Modular Congruences of Integer Squares

[phyguy321]

prove that for any integer n, n^{2} \cong 0 or 1 (mod 3), and n^{2} \cong 0,1,4(mod 5)

 

[morphism]

And what have you tried...?

 

[phyguy321]

The only thing i found was that if you can prove n\congm mod 3 than n^{2} \cong m^{2} mod 3

but i couldn't prove n \cong 0 mod 3 so i gave up

 

[morphism]

You only have to consider n^2 (mod 3) when n=0,1,2. A same type of comment applies mod 5. (Why?)