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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.

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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.

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

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

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