0

I have trouble proving the above statement.

Here is my approach.

First, I prove that if $n$ and $k$ are not relatively prime, it implies that $[k]_n$ is not invertible.

If $n$ and $k$ are not relatively prime, there exist $g \neq 1$ such that $nx+ky=g$ for some integers $x$ and $y$. (Bézout's identity)

Multiply the equation by, $n/g$

$(n^2x)/g+(nky)/g=n$

We can simplify this by taking the $(mod \hspace{0.1cm} n)$

$(nky)/g=0(mod \hspace{0.1cm} n)$

I don't understand how to proceed from here!. I sense that it has something to do with $k=0$ is the only solution here.

The converse proof is fairly straightforward (I assume.)

Here is my approach to prove the converse, if $k$ and $n$ are relatively prime then $[k]_n$ is invertible.

Again by the Bézout's identity (if $k$ and $n$ are relatively prime), there exist integers $x$ and $y$ which satisfy $nx+ky=1$. We can simplify this to $k\times(y+nx/k)=1$. It is clear that $(y+nx/k)$ is the inverse of $k$.

Bill Dubuque
  • 282,220
slhulk
  • 293

3 Answers3

1

$[k]_n$ is invertible mod $n$ $\iff$ there is an integer $y$ such that $ky\equiv 1\pmod n$

$\iff$ there are integers $x,y$ such that $ky+xn=1\iff k$ and $n$ are relatively prime

J. W. Tanner
  • 63,683
  • 4
  • 43
  • 88
1

$$\gcd(k,n)=1 \iff$$$$ \exists x,y \in \mathbb{Z}. k\cdot x + n \cdot y = 1 \iff$$$$ \exists[x]_n \in \mathbb{Z}_n. [k]_n\cdot[x]_n=[k\cdot x]_n =[k\cdot x]_n + [n \cdot y]_n = [k\cdot x + n \cdot y]_n= [1]_n \iff$$$$ [k]_n \text{ is invertible in } \mathbb{Z}_n$$

Riccardo
  • 1,304
  • 10
  • 21
1

By Bezout's lemma, $(k,n)=1\iff \exists a,b :ak+bn=1\iff \exists a: ak\cong1\pmod n\iff k$ is invertible $\pmod n$.