Question

Use euclid's division lemma to show that the square of any positive integer is either of the form 3m or m+1 for some integer m.

Answer 1

Let take a as any positive integer and b = 3. 

Then using Euclids algorithm we geta = 3q + r here r is remainder and value of q is more than or equal to 0 and r = 0, 1, 2 because 0 < r < b and the value of b is 3 So our possible values will 3q+0 , 3q+1 and 3q+2 

Now find the square of values 

Use the formula (a+b)² = a² + 2ab +b² to open the square bracket 

(3q)²�= 9q² if we divide by 3 we get no remainder

we can write it as 3*(3q²) so it is in form of 3m here m = 3q² 

(3q+1)² = (3q)² + 2*3q*1� + 1²

=9q² + 6q +1 now divide by 3 we get 1 remainder

so we can write it as 3(3q² + 2q) +1 so we can write it in form of 3m+1 and value of m is 3q² + 2q here 

(3q+2)² = (3q)² + 2*3q*2 + 2² 

=9q² + 12q +4 now divide by 3 we get 1 remainder 

so we can write it as 3(3q² + 4q +1) +1 so we can write it in form of 3m +1 and value of m will 3q² + 4q +1

Square of any positive integer is either of the form 3m or 3m + 1 for some integer m.

hence proved ...

please mark it as brainliest...and thank me

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.