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Use Euclid’s division lemma to show that the square of any positive integer is either of the form $$3m \;or\;3m+1$$ for some integer $$m.$$
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## QuestionMathsClass 10

Use Euclid’s division lemma to show that the square of any positive integer is either of the form $$3m \;or\;3m+1$$ for some integer $$m.$$

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## Solution

According to Euclid's Division Lemma if we have two positive integers $$a$$ and $$b$$, then there exist unique integers $$q$$ and $$r$$ which satisfies the condition $$a = bq + r$$ where $$0 ≤ r < b$$
So, By Euclid’s division lemma, we have $$a=bq+r \cdots(i)$$
On putting $$b=3$$ in $$Eq.(i)$$
We get $$a=3q+ r,$$ $$[0\leqslant r <3$$ i.e.$$r=0,1,2]$$
If $$r=0$$ $$\Rightarrow$$ $$a=3q$$ $$\Rightarrow$$ $$a^{2}=9q^{2}\cdots\cdots(ii)$$
If $$r=1$$ $$\Rightarrow$$ $$a=3q+1$$ $$\Rightarrow$$ $$a^{2}=9q^{2}+6q+1\cdots\cdots(iii)$$
If $$r=2$$ $$\Rightarrow$$ $$a=3q+2$$ $$\Rightarrow$$ $$a^{2}=9q^{2}+12q+4\cdots\cdots(iv)$$
From $$Eq.(ii),$$ $$9q^{2}$$ i.e. $$3(3q^2)$$ is of the form $$3m$$, where $$m=3q^{2}$$ .
From $$Eq. (iii),$$ $$9q^{2}+6q+1 \$$ i.e. $$3(3q^{2}+2q)+1$$ is of the form $$3m+1$$ ,where $$m=3q^{2}+2q$$ .
From $$Eq. (iv)$$ $$9q^{2}+12q+4$$ i.e. $$3(3q^{2}+4q+1)+1$$ is of the form, $$3m+1$$ , where $$m=3q^{2}+4q+1.$$
$$\therefore$$ The square of any positive integer is either of the form $$3m$$ or $$3m+1$$ for some integer $$m$$