Question

prove that the square of any integer a is either of the form 3k or 3k+1 for some integer k. [hint by division algorithm, a must be of the form 2q, 3q +1, or 3q+2]

Answer 1

any integer can be written as 3q, 3q +1, or 3q+2....q being an integer

case 1 when n=3q
clearly n^2=9q^2=3(3q^2)

Answer 2

when n=3q +1
n^2=9q^2+6q+1=3(3q^2+2q)+1

Answer 3

when n=3q +2
n^2=9q^2+12q+4=3(3q^2+4q+1)+1 (form 3k+1)

Answer 4

the square of any integer a is either of the form 3k or 3k+1