Show that if $\gcd(a, b) = d$ and $\gcd(a, c) = f$ then $\gcd(a, bc) = \gcd(a, df)$

Asked Dec 09 '15 at 18:29

Active Dec 09 '15 at 22:11

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Here is my solution:

Let $m = \gcd(a, bc), n = \gcd(a, df)$. Since $d|b , f|c$ so $b = kd, c = k'f$ $\Rightarrow bc = kk'df \Rightarrow df|bc$
$n|a, n|df, df|bc \Rightarrow n|bc \Rightarrow n|\gcd(a,bc) \Rightarrow n|m$

But I can't go on any more to show $m|n$!! Please help me complete the solution.

edited Dec 09 '15 at 22:11

Lisa

629

asked Dec 09 '15 at 18:29

Hamid Reza Ebrahimi

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Note that $m=\gcd(a,bc)$ divides $\gcd(a,b)* \gcd(a,c)=df$. Check this for a proof. – Nizar Dec 09 '15 at 18:37
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@Nizar Thank you,that was so helpful! – Hamid Reza Ebrahimi Dec 09 '15 at 19:51

1 Answers1

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if $m|bc$ then there exist $m'$ and $m''$ with $m = m'm''$ such that $m'|b$ and $m''|c$. Since $m = m'm''|a$ we know that $m'|a$ and $m''|a$. Hence $m'|\gcd(a,b) = d$ and $m''|\gcd(a,c) = f$, so $m = m'm''|df$ and $m|\gcd(a, df) = n$.

edited Dec 09 '15 at 22:09

Lisa

629

answered Dec 09 '15 at 18:35

Kerr

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