A cyclic system of congruences

Since the problem is symetric under any permutation of the variables $a,b,c$, we can assume wlog that $0< c\leqslant b\leqslant a$
Then the second equation from the system can be written as follow:
$$\exists k\geqslant0/ b=c+ka$$where $k$ is non-negative since $b\geqslant c$ and $a$ is a positive integer.

Now assume that $k\geqslant 1$, then one gets $b=c+ka\geqslant c+a>a$ but this is absurd since $a\geqslant b$. Hence $l=0$ and $b=c$.

Then the first (and last) congruence can be rewritten as $b\mid a$.

Reciprocally, any triple of the form $(bd, b, b)$ with $d,b$ positive integers, is a solution of this system.