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Answer :

Let a be any positive integer and b = 3.

Then a = 3q + r for some integer \(q \ne 0\)

And r = 0, 1, 2 because \( 0 \leq r < 3\)

\(\therefore \) a = 3q or 3q + 1 or 3q + 2

\(\Rightarrow \) \(a^2 = (3q)^2\) or \((3q + 1)^2 \) or \((3q + 2)^2\)

We know,

\(\Rightarrow \) \(a^2 = 9q^2 \) or \(9q^2 + 6q + 1 \) or \(9q^2 + 12q + 4\)

\(\Rightarrow \) \(a^2 = 3 × (3q^2)\) or \(3 × (3q^2 + 2q)\) or \(3 × (3q^2 + 4q + 1) + 1\)

\(\Rightarrow \) \(3k_1\) or \(3k_2 + 1\) or \(3k_3 + 1\)

Where \(k_1,k_2,k_3\) are some positive integers.

Hence, it can be said that the square of any positive integer is either of the form 3m or 3m + 1.

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