4
$\begingroup$

This is a follow-up to a question posted recently. Let $s_n = \sum_{r=1}^{n} \frac{1}{r(r+1)},$ where we take $s_0 = 0$.

The problem I am interested in is this: For fixed $k \geq 2$, find all solutions $(m,n)$ in nonnegative integers to the Diophantine equation $s_m - s_n = \frac{1}{k}.$

Current state of knowledge (see the original question):

  1. $s_m - s_n$ can be expressed as $s_m - s_n = \frac{m-n}{(m+1)(n+1)},$

  2. At least some of the solutions are given by taking each divisor $a$ of $k$ such that $a > 1$ and setting $m = (a-1)k-1,$ $n = \frac{(a-1)k}{a} - 1.$

1 Answers 1

4

We are to solve $\displaystyle \frac{1}{x} - \frac{1}{y} = \frac{1}{k}$

This can be rearranged to

$(k+y)(k-x) = k^2$

So each divisor \displaystyle a > k of $\displaystyle k^2$ will give a solution and all solutions are gotten by that. So just by taking divisors of $\displaystyle k$, you might be missing some solutions for many values of $\displaystyle k$

For instance for $\displaystyle k=15$ you will be missing the solution $\displaystyle m = 9, n = 5$

  • 0
    Excellent! For example, the construction I describe misses the solution (in your formulation) $x = 2$, $y = 3$ in the case $k = 6$, which comes from the factorization of $36 = 9 \cdot 4$.2010-10-19