MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I posted this question on Math.SE and it has been a while that no non-trivial hint has been suggested. I hope MOers will have something to say or will solve it entirely. Here is the question copied verbatim:

I stumbled upon this number theory problem while I was solving another problem. Here is the equation: $$3^kn + 3^{k-1} + 2^m(3^{k-1} + 2h) = 2^{m+l}n$$ where $k \geq 3, h,l,m,n\in\mathbb{N}$, $n$ is odd, not a multiple of $3$ and $n\geq 7$. My impression is that it does not have a solution. However, I have not progressed on the problem anymore than that. Could you please help?


share|cite|improve this question
The equation implies $3n\equiv2^m-1\pmod{2^{m+1}}$, but that's about the only nontrivial restriction (given $n,m,k$ satisfying this congruence, you can complete it to a solution (I think)). For example, $m=2$, $n=1$, $k=3$, $h=7$, $l=5$ solves the equation. – Emil Jeřábek Mar 17 '11 at 15:36
Collatz strikes again. Gerhard "Hates Wool Over His Eyes" Paseman, 2011.03.17 – Gerhard Paseman Mar 17 '11 at 15:40
@Emil, thanks for your comment. That's interesting but I want nontrivial solution for $n\geq 6$. I will include it in the question. – Chulumba Mar 17 '11 at 15:46
@Gerhard, could you please tell me why you said that? – Chulumba Mar 17 '11 at 15:50
If you take $n$ such that $3n+1=2^m$, your equation simplifies to $3^{k-1}+h=2^{l-1} n$, which has plenty of solutions because you just set $h$ to be what you want. By the way, your equation doesn't look very natural, so it would be good if you could give us some more motivation. – François Brunault Mar 17 '11 at 16:13

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.