# Modular arith 2

1. Oct 2, 2008

### phyguy321

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

2. Oct 2, 2008

### morphism

And what have you tried...?

3. Oct 3, 2008

### phyguy321

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

but i couldnt prove n $$\cong$$ 0 mod 3 so i gave up

4. Oct 4, 2008

### morphism

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