Prove existence of multiplicative inverse modulo p when p is prime

The question is

Suppose p is a prime number and p does not divide a. Prove that the congruence $ax \equiv 1 \mod p$ has a solution

(Without using Fermat's Theorem), I can intuitively tell this is true because since $p$ is prime, every $x < p$ is co-prime to it, thus $$x \mod p$$ Will produce a unique integer for each $x \in \{0, 1, 2, .... p-1\}$, before it cycles over again at $p$. Clearly, some $x$ must have that $x \mod p = 1$. However, I don't know how to prove this for $ax$, mainly because I can't prove the ramifications it can have on the mod cycle. What I do know is that if for some $x$, we have: $$ax \equiv ay \mod p$$ Then since $p$ doesn't divide $a$ (by question assumption), it must divides $x - y$, implying that the solution for $ax \equiv 1 \mod p$ only depends on $x$, however, I don't know how to show this rigorously. Could anyone enlighten me?

Solution 1:

Let's consider the numbers $-1,a-1,2a-1, \ldots (p-1)a-1$. These must have different remainders $\mod p$, since otherwise $ma \equiv na \mod p \implies p|a(m-n)$, but $p \nmid a$ and $0<|m-n|<p$, so this can't happen.

Therefore, they must leave $p-1$ different remainders $\mod p$. There are $p$ of these numbers, and $p$ possible remainders, so one of these must leave the remainder $0 \mod p$. That value of $x$ solves the equation.