I need help with this question.

(a) Use the quotient-remainder theorem with divisor equal to 3 to prove that the square of any integer has the form 3k or 3k + 1 for some integer k. Proof: Suppose n is any integer. By the quotient-remainder theorem with d = 3, we know that n = 3q, or n = 3q + 1, or n = 3q + 2 for some integer q. We must show that regardless of which of these happens to be the case, the conclusion of the given statement follows (that is, n has the required form). Case 1 (n = 3q for some integer q): In this case, 12 by substitution = 3 by algebra. In terms of q, let k = Then k is an integer because it is a ---Select--- | of integers. Hence, there is an integer k such that n = 3k. Case 2 (n = 3q + 1 for some integer q): In this case, n2 = by substitution S 92 + + = 3 92 + + by algebra. In terms of q, let k = Then k is an integer because sums and products of integers are integers. Hence, there is an integer k such that n2 = 3k + ? )