Solution: Let a is any integer. According to division algorithm, every a is of the form 3q, 3q + 1, 3q + 2 i.e., a = 3q or 3q + 1, 3q + 2.

Now, for a = 3q

⟹ a^{2} = 9q^{2}

⟹ a^{2} = 3(3q^{2})

Let, 3q^{2} = k

∴ a^{2} = 3k

Again, for a = 3q + 1

** ⟹ a**^{2} = (3q +1)^{2}

** ⟹ a**^{2} = 9q^{2} + 6q + 1

** ⟹ a**^{2} = 3(3q^{2} + 2q) + 1

Let, 3q^{2} + 1 = k

∴ a^{2} = 3k + 1

And for a = 3q + 2

⟹ a^{2} = (3q + 2)^{2}

⟹ a^{2} = 9q^{2} + 12q + 4

⟹ a^{2} = 9q^{2} + 12q + 3 + 1

⟹ a^{2} = 3(3q^{2} + 4q + 1) + 1

Let, 3q^{2} + 4q + 1=k

∴ a^{2} = 3k + 1

*So, the Square of any integer is either of the form 3k or 3k + 1. * (Proved)

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