n$ values. Doing this, we are then able to find our golden triples within the $-35 \le x,y,z \le 35$ range to be: $(2/3, 1, 1/2)$, $(2/3, 25/36, 1/180)$, $(2/5, 3/5, 1/2)$, and $(1/14, 10/21, 25/42)$. Each of these distinct $s(x, y, z)$ sums can be added to get $t = 875 / 252$. Finally to get $u + v$, the sum of the numerator and the denominator, we get $875 + 252 = 1127$.

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