Complexity of finding these original parameters
You can compute $m,s$ from the equations: $s = (c-1)/2$ and $m = (b-1)/s$; both must be integral. You can then check that indeed $m(s+1) = a$. If this holds, then you have found the unique solution $m,s$, otherwise there is no solution.
More generally, this method allows you to find the unique real solution to your system of equations, if any exists.

Note that $m = (b-1)/s = 2(b-1)/(c-1)$. Hence $s,m$ are integers if $s$ is odd and $c-1$ divides $2(b-1)$. If they are integers, the final equation is satisfied if
$$
a = ms + m = b-1 + \frac{2(b-1)}{c-1} = (b-1) \frac{c+1}{c-1}.
$$
Another way to state this condition is
$$
a(c-1) = (b-1)(c+1).
$$