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

Let a be any integer then,
a =3n or 3n+1 or 3n+2
where n is any integer

Answer 2

when a=3n then
a^2 = (3n)^2 = 3(3n^2) , which is in the form 3k

when a=3n+1 then
a^2= (3n+1)^2 = 9n^2 +6n+1=3(3n^2+n) +1, which is in the form 3k+1

when a=3n+2 then
a^2=(3n+2)^2 = 9n^2 + 12n + 4 = 3(3n^2 +4n +1) +1, which is in the form 3k+1

Answer 3

Thus, the square of any integer a is either of the form 3k or 3k+1 for some integer k

Answer 4

got it?