Using the same convention as yours, I tried to break c_10 in terms of c_9 as follows: (a^10-b^10) = (e a + f b) * (a^9-b^9) Multiplying out the RHS, e = f = 1

Using the same convention as yours, I tried to break c_10 in terms of c_9 as follows: (a^10-b^10) = (e a + f b) * (a^9-b^9) Multiplying out the RHS, e = f = 1 are the best choices, but there remained a residual. At a second glance though, this residual was proportional to (a^8-b^8). Problem solved.