Congruence Class of mod p where p is a prime number If $n =p$ is a prime number and if $[0]\neq[a]$ is in $J_p$ ,
then there is an element $[b]$ in $J_p$ such that $[a][b] =
[1]$. ($J_p$ being the set of congruence classes mod $p$)
This was a remark in the book Topics of Algebra by I.N. Herstein in the congruence modulo section of the topic Integers. He didn't prove it. 
I tried to prove it but I have no clue as to how I'm supposed to approach this, all I could do is simplify this to :
Prove, for all $a<p$ there exists a $b<p$ such that $ab=np + 1$ where $p$ is prime and $n (<a,b)$ is any integer.
Could anyone show how this is true or maybe help me out with how to approach it.
 A: Let $p$ be prime and let $a, b$ be integers less than $p$. We know from Fermat’s little theorem that $$a^{p-1}=1 (mod p)$$
Thus, we have that $b=a^{p-2} (mod p)$ satisfies this congruence.
A: You can use the existence of a Bézout identity: since $[a]\neq[0]$, $p$ does not divide $a$, in particular $a$ is prime to $p$ and there exist integers $b$, $q$ such that $ab+pq=1$, for the classes in $J_p$, this implies $[a][b]+[p][q]=[1]$, but since $[p]=[0]$, this yields $[a][b]=[1]$.  
A: If $a=1$ then the result is immediate, so suppose $1<a<p$.
Consider the first $p$ multiples of $a$, which are $a, 2a, 3a, \dots, pa$. Suppose two of these leave the same remainder modulo $p$ - say $ma=na \mod p$ with $m>n$. Then $(m-n)a$ is a multiple of $p$. But since $1<a<p$, this would give a factorization of $p$ which is not the trivial factorization $1 \times p = p$. Since $p$ is prime a non-trivial factorization does not exist. Therefore the first $p$ multiples of $a$ must all leave different remainders modulo $p$.
But there are only $p$ different remainders modulo $p$, so one of the multiples $ka$ with $1<k<p$ must leave remainder $1$ modulo $p$.
