Consider $n, k ∈ \mathbb{N}$. Prove that $\gcd(n, n + k)\mid k$ 

Consider $n, k\in\mathbb N$. Prove that $\gcd(n, n + k)\mid k$


Here is my proof. Is it correct?
A proposition in number theory states the following:

$$\forall a, b \in \mathbb{Z}, \ \gcd(n,m)\mid an+bm$$

A corollary (consequence) to this is that if $an+bm=1$ for some $a, b \in \mathbb{Z}$, then $\gcd(n,m)=1$. 
Again in our problem $(n+k)-n=k$, therefore we can use this to see that $\gcd(n,n+k)=k$.
 A: I would do it simpler. We know that $\gcd(n, n+k) | n$ and $\gcd(n, n+k)|(n+k)$. From that, it is pretty clear that $\frac{n+k}{\gcd(n, n+k)} = \frac{n}{\gcd(n, n+k)} + \frac{k}{\gcd(n, n+k)}$. The left side is an integer by definition. $\frac{n}{gcd(n, n+k)}$ is also an integer by definition. Therefore, $\frac{k}{\gcd(n, n+k)}$ is also an integer. Result follows.
A: Even easier $d=\gcd(n,n+k)$


*

*if $d=1$ then $d \mid k$

*if $d>1$ then $d \mid n$ and $d \mid n+k$ or $d\cdot q_1=n$ and $d\cdot q_2=n+k$. Altogether $d\cdot q_2 = d\cdot q_1 +k$ or $d(q_1-q_2)=k \Rightarrow d \mid k$

A: Proof: Let $m=gcd(n,n+k)$ for $n \in \mathbb{Z}$ and $k \in \mathbb{N}$. Then $m \mid n$ and $m \mid (n+k)$ and so $n=mx$ and $(n+k)=my$ for $x,y \in \mathbb{Z}$. Observe that $(n+k)-n=k=my-mx=m(y-x)$. Since $y-x$ is an integer, $m \mid k$. Thus, $gcd(n,n+k) \mid k$.
A: We want to show that $\gcd (a, a + k) | k$
Let $\gcd (a, a + k) = d$ say, where $d$ is an integer
$d| a     \implies  a     = md$    , $m$ is an integer
$d| a + k  \implies  a + k = nd$ , $n$ is an integer
$a + k - a = nd - md $
$k = d (n - m)$
$\implies d | k$
$\implies \gcd(a,a + k) | k$    [as required]
