is this proof and approach correct? $(n$ is a power of $2) \iff \neg(n$ has odd divisors other than the trivial $\pm1)$ It is a bicondition and thus I have to prove the both directions. I want to use a direct proof and a proof by contraposition. The direct - in short - is just that The prime factorization of  $n=2^x$  is  $2^x$  and is unique so no other divisors that are not factored - only by power of $2$.
The contraposition is
Assume the opposite, that $n = 2^m$ ($m$ a positive integer) is divisible by the odd number $2D + 1$, where $D$ is a positive integer. That is, $2^m = (2D + 1)(Q)$, where $Q$ is the positive integer quotient.
Since the left side is an even number, $Q$ must be an even number too because the product of two odd numbers is odd. So $Q = 2R$, for a positive integer $R$. Therefore,
$2^m = (2D + 1)(2R)$.
dividing both sides by $2$ yields
$2^{m-1} = (2D + 1)(R)$
Repeat this process until either the power of $2$ on the left side becomes $1$, or the quotient on the right side becomes $1$.
But then the left side will be even but right side will be odd. A contradiction. Therefore the original statement must be true.
Is this proof and approach correct? Any feedback is much appreciated.
 A: Let me summarize the discussion below the question, and continue in a formal answer.
You have a descent argument which inevitably reaches one of two cases:

*

*(the $R=1$ case) $2^k=2D+1$ for some $1\leq k\leq m$.

*(the exponent $1$ case) $2=(2D+1)R$ for some positive integer $R\leq Q$.

Your current proof describes why the first case is a contradiction, but overlooks the second case. For the second case, you want to argue this is impossible using what you know about the parameters involved. In particular, the assumption is that $n$ has a nontrivial odd divisor. So $2D+1$ is not $1$. Note that this fact has not yet been used anywhere in the proof. Indeed, the resolution of case 2 is precisely where this assumption becomes relevant. Specifically, if $2D+1$ is not $1$, then what does this say about $D$? More to the point, what does this say about the number $(2D+1)R$, which is supposedly equal to $2$?
A: Your proof might be okay, but it's a lot simpler than that.
$n$ has a unique prime factorization.  Either that prime factorization contains prime divisors other than $2$ or it does not.
If it does then $n$ is not a power of $2$ and those other prime divisors are all non trivial odd divisors.
If it does not then $n$ is a power of $2$ and all divisors are of the form $2^j$ and are either even or trivial.
So either $n$ is power of $2$ with no non-trivial odd divisors, or $n$ is not power of $2$ and had non-trivial odd divisors.  So $n$ is a power of $2\iff $ $n$ has no non-trivial odd divisors.
(Might be worth noting $[(P \land Q) \lor (\lnot P \land \lnot Q)]\iff (P \leftrightarrow Q).$
(Also might be worth noting that if $n = 1=2^0$ it falls under the category: $1$ is a power of $2$; the unique prime factorization of $1$ contains no prime divisors other than $2$ (and it doesn't contain $2$ either-- it has no prime divisors) and has not non-trivial odd divisors.)
