Is $f:\mathbb Z\times \mathbb Z\rightarrow\mathbb Z$, $f((m,n))=3n-4m$ injective/surjective? $f:\mathbb Z\times \mathbb Z\rightarrow\mathbb Z$, $f((m,n))=3n-4m$
Hi everyone, I am having some trouble trying to prove that this is subjective. 
I know that it is not injective: For example, consider $f(0,-4)=f(3,0)=-12$. We can see that $f(0,-4)=f(3,0)$ but $(0,-4)\neq (4,0)$. Thus, $f$ is not injective.
For subjective, I know I need to show that for some $b\in\mathbb{Z}$, $f(x,y)=b$ for some pair of integers $(x,y)$. I'm not sure where to go from here. Any help would be appreciated. Thank you.
 A: It is enough to see that $ f(2,3)=1$ and therefore every integer $b$ can be obtained using $f(2b,3b)$. 
A: Set $n:=m+1$;
$f(n,m)=3(m+1)-4m=3 -m$;
Surjective.
A: The linear Diophantine equation $ax+by=c$ has a solution if and only if $\gcd(a,b)=d$ divides $c$. Hence we may write $a=de$, $b=df$, where $\gcd(e,f)=1$. If $x_0$, $y_0$ and $x_1$, $y_1$ are two integer solutions then
$$ax_0+by_0=ax_1+by_1=c$$
or equivalently
$$ex_0+fy_0=ex_1+fy_1=\frac{c}{d}$$
Hence $d\mid c$ since the LHS are integers and
$$e(x_0-x_1)=f(y_1-y_0)$$
Now $e\nmid f$ since they are coprime, so $e\mid (y_1-y_0)$, and by the same reason $f\mid (x_0-x_1)$. So $x_0-x_1=fg$, and $y_1-y_0=eg$, for some integer $g$.
\begin{align}
x_0&=x_1+fg=x_1+\frac{b}{d}\cdot g\\
y_0&=y_1-eg=y_1-\frac{a}{d}\cdot g
\end{align}
Hence
\begin{align}
ax_0+by_0&=a\left(x_1+\frac{b}{d}\cdot g\right)+ b\left(y_1-\frac{a}{d}\cdot g\right)\\
&=ax_1+by_1+a\left(\frac{b}{d}\cdot g\right)- b\left(\frac{a}{d}\cdot g\right)=c
\end{align}
where the last two terms cancel, and so we may choose any integer for $g$ to form an infinite number of solutions.
Since in $3n-4m=k$, $\gcd(3,4)=1$, $1$ divides every integer, and so $k$ takes every value in $\mathbb{Z}$ proving $f$ is surjective.
