Use Euclid’s division lemma to show that the square of any positive integer is of the form 3p, 3p + 1
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If a and b are two positive integers, then,
a = bq + r, 0 r b Let b = 3
Therefore, r = 0, 1, 2
Therefore, a = 3q or a = 3q + 1 or a = 3q + 2
If a = 3q a2 = 9q2 = 3(3q2) = 3m where m = 3q2
If a = 3q + 1 a2 = 9q2 + 6q + 1 = 3(3q2 + 2q) + 1 = 3m + 1 where m = 3q2 + 2q
If a = 3q + 2 a2 = 9q2 + 12q + 4 = 3(3q2 + 4q + 1) + 1= 3m + 1, where m = 3q2 + 4q + 1
Therefore, the square of any positive integer is either of the form 3m or 3m + 1.
HOPE ITS HELPFUL ^__^
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