coprime divisibility: is $ax-1$ divisible by $n$?

Question

Suppose that $a,n \in \Bbb Z$ are coprime. Show that there is an integer $x$ such that $ax−1$ is divisible by $n$.

I know that $\gcd(a,n)=1$ and feel like that will be used in the proof of this, but the fact that there are no numbers is making it complicated. Do I have to work out the gcd backwards?

Immediate consequence of the Bezout identity for $,\gcd(a,n)= 1.,$ Write it down and isolate the $,ny,$ term on one side of the equation and observe what that implies about divisibility by $,n\ \ $ — Bill Dubuque, Mar 04 '19 at 22:43
Duplicate of If $q$ is coprime to $a$ then $a\mid (nq-1),,$ i.e. $q$ is invertible mod $a$ — Bill Dubuque, Mar 04 '19 at 23:23

score 0 · Answer 1 · answered Mar 04 '19 at 22:51

0

$\gcd(a, n) = 1 \Longrightarrow \exists x, y \in \Bbb Z, \; ax + ny = 1; \tag 1$

$ax + ny = 1 \Longrightarrow ax - 1 = -ny \Longrightarrow n \mid ax - 1. \tag 2$

answered Mar 04 '19 at 22:51

Robert Lewis

71,180

thank you! that has really helped – harriet Mar 04 '19 at 22:57
@harriet: glad to be of service. And thanks for the "acceptance"! Cheers! – Robert Lewis Mar 04 '19 at 23:18

score 0 · Accepted Answer · edited Mar 04 '19 at 23:42

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Let $I=a\mathbb{Z}+n\mathbb{Z}$ be an ideal in $\mathbb{Z}$. Now, $\mathbb{Z}$ is a PID; i.e., we have $I=d\mathbb{Z}$ for some $d$. But since $a,n\in I$, we have that $a,n\in d\mathbb{Z}$; i.e., $d$ divides both $a$ and $n$. But $gcd(a,n)=1$, so $d=\pm 1$. This proves that $I=\mathbb{Z}$; in particular, $1\in I$, so $1=ax+ny$ for some $x,y\in\mathbb{Z}$. This is what you wanted to prove.

edited Mar 04 '19 at 23:42

J. W. Tanner

60,406

answered Mar 04 '19 at 22:58

Julian Mejia

4,438

coprime divisibility: is $ax-1$ divisible by $n$?

2 Answers2