Necessary and Sufficient Conditions for the pair of integers $\{m,n\}$ to generate $\mathbb{Z}$ Let $m,n \in \mathbb{Z}$ be two non zero numbers. I need to find necessary and sufficient conditions on $m$ and $n$ for which the pair $\{m,n\}$ generates the additive group $\mathbb{Z}$.
I made an attempt at a proof, and would like to know if 1) It's correct, and 2) if there's anything I can do to make it better.
First of all, I asserted that a necessary and sufficient condition would be for $\mathbf{\gcd(m,n)=1}$.
Now, here's my proof:

$(\implies)$ Suppose that $\gcd(m,n)=1$. Then, by Bezout's Identity, $\exists$ integers $p,q$ such that $mp+nq = \gcd(m,n)=1$.
Let $k$ be an arbitrary element of $\mathbb{Z}$. Then, we can produce $k$ by multiplying both sides of $mp+nq = 1$ through by $k$ to yield $m(pk)+n(qk)=k$.
Defining $x=pk \in \mathbb{Z}$ and $y = qk \in \mathbb{Z}$, we see that $\forall k \in \mathbb{Z}$, $\exists x, y \in \mathbb{Z}$ such that $mx+ny=k$.
So, $<m,n>=\mathbb{Z}$.
$(\Longleftarrow)$ Suppose $<m,n>=\mathbb{Z}$. Then, $\forall k \in \mathbb{Z}$, $\exists x,y \in \mathbb{Z}$ such that $k=mx+ny$.
Therefore, we must have $\gcd(m,n)|k$.
(The part I'm really not sure about) However, $k$ is an arbitrary integer, and thee only integer that is a divisor of any arbitrary integer is $1$. 
Thus, $\gcd(m,n)=1$.

Thank you!

You have that, for $k+1\in\mathbb{Z}$ exist $x',y'\in\mathbb{Z}$ such that $mx'+ny'=k+1$ you can use $mx'+ny'-(mx+ny)=k+1-k=1$ thus $m(x'-x)+n(y-y')=1$ then you have $gcd(m,n)=1$.
