 Compute remainders upon division by 3 as "n" takes successive values 0, 1, 2, ….For the first of these polynomials, the sequence of remainders is 1, 2, 2, 1, 2, 2, …, while for the second, it is 2, 0, 0, 2, 0, 0, ….This implies that in the sequence of values taken by the second polynomial, two out of every three are divisible by 3, and hence certainly not prime, while in the sequence of values taken by the first polynomial, none are divisible by 3.