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.
