Proof verification that $\gcd(a,b)=\gcd(a, b- k \cdot a )$

Question

We wish to prove that $\gcd(a,b)=\gcd(a, b- k \cdot a )$, where $a,b, k$ are integers.

For any divisor $d|a$ and $d|b$ then we also have that for all integers $x,y$ we have $$d|(ax+by)$$ now let $x=-k$ and $y=1$ then $d|(b-ka)$. I still have to make the step to the gcd now, since I prove this for general divisors, is my proof also valid for the gcd? this only would mean that $d$ is the greatest such integer.

score 1 · Accepted Answer · answered Sep 28 '18 at 14:20

1

Since the two sets of divisors are the same, their maximum is too.

answered Sep 28 '18 at 14:20

J.G.

115,835

Since it holds for arbitrary $d$, right? – Sep 28 '18 at 14:33
1

@WesleyGroupshaveFeelingsToo Yep. – J.G. Sep 28 '18 at 14:35

score 1 · Answer 2 · answered Sep 29 '18 at 21:20

What you proved so far is $d \mid \gcd(a,b) \implies d \mid \gcd(a, b-ka)$

If you can prove $d \mid \gcd(a,b-ka) \implies d \mid \gcd(a,b)$ it will follow that $\gcd(a,b)=\gcd(a, b-ka)$.

Proof. \begin{align} d \mid \gcd(a,b-ka) &\implies (d\mid a) \ \wedge \ (d\mid b-ka) \\ &\implies (d\mid a) \ \wedge \ (d \mid b - ka + ka) \\ &\implies (d\mid a) \ \wedge \ (d \mid b) \end{align}

Proof verification that $\gcd(a,b)=\gcd(a, b- k \cdot a )$

2 Answers2