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

Question

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.

Answer 1

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 $c$)

Answer 2

Hint:

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