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I'm very new to elementary number theory proofs and have been trying to figure out how to prove these seemingly straightforward identities with divisibility with no success.

For some integers $a,b,c \in \mathbb{Z}$

1) If ${a\mid bc}$ and ${a \not\mid b}$ then ${a\mid c}$

2) If ${a\mid c}$, ${b\mid c}$ and ${\gcd(a,b) = 1}$ then ${ab\mid c}$

For 1), example, if 2 divides 3a, then 2 clearly divides a because 2 does not divide 3... not sure how to formalize
For 2), I think it is somewhat related to the divisibility rule (https://en.wikipedia.org/wiki/Divisibility_rule). Let's say $6\mid 12$ and $3\mid 12$ is true but ${6\cdot 3\mid 12}$ is not and that relates to the fact that 6 is a multiple of 3. However, suppose ${2\mid a}$ and ${3\mid a}$. 2 and 3 are not multiples of each other and so the smallest number that is divisible by 2 and 3 must be a multiple of 2 and 3 (6 being the smallest), hence ${a}$ is divisible by 6. Is there a way to formalize this a better way?

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The first claim is false: $4$ divides $(2\cdot 6)$ and $4$ does not divide $2$, do not imply that $4$ divides $6$.

For the second one, recall that if $\gcd(a,b)=1$ then, by the Bezout's identity, there are integers $m$ and $n$ such that $am+bn=1$. Moreover, $c=ra=sb$ for some integers $r$ and $s$. Hence $$c=c(am+bn)=cam+cbn=sbam+rabn=(ab)(sm+rn).$$

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  • $\begingroup$ I was going to respond to your hint but it seems like you gave the full answer for 2). Thanks! $\endgroup$ – CloudIcarus Oct 10 '17 at 8:10
  • $\begingroup$ @CloudIcarus Well done. I edit my answer with the full solution so you can check your work. $\endgroup$ – Robert Z Oct 10 '17 at 8:13
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1) As pointed by Robert, your claim is false. You should replace it by this statement:

If ${a|bc}$ and $gcd(a,b)=1$ then ${a|c}$

2) You have the right intuition.

Hint:

You can write the prime factorization of $a$,$b$ and $c$...

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  • $\begingroup$ I looked at your replacement statement but still not able to show how it is true. That said, I think it seems similar to the statement " if ${a|bc}$ then ${a|b}$ or ${a|c}$ " which I have a hunch about but not sure either. $\endgroup$ – CloudIcarus Oct 10 '17 at 8:43
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    $\begingroup$ Oh, wait. Just got it. By Bezout's Theorem, ${ax + by = 1}$ for some ${x, y \in \mathbb{Z} }$, then ${cax + cby = c}$. Since we know ${bc = ap}$ for some ${p \in \mathbb{Z} }$, then ${cax + (bc)y = cax + apy = a(cx + py) = c}$. Hence ${a|c}$ $\endgroup$ – CloudIcarus Oct 10 '17 at 9:04

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