Show that $9\mid a^2$ if given that $6\mid a$

Question

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?

Answer 1

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!

Answer 2

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, $9\mid a^2.$

Answer 3

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.