#### Question

Prove that the square of any positive integer is of the form 3m or, 3m + 1 but not of the form 3m +2.

#### Solution

By Euclid’s division algorithm

a = bq + r, where 0 ≤ r ≤ b

Put b = 3

a = 3q + r, where 0 ≤ r ≤ 3

If r = 0, then a = 3q

If r = 1, then a = 3q + 1

If r = 2, then a = 3q + 2

Now, (3q)^{2} = 9q^{2}

= 3 × 3q^{2}

= 3m, where m is some integer

(3q + 1)^{2} = (3q)^{2} + 2(3q)(1) + (1)^{2}

= 9q^{2} + 6q + 1

= 3(3q^{2} + 2q) + 1

= 3m + 1, where m is some integer

(3q + 2)^{2} = (3q)^{2} + 2(3q)(2) + (2)^{2}

= 9q^{2} + 12q + 4

= 9q^{2} + 12q + 4

= 3(3q^{2} + 4q + 1) + 1

= 3m + 1, hwrer m is some integer

Hence the square of any positive integer is of the form 3m, or 3m +1

But not of the form 3m + 2

Is there an error in this question or solution?

Solution Prove that the Square of Any Positive Integer is of the Form 3m Or, 3m + 1 but Not of the Form 3m +2. Concept: Euclid’s Division Lemma.