Show that the square of any positive integer is of the form 3m or 3m+1

1.	Show that the square of any positive integer is of the form 3m or 3m+1?
Answer» Let a be any odd positive integer.We apply the division lemma witha and b = 3 .Since 0 ≤ r < 3 , the possible remainders are 0 , 1 and 2 .That is , a can be 3q , or 3q + 1 , or 3q + 2 , where q is the quotient .Now ,a² = ( 3q )² = 9q²Which can be written in the form= 3 ( 3q² )= 3m , [ since m = 3q² ]It is divisible by 3 .Again ,a² = ( 3q + 1 )²= 9q² + 6q + 1= 3( 3q² + 2q ) + 1= 3m + 1 [ since m = 3q² + 2q , 3( 3q² + 2q ) is divisible by 3 ]Lastly ,a² = ( 3q + 2 )²= 9q² + 12q + 4= ( 9q² + 12q + 3 ) + 1= 3( 3q² + 4q + 1 ) + 1Which can be written in the form3m + 1 , since3( 3q² + 4q + 1 ) is divisible by 3.Therefore ,The square of any positive integer is either of the form 3m or 3m + 1 for some integer m . Let 3 be divisor and solve it

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