Problem 180


Problem 180


Rational zeros of a function of three variables

For any integer n, consider the three functions

f1,n(x,y,z) = xn+1 + yn+1 − zn+1
f2,n(x,y,z) = (xy + yz + zx)*(xn-1 + yn-1 − zn-1)
f3,n(x,y,z) = xyz*(xn-2 + yn-2 − zn-2)

and their combination

fn(x,y,z) = f1,n(x,y,z) + f2,n(x,y,z) − f3,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 ≤ k and there is (at least) one integer n, so that fn(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

f1,n(x,y,z) = xn+1 + yn+1 − zn+1
f2,n(x,y,z) = (xy + yz + zx)*(xn-1 + yn-1 − zn-1)
f3,n(x,y,z) = xyz*(xn-2 + yn-2 − zn-2)

以及它们的组合

fn(x,y,z) = f1,n(x,y,z) + f2,n(x,y,z) − f3,n(x,y,z)

我们称(x,y,z)为k阶黄金三元组,如果x,y和z都是形如a / b的有理数,其中
0 < a < b ≤ k,且(至少)存在一个整数n使得 fn(x,y,z) = 0。

令s(x,y,z) = x + y + z。
对于所有不同的35阶黄金三元组(x,y,z),s(x,y,z)的和记为t = u / v。
所有的s(x,y,z)以及t都是最简分数。

求u + v。