Solution:
\( \displaylines{\text{If }\left(a,b\right)\text{ was a generator for the group, then }\left\lbrace ak,bk:k\in\mathbb{Z}\right\rbrace\text{ would be all of }\mathbb{Z}\times\mathbb{Z}.\\ \text{However in order to not miss any elements, }a\text{ and }b\text{ would have to be either 1 or -1}.\\ \text{In either of these cases, we only generate the diagonals, not all of }\mathbb{Z}^2.\\ \\ \text{(However two generators would yield the entire group, such as (0,1) (1,0) with the set }\left\lbrace m\left(0,1\right)+n\left(1,0\right)\right\rbrace=\mathbb{Z}^2.)} \)