Use Euclid’s Division Lemma to show that the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.

Solution :

By Euclid’s Division Algorithm, we have

a = bq + r       …………..(i)

On putting b = 3 in (1), we get

a = 3q + r,      [0 \(\le\) r < 3]

If r = 0   a = 3q  \(\implies\)  \(a^2\) = \(9q^2\)                       …..(2)

If r = 1  a = 3q + 1   \(\implies\)  \(a^2\) = \(9q^2 + 6q + 1\)                       …..(3)

If r = 2   a = 3q + 2    \(\implies\)  \(a^2\) = \(9q^2 + 12q + 4\)                       …..(4)

From (2), \(9q^2\) is a square of the form 3m, where m = \(3q^2\)

From (3), \(9q^2 + 6q + 1\)  i.e. , 3(\(3q^2 + 2q\)) + 1 is a square which is of the form 3m + 1 where m = \(3q^2 +2q\)

From (4), \(9q^2 + 12q + 14\)  i.e. , 3(\(3q^2 + 2q + 1\)) + 1 is a square which is of the form 3m + 1 where m = \(3q^2 +4q + 1\)

Therefore, the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.

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