Question

Which remains can occur when dividing \[n^2\] with 3, let \[n \in \mathbb{N} \]

Answer 1

Remains? Remainder?

Answer 2

Sorry some difficulities with the language. What i mean is if you divide for example 4 with 3 you get the remainder 1..

Answer 3

all I number it present as this a/3 and a can be (0,1,2,3...n)

Answer 4

Ok I think i get what you say but what if I want to be more general. Let's say that n is an interger which can be written on the form m, m+1,m+2..., where 3|m then

n = m -> m^2
n= (m+1) -> m^2+2m+1
n= (m+2) -> m^2+4m+2

since 3|m then 3|m^2+2m+1 and 3|m^2+4m+2 so every integer should be devisible by 3, am I on the right track?

Answer 5

yes you got it by the way (m+2)^2=m^2+4m+4

Answer 6

Great! Oh, of course it's, my bad ;)

Answer 7

Sorry, but it's still not a clear question. You want the remainder of n^2 / 3 in terms of the remainder of n/3?

Answer 8

We did answer the remain will be all Interger #

Answer 9

Then you divide n^2/3, which remainders can be left over like the remainder in the euclidean algorithm

Answer 10

Well you can only get remainders of 0, 1, and 2... that's obvious.

Answer 11

What about the 4 I got above when n= (m+2) -> m^2+4m+4 ?

Answer 12

@wio could you please explain why the remainders only can be 0,1,2 and why it's so obvious, would really appreciate it :)

Answer 13

the 4 because I mean n=(m+2) but we have to find n^2 mean (m+2)^2= m^2+4m+4 I just want to corrct the formula you typed previous

Answer 14

Okey I get it, thanks a lot! :)

Answer 15

@frx If you have some number, 3m+4, then you have 3m+3+1, which is 3(m+1) + 1... remainder 1!