use euclid's division lemma to show that the square of any positive integer is either of the form 3m, 3m+1

Answer 1

all Z numbers can be written in these 3 forms:
N=3m Or N=3m+1 Or N=3m+2
now the square of n can befound in these 3 forms :
1- N=3m then N^2 = 9m^2 then N=(3)*(3m^2) so it is divisable by 3
2- N=3m+1 then N^2 = 9*m^2 + 6*m + 1 and N= 3*(m^2 + 2*m)+1 and it can be written as 3M+1
3- N=3m+2 and N= 9m^2+12m+4 then N= 3*(3m^2+4m+1)+1
and it is N=3M+1
so the square of no number in Z have the remainder 2 when divided by 3

Answer 2

If n is any positive integer, then n == k (mod 3) to one and only one k in {0, 1 ,2} (== means congruent to).

Therefore, according to the properties of congruences,

n^2 == k^2 (mod) 3

If k =0 (n is a multiple of 3) then n == 0 (mod 3), so that n = 3m for some positive integer m.

If k =1, then n^2 == 1 (mod 3), so that n^2 = 3m + 1 for some integer m >=0.

If k =2, the n^2 == 4 (mod 3) Since 4 == 1 (mod 3), it follows
n^2 == 1 mod (3), so that n^2 = 3m + 1 for some integer m >=0.

The proof is now complete.

What I did was, in the end, exactly what Saman did in the previous answer. But when you're aware of the propeties of congruences, things get a bit simpler.