Diofuntime Equation

Dividing by $y^2$ and solving the resulting quadratic we find $\frac{x}{y}=\frac{2k+1 \pm \sqrt{4k+1}}{2k}$ .

Since $x,y$ are positive integers, this has to be rational, so $4k+1$ is an odd perfect square, and $k=a(a+1)$ for some positive integer $a$ .

This condition is necessary, and it is also sufficient: any such $k$ will give a positive rational result that can be expressed (by definition) as the ratio of two integers $x,y$ .

So $\sum_{k\in G} \frac1k=\sum_{a=1}^{\infty} \frac{1}{a(a+1)}=\sum_{a=1}^{\infty} \left( \frac{1}{a}-\frac{1}{a+1} \right)=\boxed1$

The given equation $ky^2 - (2k+1)xy + kx^2 = 0$ can be seen as a quadratic equation for $y$ . Solving $y$ in terms of $x$ , we have $y = \dfrac {2k+1+\sqrt{4k+1}}{2k}x$ . For $y$ to be an integer, $4k+1$ , which is odd, must be a square of an odd integer. :Let $4k + 1 = (2n+1)^2$ , where $n$ is an non-negative integer. Then,

$\begin{aligned} 4k + 1 & = 4n^2 + 4n + 1 \\ \implies k & = n(n+1) \end{aligned}$

Therefore,

$\begin{aligned} \sum_{k\in G} \frac 1k & = \sum_{n=1}^\infty \frac 1{n(n+1)} & \small \color{#3D99F6} \text{See note.} \\ & = \sum_{n=1}^\infty \left(\frac 1n - \frac 1{n+1} \right) \\ & = \boxed 1 \end{aligned}$

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Note that for $4k+1 = (2n+1)^2$ $\implies k = n(n+1)$ for $n \in \mathbb N$ , Then

$\begin{aligned} y & = \frac {2k+1+2n+1}{2k}x \\ \implies \frac yx & = \frac {k+n+1}k = \frac {n(n+1)+n+1}{n(n+1)} = \frac {n+1}n \end{aligned}$

This means that there is a solution for every $n \in \mathbb N$ and therefore a $k$ for every $n$ , hence $\displaystyle \sum_{k\in G} \frac 1k = \sum_{n=1}^\infty \frac 1{n(n+1)}$ is justified.

$\text{Solution by Sam Zhou:}$

$kx^{2}-(2k+1)xy+ky^{2}=0$

$\text{Rearranging we get } k=\frac{xy}{(x-y)^2}$

$1+4k=1+4 \cdot \frac{xy}{(x-y)^2}= \frac{(x-y)^2+4xy}{(x-y)^2}=(\frac{x+y}{x-y})^2 \text{, so } 1+4k \text{ must be the square of an integer.}$

$m^2 \equiv 1 \pmod{4} \text{ for odd } m \in \mathbb{Z^{+}} \text{ , so } 1+4k \text{ must be the square of an odd integer.}$

$\text{Let } k_{n}<k_{n+1}.\, \forall \, n\in\mathbb{Z^{+}}, \exists \,\, k_n \, \text{s.t.} \, 1+4k_n=(2n+1)^2 \text{ and there is always a pair } (x_n,y_n) \text{ that satisfies } \frac{x_n+y_n}{x_n-y_n}=2n+1 \text{ (i.e. } x=n+1, y=n\text{)}$

$4k_n=(2n+1)-1=4n^2+4n \Rightarrow k_n=n^2+n=n(n+1)\, \forall \,n \in \mathbb{Z^{+}}. \text{ There exist an infinite number of values for k.}$

$\displaystyle \therefore \sum_{x \in G} x^{-1} = \sum_{i=1}^n \frac{1}{k_i} = \sum_{i=1}^{\infty} \frac{1}{i(i+1)}=\sum_{i=1}^{\infty} (\frac{1}{i} - \frac{1}{i+1}) =1\, \text{ since the series is a telescoping sum.}$

$\blacksquare$

From the given conditions of the problem, k must be of the form n(n+1), where n is a positive integer. Therefore $\dfrac{1}{k}$ = $\dfrac{1}{n}$ - $\dfrac{1}{n+1}$ . This results in a telescopic sum whose value is 1.