Show that gcd (a, b) = gcd(a - kb, b) for all k in Z.
Can I get some hints for this
I have proved that if d = gcd(a, b), then d / (a-kb) and d / b, so d is a common divisor of (a-kb, b). However, I now need to show that d is unique, which forces me to assume that there exists some (c) that c is a common divisor and c / d. I don't know how to do
Hint: Let $n\in\mathbb N$ such that $n\mid b$. Prove that$$n\mid a\iff(\forall k\in\mathbb Z):n\mid a-kb.$$