# Divisibility of n by 6

1. Dec 12, 2006

### ToastMonger

I need to prove that if n is a natural number and n^2 is divisible by 6, then n is divisible by 6.

I know that I knew how to do this at one point fairly recently, if you could refresh my memory I would greatly appreciate it.

2. Dec 12, 2006

Write n out in terms of it's prime factors, and it should be obvious.

3. Dec 13, 2006

### CRGreathouse

Suppose n was not divisible by 2. Then n^2 could not be divisible by 6.
Suppose n was not divisible by 3. Then n^2 could not be divisible by 6.
So n is divisible by 2 and 3, hence by 6.

4. Dec 13, 2006

### murshid_islam

we know that if p divides ab, then p divides a or p divides b.
now if 6 divides n.n, then 6 divides n.

5. Dec 13, 2006

### morphism

Here's another way to look at it.

n(n+1)(2n+1)/6 is the sum of the first n squares, and thus must be an integer. Its numerator is 2n^3 + 3n^2 + n. So if n^2 is divisible by 6, n must be as well.

6. Dec 13, 2006

### Hurkyl

Staff Emeritus
6 isn't a prime number, so we cannot apply that theorem directly...

7. Dec 13, 2006

### Gokul43201

Staff Emeritus
Why are folks writing up complete solutions to what appears to be a textbook problem?

8. Dec 13, 2006

### murshid_islam

now if 6 divides n.n, then 3 divides n.n and 3 divides n.
again, if 6 divides n.n, then 2 divides n.n and 2 divides n.
so then 3 divides n and 2 divides n. therefore, 6 divides n.

Last edited: Dec 13, 2006
9. Dec 13, 2006

### Gokul43201

Staff Emeritus
Great! That makes 3 complete solutions for the OP to copy down. Anyone else?

10. Dec 30, 2007

### fishingspree2

Hello, my math skills are crap, but can't we simply do that?
If n is divisible by 6, then it can be writed as 6x where x is an integer.

$$\left(6x\right)^{2}=6^{2}x^{2}=36x^{2}=6\left(6x^{2}\right)$$

11. Dec 30, 2007

### dodo

You could do that... if the intention was to show "if n is divisible by 6, then n^2 is too". But the original question was the other way around: "if true for n^2, then it's true for n".