Prove that $a\mid c$; given that $a\mid bc$ and $\gcd(a,b) = 1$.

Question

Please help answer the number theory problem. The proof I came across goes like this:

Let $d = \gcd(a,b)$ then $d = ma + nb$.

$ma + nb = 1$.

then

$mac + nbc = c$.

$a\mid mac$, $a\mid nbc$, therefore $a\mid c$.

Not sure how last step unfolded. Thanks.

(This is my personal question I have not posted it anywhere else, and the context of my question is my own)

Which part of the last step are you stuck on? Do you agree that $a|mac$ and that $a| nbc$? — littleO, Mar 12 '17 at 16:59
$mac+nbc = a(mc+n(\color{#c00}{bc/a}))$ and $\color{#c00}{bc/a}$ is an integer by hypothesis. Generally if $,a\mid x,y $ then $,a\mid nx+my,$ for any integers $m,n.,$ In paricular, the sum of two multiples of $a$ is also a multiple of $a\ \ $ — Bill Dubuque, Mar 12 '17 at 16:59
Possible duplicate of Prove the following: If $a \mid bc$, then $a \mid \gcd(a, b)c$. — Xam, Mar 13 '17 at 00:23
@Icycarus I know they are not the same question, but this one is a duplicate and it was flagged as duplicate of the question I linked above. — Xam, Mar 13 '17 at 18:47

score 2 · Answer 1 · answered Mar 12 '17 at 17:02

2

We know $\gcd(a,b)=1\iff au+bv=1$ and $a\mid bc\iff bc=ak$ for some $k\in\Bbb Z$. Then $$au+bv=1\implies auc+bvc=c\implies auc+akv=c\implies a(uc+kv)=c\implies a\mid c$$ Note, after the second implication we introduced $bc=ak$.

answered Mar 12 '17 at 17:02

Dave

13,568

Nice ideas for getting a full understanding Much appreciated – Newton Mar 12 '17 at 19:02

Icycarus · Accepted Answer · 2017-03-12T17:13:50.267

2

Given

$am+nb=1\\amc+nbc=c$

Since $a \mid bc$, then $bc = ka$ for some integer $k$. Substituting into your equation, we have that

$amc+ank=c$. Since there are solutions to $am+bn=1$, then there must be solutions to $a(mc+nk)=c$. Hence $a \mid c$

This is also known as Euclid's Lemma.

Another way to think of it is because $mc$ and $nk$ are integers (Since $m$ and $n$ exists), then $mc+nk$ is an integer, which means $\frac ca$ is an integer, which can only mean $a \mid c$.

edited Mar 12 '17 at 17:13

answered Mar 12 '17 at 17:05

Icycarus

1,142
1
8
21

very in depth, excellent explanation. Thanks! – Newton Mar 12 '17 at 19:16

Prove that $a\mid c$; given that $a\mid bc$ and $\gcd(a,b) = 1$.

2 Answers2