Show that $9\mid a^2$ if given that $6\mid a$ 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?
 A: You've done just fine! You correctly used what you were given, and used the definition of divisibility (by $6$), to obtain your result.
Good work!
A: What you've done is perfect.
An alternate method of proof for all those people who just love modular arithmetic:
Since $6\mid a$, then $a\equiv 6,\; 3,\; \text{or}\; 0 \pmod{9}$.  Note that $6^2 \equiv 3^2 \equiv 0^2 \equiv 0\pmod{9}$.  Thus, $9\mid a^2.$
A: Your proof is correct. It easily generalizes to the following (yours is special case $\,b=a,\,B=A)$
$$ \begin{eqnarray} && a\mid A\\ &&b\mid B\end{eqnarray}\ \ \Rightarrow\ \ ab\mid AB$$
This has a fundamental converse, namely
$$c\mid AB\ \ \Rightarrow\ \ c = ab, \begin{eqnarray} && a\mid A\\ &&b\mid B\end{eqnarray}\ \ \ {\rm for\ some}\ \ a,b$$
This is equivalent  to the uniqueness of factorizations into primes (atoms), since the special case when $\,c = p\,$ prime is $\ p\mid AB\,\Rightarrow\,p\mid A,\,$ or $\,p\mid B,\,$ which implies said uniqueness by a simple inductive proof. 
This leads to an important refinement view of unique factorization - which has the benefit of generalizing nicely to other rings (esp. noncomutative rings, as Paul Cohn showed). For further discussion see here and here.
