show that the square of any integer is of the form 9k or 3k+1.
Answers
Given : square of any integer is of the form 9k or 3k+1.
To find : Prove
Solution:
Any number can be represented as
3q , 3q + 1 , 3q + 2 where k is integer
(3q)²
= 9q²
= 9k ( as q is integer => q² is integer)
(3q + 1)²
= 9q² + 6q + 3
= 3( 3q² + 2q) + 1
3q² + 2q is an integer as q is integer
= 3k + 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
= 3k + 1
Hence square of any integer is of the form 9k or 3k+1
Learn more:
Prove that only one of the numbers n-1, n+1 or n+3 is divisible by 3 ...
https://brainly.in/question/2077338
6 less to 'n' gives 8 is represented as (a) n
https://brainly.in/question/11758724
Answer:
Use Euclid's algorithm to establish that (i) every odd integer is of the form 4k+1 or 4k+3. (ii) the square of any integer is either of the form 3k or 3k+1 (iii) the cube of any integer is of the from 9k, 9k+1 or 9k+8.