The range of the solution of congruence equation. (Wilson’s theorem)

Question

In order to prove Wilson’s theorem, we have to prove a lemma which is $$1 \le a \le p-1 \space \Rightarrow\space \exists \bar{a} \in \left\{ n \in \mathbb{N}| 1\le n \le p-1\right\} s.t \space a\cdot \bar{a} \equiv 1 (mod \space p)$$ And to prove the existence of $\bar{a}$, we have to show that there is a unique solution of the equation $$ ax \equiv 1 \space(mod\space p)$$ and it have to be $1 \le x \le p-1$. Since $(a,p) |1$, the equation must have a unique solution. How can I prove $1 \le x \le p-1$?

Answer 1

1. For prime $p,$ if $p\not |\; a$ there exists $x'$ with $p|(x'a-1)$ and since $p\not |\;x'$ there exists $x$ with $1\le x\le p-1$ with $p|(x'-x),$ and hence $p|(xa-1).$

There are infinitely many $x'$ such that $p|(x'a-1)$ but only one of them in the range $1,...,p-1.$