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Home / Elementary Number Theory / The Square of any integer is either of the form 3k or 3k + 1

# The Square of any integer is either of the form 3k or 3k + 1

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

⟹ a2 = 9q2

⟹ a2 = 3(3q2)

Let, 3q2 = k

∴ a2 = 3k

Again, for a = 3q + 1

⟹ a2 = (3q +1)2

⟹ a2 = 9q2 + 6q + 1

⟹ a2 = 3(3q2 + 2q) + 1

Let, 3q2 + 1 = k

∴ a2 = 3k + 1

And for a = 3q + 2

⟹ a2 = (3q + 2)2

⟹ a2 = 9q2 + 12q + 4

⟹ a2 = 9q2 + 12q + 3 + 1

⟹ a2 = 3(3q2 + 4q + 1) + 1

Let, 3q2 + 4q + 1=k

∴ a2 = 3k + 1

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

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