What are the steps to simplify the following modulus expression? I have no clue how to do this exactly. Is there a systematic way of doing this or you just have to do it by trial and error?
$n^2 \equiv 9 \pmod {72}$
to $n \equiv a \pmod b$?

Note that $n^2\equiv 9 \pmod {72}$ is equivalent to
$$
(n-3)(n+3)\equiv 0 \pmod {72}
$$
So it is sufficient to solve the simultaneous congruences
$$
(n-3)(n+3)=0\pmod 8
$$
and
$$
(n-3)(n+3)\equiv 0 \pmod 9
$$
The second congruence is true iff $n$ is divisible by $3$, because then both $n-3$ and $n+3$ are divisible by $3$, so their product is divisible by $9$. If $n$ is not divisible by $3$, then neither $n-3$ nor $n+3$ is divisible by $3$, so their product isn't. Thus, $n\equiv 0 \mod 3$.
In the first congruence, we similarly obtain that $n$ must be odd. We claim that this is also sufficient. Indeed, since clearly $n-3$ and $n+3$ are both even, assume that they are both $2 \pmod 4$. This is clearly impossible because they have a difference of $6$. Thus, one of $n-3$ and $n+3$ must be divisible by $4$, and the other is even, so their product is $0 \pmod 8$. Then $n\equiv 1 \pmod 2$.
Then, the set of solutions is just all $n$ that satisfies $n\equiv 1\pmod 2$ and $n \equiv 0 \pmod 3$, which is equivalent to $n \equiv 3 \mod 6$.
