Proof of the existence of inverses modulo n

Proof

1. Let $n$ be a positive integer and let $a$ be an integer. Let $d = \gcd(a, n)$ be their greatest common divisor.
2. Since $n \ne 0$, the integers $a$ and $n$ are not both $0$. So $d$ is a positive common divisor of $a$ and $n$.
3. First assume that $a$ and $n$ are coprime, so $d = 1$.
4. By Bézout's identity, there are integers $x$ and $y$ with $a x + n y = 1$.
5. Then $a x - 1 = n (-y)$, so $n$ divides $a x - 1$. By the definition of congruence, $a x \equiv 1 \pmod n$.
6. So $x$ is a modular multiplicative inverse of $a$ modulo $n$.
7. Conversely, assume that $a$ has an inverse $x$ modulo $n$, so $a x \equiv 1 \pmod n$.
8. Then $n$ divides $a x - 1$, so there is an integer $y$ with $a x - 1 = n y$. That is, $a x - n y = 1$.
9. The integer $d$ divides $a$ and $n$, so it divides $a x - n y = 1$.
10. Since $d$ is positive by step 2 and divides $1$, we get $d = 1$. So $a$ and $n$ are coprime.
11. By steps 3 to 10, $a$ has an inverse modulo $n$ if and only if $a$ and $n$ are coprime.