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Property of GCD

For any $m > 0$ , $\gcd(mb,mc) = m\gcd(b,c)$.

Please prove this. I am learning the Theory Of Numbers in Detail but I am not able to find the proof for this. It is not available in the internet also. So please help me with this problem. Please prove this without using the Euclidean Algorithm as it is deriver from this.

2 Answers 2

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Note that if $\gcd(b,c)=1$ then $b$ and $c$ are coprime and don't share any common factor. Thus $\gcd(mb,mc)=m\cdot\gcd(b,c)=m\cdot1=m$ since the only common factor is $m$.

If $b \mid c$ , this is $c=b.a$ ($b$ divides $c$) then $\gcd(b,c)=b$ and $\gcd(mb,mc)=m\cdot\gcd(b,c)=mb$

If $c \mid b$ , this is $b=c.a$ ($c$ divides $b$) then $\gcd(b,c)=c$ and $\gcd(mb,mc)=m\cdot\gcd(b,c)=mc$

If $\gcd(b,c)=d\neq 1$, this is, they share $d$ as a common factor (not coprime) but neither $b \not \mid c$ nor $c \not \mid b$ then $d \mid b$ and $d \mid c$ ($d$ divides both $b$ and $c$)

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    What if $b,c$ are not coprime, but do not divide one another? For instance, $4$ and $6$. – 2017-02-14
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    That $\gcd(b,c)=d$ and $d \mid b$ and $d \mid c$. Will update my answer. Thanks for the point ;) – 2017-02-14
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Hint:

The g.c.d. of two elements $a,b\;$ is the positive generator of the ideal $\langle a,b\rangle$.

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    While this is certainly true and one way to prove it, I don't think it's appropriate in this specific context. – 2017-02-14