# need descrete math help

by mamma_mia66
Tags: descrete, math
 P: 52 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. Please help me if you can.
 HW Helper P: 5,004 Hmmm....have you tried proof by contradiction? That is, assume that $n^2 \equiv 2 \pmod{6}$....what does that imply?
 P: 612 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.
P: 52

## need descrete math help

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.
 Mentor P: 19,710 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.
 P: 52 I don't think I know how to do it by induction. Thank you. I will try and I will come back again.
 Mentor P: 19,710 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).

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