Subjective Type

Write whether the square of any positive integer can be of the form $$3m+2$$, where $$m$$ is a natural number. Justify your answer.

Solution

By Euclid's division lemma, $$b = aq + r$$
where $$a, b, q, r$$ are $$+ve$$ integers and here $$a=3$$ then $$b = 3q + r$$ then $$0 \le r < 3$$ or $$r= 0, 1, 2$$ so $$b$$ becomes $$b = 3q, 3q + 1, 3q+2$$,

$$b= 3q$$

$$\Rightarrow (b)^2 = (3q)^22$$

$$\Rightarrow b^2 = 3 . 3q^2 = 3m$$ where, $$3q^2 = m$$

So, as $$b^2$$ is perfect square so $$3m$$ will also be perfect square.
when $$r = 1, b= 3q+1$$

$$\Rightarrow (b)^2 = (3q+1)^2$$

$$\Rightarrow b^2 = 9q^2 + 1 + 2 \times 3q$$

$$\Rightarrow b^2 = 3[3q^2 + q] + 1$$

$$\Rightarrow b^2 = 3m + 1$$ and $$m= 3q^2 + 2q$$

So, $$b^2$$ is perfect square or a number of the form $$3m+1$$ is perfect square.

when $$r = 2, b= 3q+2$$

$$\Rightarrow b^2 = 9q^2 + 4 + 2. 3q . 2$$

$$= 9q^2 + 3 + 3 \times 4q + 1$$

$$= 3[3q^2 + 1 + 4q] + 1$$

$$\Rightarrow b^2 = 3m + 1$$

Again, a number of the form $$3m + 1$$ is perfect square.

Hence, a number of the $$(3m+2)cm$$ never be perfect square. But a number of the form $$3m$$, and $$3m+1$$ are perfect squares.

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