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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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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 \( m \) 
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