Question

Use division algorithm to show that the square of any positive integers is of the form 3p,3p+1

Answer 1

Hey user here is your answer.....

----> Let a=3q {where q is an integer)

a=bq+r where r =(0,1,2)

a=3q +0

◆squarw both side

a^2=(3q)^2

◆a^2=9q^2

◆a^2=3(3q^2)

◆Where p=3q^2

◆r=1

◆a=3q+1

◆square both side

◆a^2=(3q+1)^2

◆a^2=9q^2+6q+1 ( By identity)

◆a^2= 3(3q^2+2q)+1

◆Where p= 3q^2+2q

------> So square of any positive interger is in form of 3p , 3p+1

Hope it helps you☺️

Answer 2

Given : square of any integer is of the form 3p or 3p+1.

To find :  Prove

Solution:

Any number can be represented  as

3q , 3q + 1 , 3q + 2   where q is integer

(3q)²

= 9q²  

= 3* 3q²  

= 3p    ( as q is integer => 3q²  is integer)

(3q + 1)²

= 9q² + 6q  + 1

= 3( 3q²  + 2q)  + 1

3q²  + 2q  is an integer as q is integer

= 3p + 1

(3q + 2)²

=  9q² + 12q  + 4

= 9q² + 12q  + 3 + 1

= 3( 3q²  + 4q + 1 )  + 1

3q²  + 4q + 1  is an integer as q is integer

= 3p + 1

Hence proved square of any integer is of the form 3p or 3p+1

Learn more:

Prove that only one of the numbers n-1, n+1 or n+3 is divisible by 3 ...

brainly.in/question/2077338

6 less to 'n' gives 8 is represented as (a) n

brainly.in/question/11758724

https://brainly.in/question/5835844