Let P = N^2 where N is an odd integer. What is the remainder when P is divided by 8 ?

Explanation

Any odd integer can be represented as N = 2k + 1. Thus, N^2 = 4k^2 + 4k + 1 = 4k(k+1) + 1. Since k(k+1) is the product of two consecutive integers, it is always even (2m). Substituting gives N^2 = 8m + 1. Hence, dividing by 8 always leaves a remainder of 1.

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