Let P = N² where N is an odd integer. What is the remainder when P is divided by 8?

Any odd integer N can be written as 2k+1. Then N² = (2k+1)² = 4k² + 4k + 1 = 4k(k+1) + 1. Since k(k+1) is always even (product of consecutive integers), 4k(k+1) is a multiple of 8. So N² = 8m + 1 for some integer m. Hence remainder when N² is divided by 8 is 1.