elementary number theory - Show that $9mid a^2$ if given that $6mid a$

Does this prove I made seem correct to show that if $6$ divides $a$ then $9$ divides $a^2$

If $6\mid a$, then $a = 6k$ (k is some integer).

Then $a^2 = 36k^2 = 9(4k^2)$.

Which means that $9\mid a^2$.

perhaps if not is there any other way?