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For all integers $a,b,c$,

$$\;a\mid bc \iff \frac{a}{\gcd⁡(a,b)}∣c.$$

Can anyone help me in proving above statement?

I thought I'd start with trying to prove $\Leftarrow.$

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

I know that $\gcd(a, b) = ax+by = d$ but I am lost as to how to make use of that to proceed from here.

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Pratik Gaikwad
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  • @CarlSchildkraut,not much. I am trying to go other way. so far, I am trying to work on a/(gcd(a,b) | c. by inserting gcd(a,b) = ax+by = d. But don't know how to proceed from there and show that it is necessary condition. – Pratik Gaikwad Sep 25 '16 at 20:04
  • For ($\Rightarrow$) it is equivalent to prove that $a \vert (acx+bcy)$.

    For ($\Leftarrow$) it is equivalent to prove that $a \vert \gcd(ac,bc)$

     – Maman Sep 25 '16 at 20:57

2 Answers2

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$ a\mid bc \iff a\mid ac,bc\iff a\mid (ac,bc)=(a,b)c\iff a/(a,b)\mid c\ $ by gcd Distributive Law

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Bill Dubuque
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Let $a,b,c$ integers. We will take them $\ne0$.

For ($\Rightarrow$) we denote $\gcd(a,b)=d$ which gives with the Bachet-Bézout identity the existence of $x,y\in \mathbb{Z}$ that $d=ax+by$. As we want to prove that $a\vert dc$ we multiply the identity by $c$. We obtain : $dc=axc+byc$. But by hypothesis we have $a\vert bc$ so there exists $k\in\mathbb{Z}$ such that $ak=bc$ Finally it gives that $dc=axc+byc=axc+aky=a(xc+ky)$ with $(xc+ky) \in \mathbb{Z}$. So $a\vert dc$.

For ($\Leftarrow$) suppose that $a\vert \gcd(a,b)c$. By properties of $\gcd$ we can suppose that $a\vert \gcd(ac,bc)$. Let denote $d=\gcd(ac,bc)$. By definition there exists $k\in \mathbb{Z}$ such that $dk=ac$ and $dk=bc$. Then by hypothesis there exists $l\in \mathbb{Z}$ such that $al=d$. So we obtain by multiplying by $k$ : $alk=dk=bc$ with $lk\in \mathbb{Z}$. So $a\vert bc$.

PS : By definition, $\frac{a}{\gcd(a,b)}$ is always an integer so we obtain easily the form of your statement !

Maman
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