# 3|n^2 implies 3|n

1. Mar 25, 2012

### cubicmonkey

I'm completely lost on this one. I need this to be able to solve that the square root of three is irrational. So it's a proof within a proof, but I like this way best. Please help me out.
This is what I need to know how to prove.

3|n^2 implies 3|n where n is some integer.

2. Mar 25, 2012

### HallsofIvy

Staff Emeritus
3 is a prime number. If 3 were not a prime factor of n, it could not be a factor of n2.

3. Mar 26, 2012

### cubicmonkey

I think I understand what you're saying and that looks like a contrapositive proof, but you don't actually prove it. Could you elaborate?

4. Mar 26, 2012

### pedja

Suppose :

$$n \equiv a \pmod 3 ~\text{and}~ a \neq 0$$
then :
$$n^2 \equiv a^2 \pmod 3 ~\text{and}~ a^2 \neq 0$$
hence :
$$3 \nmid n^2$$
contradiction .

Q.E.D.

5. Mar 26, 2012

### 20Tauri

I think the Fundamental Theorem of Arithmetic (uniqueness up to order of prime factorizations) might help?

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