$a \equiv b \pmod n$ solveable - which necessary and sufficient conditions ? For given $n\in \Bbb N, n \ge2, a,b \in \Bbb Z$ give necessary and sufficient conditions such that $$a\equiv b\pmod n$$
  is solveable.


First I thought that it would be sufficient if $b$ would be a multiple of $a$, so for $c\in \Bbb Z$ we'd have $a= b \cdot c$.

For the necessary part I have no clue, I mean if the $gcd(a,n)=1$ so it's solveable for each $b \in \Bbb Z$, but this had been already be talked about in the lecture, so it's not that what it's asked for.

Any hint for which part I should pay attention to ?
The congruence holds if and only if the gcd of $a$ and $n$ is a divisor of $b$.
The congruence holds if and only if the gcd of $a$ and $n$ is a divisor of $b$.