Prove that $12 \mid m \iff$ both $6 \mid m$ and $4 \mid m$.

Question

Give a formal proof to the following theorem which I do not know where to start.

Theorem: For all natural numbers 'm', 12 divides m only if 6 divides m and 4 divides m.

Answer 1

If you really means‘only if’, it's trivial. So I'll suppose you meant ‘if’.

Hint: If $6$ divides $m$, $3$ divides $m$. Furthermore $3$ and $4$ are coprime. Then use Gauß's lemma.

Answer 2

Hint $\,\ \dfrac{m}{12}\ =\ \dfrac{m}4 - \dfrac{m}6\ $ for the more difficult direction.

The other direction is easy: $\ 6(2m) = (6\cdot 2) m = 12m = (4\cdot 3)m = 4(3m)\ $ or, invoke the transitivity of $ $ "divides", i.e. $\ 6\mid 12\mid m\,\Rightarrow\,6\mid m,\,$ etc.