Question

Use euclids division lemma to show that square of any +ve integer is either of form 3m or 3m+1 for some integer m.. plzz guys tell me...

Answer 1

X² is either  3 m or 3 m + 1          where X >= 1    and m >= 1
let z>=1

1) IF  3 | X X    =>    3 | X      =>      X = 3 z      X² =  3 * (3*z²)  
         THEN   X²  is of form 3m

2) IF  3 does not divide X X   =>  3 does not divide X    =>  X = 3 z+1  or   3z+2

  (i)  X²  = (3z+1)²  =  3 (3z²+2z) + 1  of  the form 3 m + 1

  (ii)  X² =  (3z+2)²  =  3(3z²+4z) + 4 = 3(3z²+4z+1) + 1       
            it is in form  3 m + 1

Hence proved. 
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any natural number, a square or otherwise is in form :  3 m ,  3 m + 1,  3 m + 2
But if it is a square then it is in form : 3 m  or 3m+1 only.

Answer 2

Step-by-step explanation:

let ' a' be any positive integer and b = 3.

we know, a = bq + r , 0 <  r< b.

now, a = 3q + r , 0<r < 3.

the possibilities of remainder = 0,1 or 2

Case I - a = 3q

a² = 9q² .

= 3 x ( 3q²)

= 3m (where m = 3q²)

Case II - a = 3q +1

a² = ( 3q +1 )²

=  9q² + 6q +1

= 3 (3q² +2q ) + 1

= 3m +1 (where m = 3q² + 2q )

Case III - a = 3q + 2

a² = (3q +2 )²

= 9q² + 12q + 4

= 9q² +12q + 3 + 1

= 3 (3q² + 4q + 1 ) + 1

= 3m + 1 ( where m = 3q² + 4q + 1)

From all the above cases it is clear that square of any positive integer ( as in this case a² ) is either of the form 3m or 3m +1.