Proving that $\gcd(a,ak+c)=\gcd(a,c)$

Question

I'm having some trouble proving the following proposition:

Let $a,c \in \mathbb Z\setminus\{0\} $ and $k \in \mathbb Z$, then:$$\gcd(a,ak+c)=\gcd(a,c)$$

If $D_a$ and $D_c$ are the set of all numbers that devide $a$ and $c$ respectively I was able to prove that, if $\lambda := \gcd(a,ak+c)$ then $\lambda \in D_a \cap D_c$, but now, I can't arrive at the conclustion that $\max(D_a \cap D_c) = \lambda$. How can I conclude this proof?

Hint: If $ax+cy=d,$ then $a(x-ky)+(ak+c)y=d.$ – Thomas Andrews Jan 14 '21 at 19:46 — Thomas Andrews, Jan 14 '21 at 19:46

score 2 · Accepted Answer · answered Jan 14 '21 at 19:46

2

Let $d = \gcd(a,ak+c)$ and $d_{1}=\gcd(a,c)$.

We have that $d|a \implies d|ak$ so $d|(ak+c)-ak=c \implies d|d_{1}$.

$d_{1}|a \implies d_{1}|ak$. Since $d_{1}|c, d_{1}|ak+c$.

Hence $d|d_{1}$ and $d_{1}|d$ with $d,d_{1} > 0.$

Therefore $d = d_{1}.$

You may also use $d \leq d_{1}$ and $d_{1} \leq d \implies d=d_{1}$ instead.

answered Jan 14 '21 at 19:46

Derek Luna

2,732
8
19

I was able to follow most of it, I just have one question. Why does $d|(ak+c)-ak=c \implies d|d_{1}$? – Eduardo Magalhães Jan 14 '21 at 19:52
Then $d|a$ and $d|c$, so $d \leq$ the greater common divisor of $a$ and $d$, which is $d_{1}$ i.e., $d \leq d_{1}$. In fact it must be that $d|d_{1}$, but maybe you don't know this result, so the inequality is sufficient. – Derek Luna Jan 14 '21 at 19:53
Then is it true that If $k|a$ and $k|c$ then $k| \gcd(a,c)$? – Eduardo Magalhães Jan 14 '21 at 19:55
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Yes, but I just said that the inequality is sufficient. You do know that the property of $\gcd (a,c)$ means that any divisor of $a$ and $c$ is less than $\gcd (a,c)$? Then the inequality is clear. It being a divisor is technically a different result (although should be known at this point). – Derek Luna Jan 14 '21 at 19:56
If there is still confusion, let me know. – Derek Luna Jan 14 '21 at 19:58
I just did the proof on my own using your process and I think I got it. Thanks – Eduardo Magalhães Jan 14 '21 at 20:03
You're welcome. – Derek Luna Jan 14 '21 at 20:03

gt6989b · Answer 2 · 2021-01-14T19:50:45.613

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HINT

Sufficient to prove $\gcd(a,c) = \gcd(a,a+c)$.

This is the basis for the Euclidean algorithm, which finds $\gcd(a,b)$ with $a>b$ as $\gcd(a,b) = \gcd(a-b,b)$.

edited Jan 14 '21 at 19:50

answered Jan 14 '21 at 19:44

gt6989b

54,422

Why is it Sufficient? – Eduardo Magalhães Jan 14 '21 at 19:46
@EduardoMagalhães apply this $k$ times to get your result – gt6989b Jan 14 '21 at 19:48

score 1 · Answer 3 · answered Jan 14 '21 at 20:07

1

By Bezout $g=au+cv=a(u-kv)+(ak+c)v$ and since $\begin{cases}g\mid a\\g\mid c\end{cases}\implies g\mid ak+c$ then $g=\gcd(a,ak+c)$

answered Jan 14 '21 at 20:07

zwim

28,563

Proving that $\gcd(a,ak+c)=\gcd(a,c)$

3 Answers3