>for any y >= n with f(y) = 0 (mod n), there's some x between 0 and n-1

There's a simpler way to see this, any such y can be represented as y = nk + x where i,j are divisor & remainder. Then f(y) = f(nk + x) = f(x) modulo n since by binomial theorem all other terms other than those with just x will be divisible by n.