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I currently have the expression $2^a+1=3^bc$, where $a$, $b$ and $c$ must be positive integers. I am looking for the possible values of $a$ in relation to $b$. I know the solution is $a=3^{b-1}(2n+1)$, where $n$ can be a positive integer, by trial and error, however I cannot seem to come up with a solid proof for this. I have tried using logarithms and different series but it has not worked so far, so I was wondering if anyone could give me a push in the right direction.

EDIT: A rephrasing of my question is: "What values of $a\in\mathbb{N}$ ensure that $2^a+1$ is some multiple of $3^b$?" - if possible giving a solution using elementary methods.

Thanks :)

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  • $\begingroup$ $c$ is not the solution you know? $\endgroup$ – Henrik - stop hurting Monica Jul 3 '17 at 19:45
  • $\begingroup$ What do you mean? I'm looking for the value of $a$ such that $b$ and $c$ are integers and $2^a+1=3^bc$ $\endgroup$ – Daniel Castle Jul 3 '17 at 19:47
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    $\begingroup$ Write $2 = (-1) + 3^1 + k\cdot 3^2$. Raise to the third power (binomial expansion) to find $2^{3^1} = (-1) + 3^2 + m\cdot 3^3$. Repeat to find something about $2^{3^n}$. $\endgroup$ – Daniel Fischer Jul 3 '17 at 19:51
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    $\begingroup$ I suggest rephrasing the question as: what constraints on $a\in\mathbb{N}$ ensure that $2^a+1$ is a multiple of a fixed power of $3$ ? $\endgroup$ – Jack D'Aurizio Jul 3 '17 at 20:05
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    $\begingroup$ You may just try to re-prove by elementary techniques that $2^a+1$ is a multiple of $3^b$ if and only if $a$ is an odd multiple of $3^{b-1}$. $\endgroup$ – Jack D'Aurizio Jul 3 '17 at 20:16
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Let $\nu_3(n)=\max\{m\in\mathbb{N}: 3^m\mid n\}$. Since $2^a+1\equiv(-1)^a+1\pmod{3}$, $3$ is a divisor of $2^a+1$ iff $a$ is odd. Since $2$ is a generator for $\mathbb{Z}/(9\mathbb{Z})^*$ we have that $9\mid(2^a+1)$ iff $a$ is odd and a multiple of $3$. By Hensel's lifting lemma we have that $2$ is a generator for $\mathbb{Z}/(3^k\mathbb{Z})^*$, hence for any $k\geq 1$ we have that $3^k$ is a divisor of $2^a+1$ iff $a$ is odd and $a$ is a multiple of $3^{k-1}$:

$$ \nu_3(2^a+1) = \mathbb{1}_{\text{odd}}(a)\cdot\left(\nu_3(a)+1\right).\tag{1} $$ In particular, if $2^a+1 = 3^b\cdot c$, then $\color{red}{a\geq 3^{b-1}}$.

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    $\begingroup$ Lifting The Exponent Lemma is relevant. $\endgroup$ – user26486 Jul 3 '17 at 20:52
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Basically it is equivalent to find some $a\in \mathcal{N}$ such that $2^a+1$ can be divided by 3. It is easy to find that all the odd numbers is a solution to $a$, but the even numbers are not. Now $b$ and $c$ can be easily obtained once you get $a$.

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  • $\begingroup$ Ok, but wouldn't that imply $b$ is 1? It also has to be divisible by $3^b$ not just $3$. I'm looking for a set of solutions for $a$ in terms of $b$ $\endgroup$ – Daniel Castle Jul 4 '17 at 0:31
  • $\begingroup$ For $b=1$, it is easy to see $a=2k+1$. For $b=2$, $a=6k+3$. For $b=3$, $a=18k+9$ $\endgroup$ – Distance Jul 4 '17 at 1:14
  • $\begingroup$ Yes, this is the result I found - however I have not found not any concrete or intuitive way of deriving it (without using maths out of my scope of understanding) $\endgroup$ – Daniel Castle Jul 4 '17 at 1:36

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