Using Eulid's divison lemma, show that the square of any positive interger is either of the form 3m or ( 3m+1) for some integer m. |

Explanation

Let n be an arbitrary positive integer.
on dividing n by 3, let q be the quotient and r be the remainder. Then, by Eulid's divison lemma, we have
n = 3q +r , where ` 0 le r lt 3`
`n^(2) = 9q^(2), + r^(2) + 6qr `
Case I where r=0
Putting r=0 in (i) , we get
` n^(2) , = 9q^(2)= 3( 3q^(2)) = 3m` where ` m= 3q^(2)` is an integer.
Case II Putting r=1 in (i) we get
` n^(2) = (9q^(2) +1 + eq) = 3 ( 3q^(2) + 2q) +1 = 3 ( 3q^(2) + 2q) =1`
= 3 m +1, where m = ` ( 3q^(2) + 2q) ` is an integer.
Case III when r =2
Putting r=2 in (i), we get
` n^(2) = (9q^(2) +4 +12q) = 3 ( 3q^(2) +4q +1)+1`
= 3m+1 , where m = `( 3q^(2) + 4q +1)` is an interger.
Hence, the square of any positive integer is of the form 3m or ( 3m+1) for some integer m