Show that the square of an odd positive integer is of the form 8q + 1, for some integer q.
Since any odd positive integer n is of the form 4p + 1 or 4p + 3.
When n = 4p + 1, then n2 = (4p + 1)2 = 16p2 + 8p + 1 ⇒ 8(2p2 + 1) + 1 ⇒ 8q + 1 where q = p(2p + 1)
If n = 4p + 3, then n2 = (4p + 3)2 = 16p2 + 24p + 9 ⇒ 8(2p2 + 3p + 1) + 1 ⇒ 8q + 1 where q = 2p2 + 3p + 1
From above results we got that n2 is of the form 8q + 1.