[dh2022]

n = 0 or 1 modulo 2. So we have to check out only two cases module 2, and these cases trivial. To prove the problem note that 6 = 2 * 3 and then is trivial to see that the polynomial is=0 modulo 2 if n=0,1 modulo 6and then check it is =0 modulo 3 for n=0,1,2 modulo 3 and you are done.