Proof for $Z^2$ not being cyclic

Question

Heres my proof for $Z^2$ not being a cyclic group can anyone confirm if this looks good.

Let $Z^2=\{ (a,b):a,b∈Z \}$

Since $Z^2$ is additive group, if it were cyclic there must be a fixed p,q such that $k(p,q) = (a,b)$ for any $ a,b∈Z$. So $(p,q) = (\frac{a}{k},\frac{b}{k})$. But if k doesnt divide a,b $(p,q)$ wont be integers and therefore not in the group so not cyclic.

Can anyone confirm if this is a valid proof

Answer 1

It is not correct. The assertion “if it were cyclic there must be a fixed $(p,q)$ such that $k(p,q) = (a,b)$ for any $a,b\in\mathbb Z$” is meaningless, since you say nothing about $k$.

Answer 2

Hint:

If $k(p,q)=(a,b)$ then $k$ must divide both $a,b$. What can you say about $(p,q)$ if you try to generate, say, $(0,1)$ and $(1,0)$?