Proof of an equivalence about powers of $2$ and divisors I found this equivalence:


  For all natural numbers $n &gt; 1$,
  $\exists k\in\mathbf N, n=2^k \iff \neg (\exists p\in\mathbf Z, (2p+1)\ne\pm 1 \wedge (2p+1)\mid n)$


in my textbook. The proof has been omitted. I will like to see the proof and run through it myself. I have tried to search on the web but could not find any.

P.-S. I have translated the equivalence into English from the original language.
This is just not true. Take e.g. $n=6$.    

$ n = 6 = 2 . 3$  but $n$ has an odd divisor other than $+1$ and $-1$.
This divisor is 3.   

Probably in the statement should have been written $∃k \in N,n=2^k$ instead of $n=2k$.    

Almost certainly it's a typo, that typo.     

Proof:  

1) If $n$ has no odd divisor other than $+1$ and $-1$, then n has no prime odd divisor. Then since n>1, it means that only the prime $2$ divides $n$. So the left side follows.   

2) If $n=2^k$ then obviously it has no odd prime factor. So it cannot have an odd divisor other than $+1$ and $-1$.    

All in all, this is too obvious if you have already studied the fundamental theorem of arithmetic. 

Now... I don't know if you know that theorem or not. I used it in this proof (in fact it makes the proof trivial).  

Maybe you're looking for a proof which does not use that theorem, not sure.
Some one has edited the equivalence.
A moderator maybe? I find this very odd and I'm not sure what to do.
The original equivalence was
$\exists k\in\Bbb N: n=2^k\iff\neg(n\text{ has other odd divisors than}\pm 1)$