Proving that if $n$ is odd and $\gcd(m, n) = 1$, then $\gcd(2m + n, 2n) = 1$ I've been trying to crack this one for the last little while. I've tried a few approaches, but none have bore any fruit.

Let $n > 0$ be an odd integer. Prove that if $\gcd(m, n) = 1$, then $\gcd(2m + n, 2n) = 1$.  

 A: It can easily be deduced that $d(=gcd(2m+n,2n))$ will be odd because if it is not then $2|(2m+n) \Rightarrow 2|n$ which is not true.
Now ($d|2n \Rightarrow d|n$ and $d|(2m+n) \Rightarrow d|m$) $\Rightarrow d|gcd(m,n)\Rightarrow d|1\Rightarrow d=1$
A: Since we have that $(m,n)=1$, there are $x,y$ so that
$$
mx+ny=1\tag{1}
$$
Then,
$$
\begin{align}
(\color{#C00000}{2m+n})2x+\color{#C00000}{2n}(2y-x)&=\color{#00A000}{4}\\
(\color{#C00000}{2m+n})n-\color{#C00000}{2n}m&=\color{#00A000}{n^2}
\end{align}\tag{2}
$$
Since $n$ is odd, there are $w,z$ so that
$$
2w+nz=1\tag{3}
$$
Squaring $(3)$ gives
$$
\color{#00A000}{4}(w^2+wnz)+\color{#00A000}{n^2}z^2=\color{#0000FF}{1}\tag{4}
$$
Combining $(2)$ and $(4)$, we get an integral combination of $\color{#C00000}{2m+n}$ and $\color{#C00000}{2n}$ that equals $\color{#0000FF}{1}$:
$$
(\color{#C00000}{2m+n})\left(2x(w^2+wnz)+nz^2\right)+\color{#C00000}{2n}\left((2y-x)(w^2+wnz)-mz^2\right)=\color{#0000FF}{1}\tag{5}
$$
Therefore, $(\color{#C00000}{2m+n},\color{#C00000}{2n})=\color{#0000FF}{1}$.
A: Hint: Suppose $k\mid 2n,2m+n$. Could $k$ be even? What can you deduce? 
A: d | 2m + n (this is odd),
d | 2n (this is even).
Therefore d = 1.
Justification:
$$d = p_1^{\alpha_1} * ... * p_n^{\alpha_n}$$
$d | 2n$, wlog we can assume $p_1 = 2$.
Therefore it follows $d = 2^{\alpha_1} p_2^{\alpha_2} * ... p_n^{\alpha_n}$.
$$d | 2m + n = 2m + 2k + 1 = 2(m + k) + 1 (odd)$$
Therefore d doesn't have any factors of 2. It follows that:
$$\alpha_1 = \alpha_2 = \ldots = \alpha_n = 0$$
Whence we have our result d = 1.
