Why is $\gcd(a,b)=\gcd(b,r)$ when $a = qb + r$?
Prove: $\gcd(a, b) = \gcd(a+bq, b)$
By the division algorithm we know:
$a = b\cdot q + r \iff r = a -b\cdot q$
Thus, we see it is equivalent to prove $\gcd(a, b) = \gcd(b, r)$
To me the claim seems straight forward, since the term $a+bq$ in the $\gcd(a+bq, b)$ is "dependent" on the $a$. That is, $\gcd(bq, b) = b$, so adding an $a$ really determines the $\gcd(a+bq, b)$.
How can I formalize this and use more mathematical terms?