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I'm trying to prove that $a$ divides $bc$ if and only if

$$\frac{a}{\gcd(a, b)} \mid c$$

I go in the right direction first (i.e. if $a$ divides $bc$ then $\frac{a}{\gcd(a, b)} \mid c$):

We want to show that $\frac{a}{\gcd(a, b)} \cdot f \ = c$ for some $f \in \mathbb{Z}$. In other words, we want to show $a \cdot f = c \cdot \gcd(a, b)$ for some $f \in \mathbb{Z}$.

We know $a$ divides $bc$; this means that $ax = bc$ for some $x \in \mathbb{Z}$. We know that $\gcd(a, b)$ divides $a$ and $\gcd(a, b)$ divides $b$ by definition. This means that $\gcd(a, b) \cdot y = b$ for some $y \in \mathbb{Z}$.

That is, $ax = \gcd(a, b) \cdot y \cdot c$ for some $x, y \in \mathbb{Z}$.

I'm almost there... just need to divide $y$ by both sides; but how do I prove that $\frac{x}{y}$ is an integer?!

Thanks.

Anne Bauval
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r123454321
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    Do you know Bezout's lemma, that $gcd(a,b) = ax + by$ for some integers $x$ and $y$? It's often a powerful tool in this sort of problem. Apply it to your desired equality: $af = c \gcd(a,b)$, and you might see how to find your $f$. – Barry Smith Sep 21 '14 at 18:42
  • If you write \gcd with a backslash, that not only prevents italicization but results in proper spacing in things like $a\gcd(b,c)$, between $a$ and $\gcd$. It is standard usage. I edited accordingly. Also, I used \mid where appropriate. – Michael Hardy Sep 21 '14 at 18:54
  • @MichaelHardy: Thanks.

    BarrySmith: The farthest I'm able to get with this is that:

    WTS: $a(f-cx) = bcy$. Since we know that $az = bc$ for some $z \in \mathbb{Z}$, can I simply multiply both sides by $y$? i.e. $azy = bcy$, and I'm done?

     – r123454321 Sep 21 '14 at 19:03
  • Essentially. You might be more convinced if you find a formula for $f$ in terms of $c$, $x$, $y$, and $z$. Since your goal is to show there exists an $f$ with a certain property, you can then substitute in your $f$ and check that it works. – Barry Smith Sep 21 '14 at 19:14
  • Got it, thanks. In regards to the converse, I know that $ay = c(am + bn)$ for some $y, m, n \in \mathbb{Z}$, and want to show that $ax = bc$ for some $x \in \mathbb{Z}$. Am I correct in saying that this is equivalent to wanting to show that $axn = bcn$ for some $x, n \in \mathbb{Z}$? From here, we simplify $ay = c(am + bn)$ to $a(y - cm) = bcn$, and get that $y - cm = xn$. But how do we show that x an integer? – r123454321 Sep 21 '14 at 21:11
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    Heh, for this direction I think it is clearer to use the definition of GCD rather than Bezout's Lemma. You assume that $ax=b \gcd(b,c)$ for some integer $x$, and you want to show that $ay = bc$ for some integer $y$. If you multiply the first equality through by an appropriate integer, you will get one that looks like the second. – Barry Smith Sep 22 '14 at 17:58
  • Does this answer your question? If $ a \mid bc $ then $\frac{a}{\gcd(a,b)} \mid c$? – Anne Bauval Jul 23 '23 at 16:07

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