Prove that the $GCD(2^{k}, n) = 1$ given that $GCD(2, n) = 1 $ and n is odd Not too sure how to approach this problem, 
If  $GCD(2, n) = 1 $, this implies that $2x + ny = 1$ for some integer $x,y$. How do we show the statement given this information? (If it is in fact possible)
Thanks
 A: Hint: I wouldn't use Bezout's here (although you probably can). Assume $\gcd(2^k,n)\neq 1$. Then there exists a prime $p$ that divides both $2^k$ and $n$. What options are there for that prime?
A: If you want to use Bezout's identity, first, I would note that $\gcd(2, n)=1$ means $n$ is not a multiple of $2$, so $n$ is odd and $n=2x+1$ for some $x\in \Bbb Z$. Now, just use induction.
Base Case: $\gcd(2^1, n)=1\rightarrow$ This is given.
Induction: Assume $\gcd(2^{k-1}, n)=1$. Now, we will prove $\gcd(2^k, n)=1$.
Note that:
$$2^{k-1}n-x2^k=2^{k-1}(2x+1)-2^kx=2^kx+2^{k-1}-2^kx=2^{k-1}$$
Thus, by Bézout's identity, $\gcd(2^k, n)$ is a divisor of $2^{k-1}$. Furthermore, $\gcd(2^k, n)$ is a divisor of $n$. Therefore, $\gcd(2^k, n) \leq \gcd(2^{k-1}, n)=1$, so since the GCD must be positive, $\gcd(2^k, n)=1$.

Here's a more intuitive proof using the Fundamental Theorem of Arithmetic. First, all divisors of $2^k$ are in the form of $2^l$ for some non-negative $l\in \Bbb{Z}$. I will prove this by proving the contrapositive: If $a$ is not in the form $2^l$, then it is not a divisor of $2^k$ for $k \geq 1$.
Let's say $a\neq 2^l$ for all $l\in \Bbb{Z}$. Note that this means $a\neq 1$ since $a\neq 2^0$. Thus, by the Fundamental Theorem of Arithmetic, $a=\Pi_{i=1}^m p_i^{e_i}$ where $m$ is the number of prime factors of $a$ and $p_i$ are all the prime factors of $a$. If $2$ is the only prime factor of $a$, then $a=2^{e_i}$, which contradicts the assumption. Thus, $a$ has some prime factor other than $2$. However, by the Fundamental Theorem of Arithmetic, prime factorization is unique, so the only prime factor of $2^k$ is $2$. Thus, $a$ has some prime factor that $2^k$ does not have, meaning $a$ is not a divisor for $2^k$.
Therefore, all divisors of $2^k$ are in the form of $2^l$ for some non-negative $l\in \Bbb Z$.
Now, this means $\gcd(2^k, n)=2^l$ for some $l\in \Bbb Z$ since $\gcd(2^k, n)$ is a divisor of $2^k$. However, if $\gcd(2^k, n)$ is even, since $\gcd(2^k, n)$ is a factor of $n$, that would also imply $n$ is even, which contradicts $\gcd(2, n)=1$. Thus, $\gcd(2^k, n)=2^l$ is odd, so $l=0$. This means $\gcd(2^k, n)=1$, which is what we set out to prove.
A: In some integral domain, let $p$ be a nonzero element satisfying the the conclusions of Euclid's lemma (i.e., $p$ is a nonunit such that $p\mid b$ or $p\mid c$ whenever $p\mid bc$). We show that if $p$ is coprime to $n$ then $p^k$ is coprime to $n$.
This holds for $k=0$, so assume inductively that $p^k$ is coprime to $n$, and consider any common divisor $d$ of $p^{k+1}$ and $n$. Then by definition there exists an $a$ such that $$ad=p^{k+1}=p^k\cdot p\text{.}$$
Therefore $p\mid ad$, and since $p$ satisfies the conclusions of Euclid's lemma, either $p\mid a$ or $p\mid d$. But if $p$ were to divide $d$, we would have a nonunit common divisor of $p$ and $n$, which contradicts the hypotheses. Therefore $p\mid a$, so by definition there exists a $b$ such that $a=pb$. Then
$$pbd=p^k\cdot p$$
so that $bd=p^k$, i.e., $d\mid p^k$. But since $d\mid n$, we have that $d$ must be a unit since $p^k$ and $n$ are coprime by the inductive assumption. Since $d$ is an arbitrary common factor of $p^{k+1}$ and $n$, we have shown that $p^{k+1}$ and $n$ are coprime whenever $p^k$ and $n$ are, so by induction we are done.
A: Take the $k$'th power of the Bezout identity and use the Binomial Theorem to expand it
$$n=2j\!-\!1\,\Rightarrow\, \color{#c00}{\bf 1}^{\large k}\! = (2j\!-\!n)^{\large k}\! = \color{#0a0}{2^{\large k}} j^{\large k} \!+\! \color{#0a0}n\,(\cdots)\ \ \ {\rm so}\ \ \ d\mid \color{#0a0}{2^k,n}\Rightarrow\, d\mid\color{#c00}{\bf 1}\qquad $$ 
A: If ( 2 and n ) have  greatest common divisor only 1 ,then obviously  ( 2*2*2.... ) and n will have greatest common divisor 1 So , gcd ( 2 ^ k , n) =1 
