This is anything but comprehensive. Sorry about that.
For three variables we can find solutions from copies of the Klein Four group inside $\Bbb{Z}_n^*$. Those exist if and only if one of the following conditions holds:
- $n$ is divisible by two distinct odd primes,
- $4\mid n$, and $n$ is divisible by an odd prime, or
- $8\mid n$.
The resulting solutions have the further property that $a,b,c$ are all coprime to $n$. We can introduce an auxiliary common factor $q$ to all of $a,b,c,n$ to get a solution modulo $nq$ provided that $\gcd(q,n)=1$ by lifting a solution modulo $n$ while making all the variables divisible by $q$. This is straightforward by Chinese remainder theorem. For example with $n=12, \{a,b,c\}=\{5,7,11\}, q=5$ we arrive at the solution $A=5, B=55, C=35$ modulo $nq=60$. Observe that $5$ is a factor of all of $A,B,C$, and $A\equiv a$, $B\equiv b$, $C\equiv c$ modulo $12$. This latter trick does not expand the range of available moduli $n$, so it does not really advance our cause. I brought it up, because grouping solution by the gcd may lead to a classification of all three variable solutions.
We also have the following mechanism for getting solutions with more than three variables. I find it a bit curious, which is why I post this.
Let $L$ be the Pisano period modulo $m$. That is, we define the sequence $(F_i)_{i\ge0}$ by declaring $F_0=0$, $F_1=1$ and then $F_{i+2}\equiv F_{i+1}+F_i\pmod m$. We necessarily end up with a periodically repeating sequence $(F_i)_{i\ge0}$ in
$\Bbb{Z}_m$, where $L$ is the shortest period: $F_i=F_{i+L}$ for all $i$. Then let us choose $n$ and $g$ such that $g$ has multiplicative order $m$ in the group $\Bbb{Z}_n^*$. We then get a solution with $L$ variables modulo $n$ by declaring
$$a_i\equiv g^{F_i}\pmod n$$
for all $i$.
For example, when $m=4$, the sequence $(F_i)_{i\ge0}$ repeats with the period
$(0,1,1,2,3,1)$ of length $L=6$. The residue class $g=2$ has order four in $\Bbb{Z}_5^*$, so we get a solution with six variables modulo $n=5$:
$$a_0=1, a_1=2, a_2=2, a_3=4, a_4=3, a_5=2.$$
While that felt cool at first, do observe that this mechanism is a bit boring in the following sense. We can equally well:
- Select random residue classes $a_0,a_1$ modulo $n$, both coprime to $n$.
- Continue recursively: $a_{i+2}=a_ia_{i+1}$ for all $i$.
- And stop when we reach the end of the period. Only finitely many alternatives to consecutive pairs $(a_i,a_{i+1})$, so eventually we reach a period (by the pigeon hole principle), and have a solution with a somewhat unpredictable number of variables.
This is unsatisfactory in the sense that we don't have control of the number of variables we end up with :-(.
