Number Problem 19

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If the natural number N is not a multiple of 3, prove that N^2 is not also a multiple of 3.

Solution: N is not a multiple of 3.

Let N be 3n + a, where a is 1 or 2.

N^2 = (3n + a)^2 = 9n^2 + 6an + a^2

= 3n (3n + 2a) + a^2

= a multiple of 3 + a^2

a^2 is not divisible by 3 since a = 1 or 2

Therefore, N^2 is not a multiple of 3