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Sorry some difficulities with the language. What i mean is if you divide for example 4 with 3 you get the remainder 1..
all I number it present as this a/3 and a can be (0,1,2,3...n)
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?
yes you got it by the way (m+2)^2=m^2+4m+4
Great! Oh, of course it's, my bad ;)
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?
We did answer the remain will be all Interger #
Then you divide n^2/3, which remainders can be left over like the remainder in the euclidean algorithm
Well you can only get remainders of 0, 1, and 2... that's obvious.
What about the 4 I got above when n= (m+2) -> m^2+4m+4 ?
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
Okey I get it, thanks a lot! :)