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Need descrete math help

by mamma_mia66
Tags: descrete, math
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mamma_mia66
#1
Nov9-08, 07:52 PM
P: 52
1. The problem statement, all variables and given/known data
Show that n2 [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.
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gabbagabbahey
#2
Nov9-08, 07:59 PM
HW Helper
gabbagabbahey's Avatar
P: 5,003
Hmmm....have you tried proof by contradiction? That is, assume that [itex]n^2 \equiv 2 \pmod{6}[/itex]....what does that imply?
VeeEight
#3
Nov9-08, 08:02 PM
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.

mamma_mia66
#4
Nov9-08, 08:11 PM
P: 52
Need descrete math help

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.
Mark44
#5
Nov9-08, 08:57 PM
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P: 21,215
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.
mamma_mia66
#6
Nov10-08, 07:24 AM
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.
Mark44
#7
Nov10-08, 02:13 PM
Mentor
P: 21,215
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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