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Elementary math proof

  1. May 7, 2016 #1
    1. The problem statement, all variables and given/known data
    . Disprove the following statement: There exists integers a, b, c, none divisible by 7, such that 7|a^3 + b^3 + c^3
    2. Relevant equations


    3. The attempt at a solution
    if 7|a^3 + b^3 + c^3, then a^3 + b^3 + c^3 is congruent to 0(mod 7)

    if a,b,c are none divisible by 7 then I just work out the cases for 1,2,3,4,5,6 and show that there is no way to get to a^3 + b^3 + c^3 is congruent to 0(mod 7).

    Is that right?

    Is there an easier way to do it cause mine is very inefficient.
     
    Last edited: May 7, 2016
  2. jcsd
  3. May 7, 2016 #2

    Samy_A

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    Is the "0(mod 3)" that appears twice a typo, and do you mean "0(mod 7)"?
     
  4. May 7, 2016 #3
    sorry 0(mod7)
     
  5. May 7, 2016 #4

    Samy_A

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    Your method is not wrong.
    You don't give details on how you did it, so maybe what follows is moot.

    Let a be an integer not divisible by 7, and r = a (mod 7).
    What is the relation between a³ (mod 7) and r³ (mod 7)?
    What are the possible values for r³ (mod 7)?
     
  6. May 7, 2016 #5
    Yes I have the same approach a is congruent to 1,2,3,4,5,6 mod 7. Same thing for b and c. If I cube the congruence for each case, I show that there is no way you will get to a^3+b^3+c^3 congruent to 0(mod 7)

    a^3 congruent to 1,8,27,64,125 mod 7


    Ohhhh I got it so my last expression is equivalent to a^3 is congruent to 1 mod 7

    a^3 congruent to (1 mod 7)
    b^3 congruent to (1 mod7)
    c^3 congruent to 1 (mod7)

    a^3+b^3+c^3 is congruent to 3 mod 7
    right?
     
  7. May 7, 2016 #6

    Samy_A

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    Not quite.
    1 = 1 (mod 7)
    8 = 1 (mod 7)
    but 27 = 6 (mod 7)
    and so on.
    You also forgot 6³ = 216.

    But yes, there is a pattern.
     
  8. May 7, 2016 #7

    Samy_A

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    By the way: if you know Fermat's little theorem, then there is a more elegant solution.
     
  9. May 7, 2016 #8
    yeah I forgot that one xD, but yours state that 36 is congruent to 1 mod 7 and not 0 mod 7. If we try every case we should end up seeing that it's never congruent to 0 mod 7. Right?

    Everyone talks about fermat's little theorem. They always suggest me to use that in nearly 70% of the problems that I do XD. I will look at it.
     
  10. May 7, 2016 #9

    Samy_A

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    Not sure I understand what you say here.
    If you just inspect the values of 1³ (mod 7), 2³ (mod 7), ... , 6³ (mod 7), you will notice a clear pattern in the possible values.
    Now, you have to add the numbers (a³ + b³ + c³), and prove that that number is not divisible by 7. Once you have noticed the pattern mentioned above, you are left with a small number of cases to inspect.
     
  11. May 7, 2016 #10
    what I am saying is that

    a^3 is congruent to 6,1 mod 7
    b^3 is congruent to 6,1 mod 7
    c^3 is congruent to 6,1 mod 7

    so my point is that you can never add up these numbers to compute a multiple of 7. Examole, 6,6,6,.....6,1,1,......6,1,6
     
  12. May 7, 2016 #11

    Samy_A

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    Yes, that is correct. You basically have 4 cases: 1 1 1, 1 1 6, 1 6 6, 6 6 6, and none adds up to a multiple of 7.
     
  13. May 7, 2016 #12
    thanks!!!!!!. I will study the fermat's little theorem. It's too famous.
     
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