## anonymous 4 years ago Which remains can occur when dividing $n^2$ with 3, let $n \in \mathbb{N}$

1. anonymous

Remains? Remainder?

2. anonymous

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

3. anonymous

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

4. anonymous

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?

5. anonymous

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

6. anonymous

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

7. anonymous

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?

8. anonymous

We did answer the remain will be all Interger #

9. anonymous

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

10. anonymous

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

11. anonymous

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

12. anonymous

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

13. anonymous

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

14. anonymous

Okey I get it, thanks a lot! :)

15. anonymous

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