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 $$m = (a-1)k-1,$$ $$n = \frac{(a-1)k}{a} - 1.$$


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 $\displaystyle m = 9, n = 5$

  • $\begingroup$ 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 $36 = 9 \cdot 4$. $\endgroup$ – Mike Spivey Oct 19 '10 at 6:15

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