# Prove that if 2 divides n and 7 divides n, then 14 divides n

Okay so I have to prove this. I can write that if 2 divides n and 7 divides n, then there must be integers k and m such that $2*k=n$ and $7*m=n$

So $14*k*m=n^2$

But what to do after that?

If I say that then 14 divides $n^2$, I get bit of a circular argument, but if I write that n divides $14*k*m$, then I don't know what to do next.

Any help/suggestions?

• Some general advice: when you are asked to prove something that seems so incredibly obvious that it appears all proofs would go in circles, what you're really being asked to do is to use the explicitly allowed axioms and definitions to prove it. Don't start doubting your intuition about the patently obvious. Just examine the given definitions to establish to your satisfaction that they are sufficient, all by themselves, to formally prove the statement in question. Commented Oct 17, 2017 at 22:54

Following from what you have written, $$n = 2k=7m \implies k=\frac{7m}{2}.$$ Since $k$ is an integer and $\gcd(2,7)=1$, $m/2$ must be an integer; i.e., $m/2=r \implies m=2r$, where $r$ is an integer. Therefore, $$n=7m=7\times 2r = 14 r.$$ Q.E.D.
You can say much more: Say $a$ and $b$ are relatively prime. If $a|n$ and $b|n$ then $ab|n$.
Proof: Since $a|n$ we can write $n=ak$. Now since $b|ak$ we have, by Euclid lemma $b|k$, so $k=bl$. Thus $n=abl$ and so $ab|n$.
Write $n = 7 m$. If $m$ were odd, $7m$ would also be odd, contradiction. So $m$ is even, $m=2k$, and $n = 14k$.