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

1. wio

Remains? Remainder?

2. frx

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

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

4. frx

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

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

6. frx

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

7. wio

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

We did answer the remain will be all Interger #

9. frx

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

10. wio

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

11. frx

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

12. frx

@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

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

Okey I get it, thanks a lot! :)

15. wio

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