Ok, I just realized that the book gives a hint saying, "Since n is odd, there exists an integer u for which n = 2u + 1. There are now two cases, depending on whether u is even or odd."
So, I started out by saying:
(I re-stated the hint) We proceed by cases.
Case 1: Assume u is even. Then, there exists an integer k for which u = 2k. So, n = 2u + 1 = 2(2k) + 1 = 4k + 1. So, n^2 = (4k + 1)^2 = 16k^2 + 8k + 1.
I see the 8k + 1 part, which is what I will need, but the 16k^2 part is throwing me off on what I do from here. Any help would be appreciated.