(a) Prove that no two numbers from the list are congruent modulo $p$
1$·$a, 2$·$a, 3$·$a, ... $(p-2)·a$, $(p-1)·a$
I think I am suppose to assume there exist two number $(x,y)$ that are congruent modulo $p$, so $xa = pq+r$ and $ya = pk+r$. But I am not sure where to go from here.
(b) Prove that there exists $b∈S$ such that $b·a ≡_p1$.
For this one I am not sure how to start. I tried setting $b·a = p·q + 1$, how can I use this to prove the statement is true?
If two elements of $S$, say $ma$ and $na$ are congruent mod $(P)$, their difference, $(m-n)a$ is a multiple of $p$.
That is not possible because $p$ is prime and both $a$ and $m-n$ are less than $p$. We know that if a prime number divides a product it has to divide at least one of the factors.
The second part of the problem is that why one of the remainders is $1$.
Since there are only $p-1$ non-zero remainders and the $p-1$ elements of S have different remainders,one remainder must be $1$
That is $a$ and $m$ are multiplicative inverse, mod $(p) $ for some $1\le m \le p-1$