# Prove $gcd(a,b)=gcd(a,2a+b)$

Call $gcd(a,b)=d$. Then $d|a$ and $d|b$. And if $c|a$ and $c|b$, then $c|d$.

It's simple to show that $d$ is SOME divisor of $a$ and $2a+b$, since we already know $d|a$ and $d|b$, so it divides the linear combination $2a+b$. However, how can I show that $d$ is the GREATEST common divisor of $a$ and $2a+b$?

Edit: Let $e$ be some other common divisor of $a$ and $2a+b$. Do we know if it's true that $e|b$? If so, we're finished since anything that divides both $a$ and $b$ will also divide $d$, since $d=gcd(a,b)$.

If $x$ divides both $a$ and $2a+b$ then $x$ divides both $a$ and $b$. Therefore $x$ divides $d$, so $x \leq d$.
• How do you know $x|b$? Dec 4, 2014 at 19:17
• $b=(2a+b)-2a$ … Dec 4, 2014 at 19:17
• $b = (2a+b) - 2a$. Since $x$ divides both terms on the right, it also divides the left. Dec 4, 2014 at 19:18
Another way to put the answer by Michael: What you (Ethan) have done, shows that $\gcd(a,2a+b)\ge \gcd(a,b)$. If you put $c=2a+b$, then the same argument shows that $\gcd(a,b)=\gcd(a,c-2a)\ge\gcd(a,c)=\gcd(a,2a+b)$.