# Need descrete math help

1. Nov 9, 2008

### mamma_mia66

1. The problem statement, all variables and given/known data
Show that n2 $$\neq$$2 (mod6) for all n in Z

2. Relevant equations

3. The attempt at a solution

0 1 2 3 4 5
0 0 0 0 0 0 0
1 0 1 2 3 4 5
2 0 2 4 0 2 4
3 0 3 0 3 0 3
4 0 4 2 0 4 2
5 0 5 4 3 2 1

I did only the table for mod 6 and then I don't have an idea what to do.
I am not even sure if I understand what exactly I have to do with this problem.

Last edited: Nov 9, 2008
2. Nov 9, 2008

### gabbagabbahey

Hmmm....have you tried proof by contradiction? That is, assume that $n^2 \equiv 2 \pmod{6}$....what does that imply?

3. Nov 9, 2008

### VeeEight

I did not attempt a solution but you might want to rewrite the statement.

n2 congruent to 2 mod 6 is the same as n2 - 2 is a multiple of 6. So you might want to define f(n) = n2 - 2 and show what happens when you divide f(n) by 6.

4. Nov 9, 2008

### mamma_mia66

I think the only hint I get for this was the reminder needs to be $$\neq$$2.

I am gessing that has something to do with division Algorithm. I will try the above ideas.

5. Nov 9, 2008

### Staff: Mentor

VeeEight said that "n$$^2$$ congruent to 2 mod 6 is the same as n$$^2$$ - 2 is a a multiple of 6."

That's also the same as saying that n$$^2 - 2 \equiv$$ 0 mod 6.

This one is ripe for a proof by induction.

6. Nov 10, 2008

### mamma_mia66

I don't think I know how to do it by induction. Thank you. I will try and I will come back again.

7. Nov 10, 2008

### Staff: Mentor

Pick a value of n for which your statement is true, such as n = 2.

Assume that for n = k, your statement is true. IOW, assume that k^2 != 2 mod 6.
Now show that for n = k + 1, (k + 1)^2 != 2 mod 6, using the induction hypothesis (the thing you assumed in the previous step).