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

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.

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.

• unfortunately I meant only if – user3321427 May 20 '15 at 19:33
• How can I translate this arithmetic to a "formal" proof? – user3321427 May 20 '15 at 19:39
• @user3321427 What do you mean by a "formal" proof? Are you usng some automater theorem prover or some other rigorous formal proof system? – Bill Dubuque May 20 '15 at 20:54
• OP's problem is a direct application of what you call the universal property of lcm: $a,b\mid m\,\Leftrightarrow\, \text{lcm}(a,b)\mid m$. – user26486 May 20 '15 at 22:26
• @user31415 Indeed, but most likely that is unfamiliar to the OP (and they clarified that the question is about the easier direction). – Bill Dubuque May 20 '15 at 22:57

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.

• unfortunately I meant only if – user3321427 May 20 '15 at 19:33
• So it's trivial: a divisor of a divisor is a divisor. – Bernard May 20 '15 at 19:35