Given $(\frac{n+1}{n})^a\cdot (\frac{m+1}{m})^b = 2$ where a, b, n, and m are all positive integers, are there infinitely many solutions $(a,b,n,m)$?

  • 1
    $\begingroup$ you mean that for every a,b there's inf .valid (n,m) pair? $\endgroup$ – Dleep May 20 '15 at 2:22
  • $\begingroup$ Not necessarily. There may be a,b with no valid n,m. But I'm wondering about n,m pairs when considering all possible a,b. $\endgroup$ – Dan D May 20 '15 at 2:25
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    $\begingroup$ Seems there are only $\frac{4}{3}\cdot\frac{3}{2} = \frac{9}{8}\cdot\left(\frac{4}{3}\right)^2 = 2$. $\endgroup$ – Alexey Burdin May 20 '15 at 3:47
  • $\begingroup$ That's what I was afraid of. Those are the only two I've found as well. $\endgroup$ – Dan D May 20 '15 at 3:51

Let's try this:

$(n+1)^{a}(m+1)^{b}=2\cdot n^{a}m^{b}$

Obviously either $(n+1)$ or $(m+1)$ must be even but not both. Take $(n+1)=2^{j}\cdot r;(n+1)^{a}=2^{a\cdot j}r^{a}$

Now $n+1$ is relatively prime to $n$ so if $a\cdot j>1$ then $m=2^{k}s$ and we have

$a\cdot j-b\cdot k=1$

Thus $(a,b),(a,j),(j,k),(j,b)$ are all relatively prime.

When we ``reverse engineer'' this we find a problem. We are left with:


Thus every factor of $r^{a}$ is a factor of $s^{b}$ and vise versa; $r=s,a=b$ .

And the positive resolution of.

$a\cdot j-b\cdot k=1,a=b$ is




Your solution

  • $\begingroup$ This is excellent! Thank you! $\endgroup$ – Dan D May 27 '15 at 16:13

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