If $\gcd ( a, b) = 1$ and $a \equiv b$ mod $n $, then $\gcd (a, n) = 1$.

Question

I want to show that for $a,b,n \in \mathbb{Z}$, if $\gcd ( a, b) = 1$ and $a \equiv b$ mod $n $, then $\gcd (a, n) = 1$. My attempt is as follows:

Given : $\gcd (a, b) =1 \implies \exists x, y \in \mathbb{Z}$ such that $a*x + b * y = 1$.

I got $b= \frac{(1 - (a*x))}{y}$.

Substituting this in $a \equiv b$ mod $n $, I got $a * (x + y) \equiv 1 $ mod $n$.

However, I am not able to proceed further as I do not know how to show that $\gcd (x+y,a) =1$.

Any common factor of $a$ and $n$ divides $a+kn$ for any $k$, and is therefore a common factor of $a$ and $a+kn$. — Andreas Blass, Mar 26 '22 at 16:32
By the linked dupe $,1=(a,b) = (a,b!-!a) = (\color{#c00}{a,n}\frac{b-a}n)\Rightarrow 1 = (\color{#c00}{a,n})\ \ $ — Bill Dubuque, Mar 26 '22 at 16:32
Or: $\ \color{#c00}{n\mid a!-!b}\Rightarrow (a,\color{#c00}n)\mid (a,\color{#c00}{a!-!b})=(a,b),$ by the 3rd & 1st dupes. — Bill Dubuque, Mar 26 '22 at 17:16

score 1 · Accepted Answer · answered Mar 26 '22 at 16:14

1

We have $ax+by=1$ and $a-b=kn$ for some integers $x,y,k$.

Then $ax+ (a-kn)y=1$, i.e. $a(x+y) + (-k)n=1$.

Thus each common divisor of $a$ and $n$ is a divisor of $1$ and hence $(a,n)=1$.

answered Mar 26 '22 at 16:14

Wuestenfux

2

Please strive not to add more (dupe) answers to dupes of FAQs, cf. recent site policy announcement here. – Bill Dubuque Mar 26 '22 at 16:46

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