figure out missing details. Nice problem and nice solution.

3) Let f(n) = kc + i, k >= 2 and i is in C, then

2c >= n + c = f(f(n)) = f(kc + i) = f(i) + kc > 2c. Contradiction.

6) let f(b) = a + c, a is in C, then f(f(a)) = i + c = f(b), so f(a) = b <= c. a is in A.

On the other hand, for each a in A, b=f(a) is in C, and f(f(a)) = a + c, so b is in B.