Question

Square of any positive integer is of the form 3k or 3k+1 for some positive integer k

Answer 1

let, any positive integer=p
divide it by 3
then, possible value=3p or 3p+1 or 3p+2[r=0,1,2]
after squaring, we obtain
(3p)^2 or (3p+1)^2 or (3p+2)^2
=9p^2 or 9p^2+6p+1 or 9p^2+12p+4
=3(3p^2) or 3(3p^2+2p)+1 or 3(3p^2+4p)+4
let, 3p^2=k or 3p^2+2p=k or 3p^2+4p=k
then, 3k or 3k+1 or 3k+4

so, square of any integer is of the firm of 3k or 3k+1 or 3k+4 for some positive integer k.

hope this will help you.
plz mark me as brainliest.

Answer 2

What confuses me about this is that I think I am able to show that the square of any integer is in the form X∗k where x is any integer. For Example:

x=3q+0x=3q+1x=3q+2

I show 3k first

(3q)2=3(3q2)

where k=3q2 is this valid use of the division algorithm?

If it is then can I also say that int is in the form for example 10*k

for example

(3q)2=10∗(910q2)

where k=(910q2)

Why isn't this valid? Am I using the div algorithm incorrectly to show that any integer is the form 3k and 3k+1, if so how do I use it? Keep in mind I am teaching myself Number Theory and the only help I can get is from you guys in stackexchange.