1. The problem statement, all variables and given/known data Show that n^{2 [tex]\neq[/tex]}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.
Hmmm....have you tried proof by contradiction? That is, assume that [itex]n^2 \equiv 2 \pmod{6}[/itex]....what does that imply?
I did not attempt a solution but you might want to rewrite the statement. n^{2} congruent to 2 mod 6 is the same as n^{2} - 2 is a multiple of 6. So you might want to define f(n) = n^{2} - 2 and show what happens when you divide f(n) by 6.
I think the only hint I get for this was the reminder needs to be [tex]\neq[/tex]2. I am gessing that has something to do with division Algorithm. I will try the above ideas.
VeeEight said that "n[tex]^2[/tex] congruent to 2 mod 6 is the same as n[tex]^2[/tex] - 2 is a a multiple of 6." That's also the same as saying that n[tex]^2 - 2 \equiv[/tex] 0 mod 6. This one is ripe for a proof by induction.
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).