Prove that if $\gcd(a,n)=1$, then the integers $c,c+a,c+2a,\ldots,c+(n-1)a$ form a complete set of residues modulo $n$ for any $c$ I am guessing I need to show that the given integers equal $0,1,2,\ldots, (n-1)$ mod n taken in some order. However I am not sure on how to start, Any help ?
 A: Suppose that there are $i< j$ such that $i,j\in\{0,\ldots,n-1\}$ and
$$
c+ia\equiv c+ja\;(\text{mod } n)
$$
then $n$ divides $(j-i)a$ which is a contradiction to $0<j-i<n$ and $\text{gcd}(a,n)=1$. This means that the set of $n$ numbers $\{c,c+a,+\ldots,c+(n-1)a\}$ have distinct residues modulo $n$. But there are exactly $n$ distinct residues modulo $n$, so the conclusion follows.
A: Lemma: The integers $a, 2a, 3a, \ldots na$ are a complete residue system modulo $n$.
Proof: There are $n$ numbers in this set, so it is enough to show they are all inequivalent modulo $n$. If not then for some $0<k<j<n$ we have

$$n|(ka-ja)$$

As $0<|k-j|<n$, $n\not | (k-j)$ but also $\gcd (n,a)=1$, a contradiction to the theorem that $\gcd(a,b)=1$ implies that $a|(bc)\implies a|c$. $\square$
Theorem: Any translate of a complete residue system is also a complete residue system. I.e. if $\{x_1,\ldots, x_n\}$ is a complete residue system, so too is $\{c+x_1,\ldots, c+x_n\}$
Proof: Again, we need only show that they are all inequivalent modulo $n$. This is because

$$n|(c+x_i-(c+x_j))\iff n|(x_i-x_j),\quad 1\le i\ne j\le n$$

However, $x_i\not\equiv x_j\mod n$, so by definition, $x_i-x_j\not\equiv 0\mod n\iff n\not |(x_i-x_j)$. $\square$
Since $a, 2a, 3a, \ldots, (n-1)a$ is a complete residue system, so is $c+a, c+2a,\ldots, c+(n-1)a$.
