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Proving this modular problem?

  1. Sep 23, 2012 #1
    I have to prove "For every integer n, n^2 is congruent to exactly one 0,2,or 4mod 7? I don't even know where to start? Apparently its a problem that has 6 other problems to it too meaning a-f and my professor assigned us the last one.
     
  2. jcsd
  3. Sep 23, 2012 #2
    Hint: You don't have to do it for every n; just for n = 0, 1, 2, 3, 4, 5, 6. Why? And if you think about it a little, you really only need to test n = 0, 1, 2, and 3. Why?
     
  4. Sep 23, 2012 #3
    Hi
    Any integer is written as 7m+k where k=0,1,2,3,4,5,6. mod((7m+k)^2) = mod(k^2) for modulus 7.
    You can get possible cases in seven calculations. As SteveL27 says you can reduce number of calculations half with attention that mod((7m+k)^2)=mod((7m-k)^2)=mod({7(m-1)+(7-k)}^2)=mod((7-k)^2)=mod(k^2)

    In similar ways
    For modulus 2, Ans {0,1}
    For modulus 3, Ans {0,1}
    For modulus 4, Ans {0,1}
    For modulus 5, Ans {0,1,4}
    For modulus 6, Ans {0,1,3,4}
    For modulus 7, Ans {0,1,2,4}
    For modulus 8, Ans {0,1,2,3,4}
    For modulus 10, Ans {0,1,4,5,6,9}

    Regards.
     
    Last edited: Sep 23, 2012
  5. Sep 23, 2012 #4
    thanks for the explanation! I think I get what you guys say. I'm going to figure it out and see how far I can get.
     
  6. Sep 23, 2012 #5
    Oh that's because 4 and 6 are multiples of 2 and 3 and 5 is basically 2(2)+1.
     
  7. Sep 24, 2012 #6
    What I had in mind was that 4 = -3 (mod 7) so you dont have to test 4 since you're squaring.

    4^2 is already the same as (-3)^2 = 3^2. So you only need to check 3 but not 4. Likewise you don't have to check 5 or 6.
     
  8. Sep 24, 2012 #7

    mathwonk

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    the whole point of modular arithmetic is that every specific problem can be done in a finite amount of time by anyone no matter how clueless, just by trying all the cases.

    your predicament suggests that you lack the basic idea. hopefully it is getting clearer with this example.
     
  9. Sep 25, 2012 #8
    I proved it by using 4 cases where n=0,n=1,n=2, and n=3. And showed that each case was congruent to exactly one thing. As you use 4,5,6... it repeats so I chose not to show those in my proof.
     
  10. Sep 25, 2012 #9
    I don't see the work where you showed or gave an explanation why it repeats for 4,5,6?
     
  11. Sep 26, 2012 #10
    @bonfire09:
    Depending on the situation, you may want to show that the cases repeat. For example, showing these 3 computations should be satisfactory (though if this is for an introductory class, it might be prudent to be more rigorous than this):

    0^2 mod 7 = 0
    1^2 mod 7 = 1 = 6^2 mod 7
    2^2 mod 7 = 4 = 5^2 mod 7
    3^2 mod 7 = 2 = 4^2 mod 7

    Also, in response to your very first post, 1^2 is not in {0,2,4} mod 7 :P

    (I think you made a typo)
     
  12. Sep 26, 2012 #11
    No, Bonfire wrote 0, 1, 4, 2 as "one, 0, 4, 2". In my post, I tried to say that any proof must be complete,i.e. you can't merely ignore the 4,5,6.
     
  13. Sep 26, 2012 #12
    I know, that was what I was saying. I was responding to him, not you, sorry :/

    Also, I read that as a different syntax. I thought he was saying "congruent to exactly one 0,2,or 4mod 7" meaning that it was one of the values in the list {0,2,4}

    EDIT: This is one of those cases where you could get away with "WLOG" (Without Loss of Generality), depending on the level at which you are writing the proof.
     
  14. Sep 26, 2012 #13
    Yeah, it could be read either way, which is how some professors try to confuse things to make the student think. As for your edit, I disagree.
     
  15. Sep 26, 2012 #14
    Oh yeah here is my proof:
    Case:1 let n=0
    Since n=0 then n^2=0. Hence 0 is congruent to 0mod 7.
    Case 2:let n=1
    Since n=1 then n^2=1. Hence 1 is congruent to 1mod7.
    Case 3 let n=2
    Since n=2 then n^2=4. Hence 4 is congruent to 4mod7.
    Case 4 let n=3
    Since n=3 then n^2=9. Hence 9 is congruent to 2mod7.
    This is the rundown of my proof. There is not much to assume here except that n is an integer. And prove that each integer n^2 is congruent to exactly one of the four cases.
     
    Last edited: Sep 26, 2012
  16. Sep 26, 2012 #15
    What happens if n = 5793459834895479437?
     
  17. Sep 26, 2012 #16
    Dang! Oh i forgot to represent the integer n as an arbitrary value meaning that n=7m, 7m+1,7m+2,etc
    Where m is an integer.
     
  18. Sep 26, 2012 #17
    How does your proof address the question of whether n = 7m + 4 , 7m +5 and 7m +6 give one of the same values for n = 0 to 3?
     
  19. Sep 26, 2012 #18
    But the 4 cases you gave even if taken mod 7 do not address the case mentioned by SteveL27 since that is the case where n = 7m + 5, which is not one of the four cases you did.

    Actually you could had broken it down into one of the following 4 cases:

    n = 7m
    n = 7m +/- 1
    n = 7m +/- 2
    n = 7m +/- 3
     
    Last edited: Sep 26, 2012
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