## frx Group Title Which remains can occur when dividing $n^2$ with 3, let $n \in \mathbb{N}$ one year ago one year ago

1. wio Group Title

Remains? Remainder?

2. frx Group Title

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

3. ktnguyen1 Group Title

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

4. frx Group Title

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. ktnguyen1 Group Title

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

6. frx Group Title

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

7. wio Group Title

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. ktnguyen1 Group Title

We did answer the remain will be all Interger #

9. frx Group Title

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

10. wio Group Title

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

11. frx Group Title

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

12. frx Group Title

@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. ktnguyen1 Group Title

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. frx Group Title

Okey I get it, thanks a lot! :)

15. wio Group Title

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