
#1
Nov908, 07:52 PM

P: 52

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. 



#2
Nov908, 07:59 PM

HW Helper
P: 5,004

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




#3
Nov908, 08:02 PM

P: 612

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. 



#4
Nov908, 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. 



#5
Nov908, 08:57 PM

Mentor
P: 20,937

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. 



#6
Nov1008, 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.




#7
Nov1008, 02:13 PM

Mentor
P: 20,937

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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