# 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.

It is enough to see that $$f(2,3)=1$$ and therefore every integer $$b$$ can be obtained using $$f(2b,3b)$$.
Set $$n:=m+1$$;
$$f(n,m)=3(m+1)-4m=3 -m$$;
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.