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Stumped math problem

  1. Jun 14, 2005 #1
    I want to find three integers, a b and c, with a^3 + b^3 = 17 * c^3.
    all each the smallest integer possible
     
  2. jcsd
  3. Jun 15, 2005 #2

    matt grime

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    you could factor a^3+b^3 for a start, I suppose. 17 being prime must divide one of the factors. we can also suppose that a,b a nd c have no common factors.
     
  4. Jun 15, 2005 #3

    saltydog

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    Well I like Number Theory too. Before I start coding this into Mathematica in a brute-force search, you know, (1,1), (1,2),. . . (2,1),(2,2), . . . I suppose there is no algebraic method for doing so? And while I'm at it, why not just search for the relation involving an arbitrary prime instead of 17 and then see if any conclusion can be drawn say for the first 100 primes.
     
  5. Jun 15, 2005 #4

    Zurtex

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    Assuming your not looking for the trivial solution (0,0,0) and other trivial solutions like (1, -1, 0). Then mathematica can't find a single instance where it is true.
     
  6. Jun 15, 2005 #5

    matt grime

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    there are algebraic ways of factoring such things in number fields, though i don'y know much about them - they are of elementary (in the sense of important) interest in algebraic number theory.

    you can also get conditions on a,b and c by considering it modulo some inetger. eg doing so mod 8 the cubic remainders are 0,1,3,5,7 and we need

    a^3+b^3=c^3 mod 8

    so that the only non trivisl way to do it is to have a^3=1,b^3=7 mod 8 or a^3=3, b^3=5 mod 8 or a^3=b^3=c^3=0 mod 8
     
    Last edited: Jun 15, 2005
  7. Jun 15, 2005 #6

    saltydog

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    Why not Zurtex? I respect you and Matt's opinions but I'm stubborn. You mind if I check?
     
  8. Jun 15, 2005 #7

    matt grime

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    perhaps zurtex checked it in mathematica?

    i meant to say, that the case a^3=b^3=c^3 can be dismissed as trivial since it implies that all of a,b,c are even thus it is not a "primitive" solution, but i forgot the cases

    a^3=0 and b^3=c^3 mod 8
     
  9. Jun 15, 2005 #8

    matt grime

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    looking mod 7 i think we find that 7 divides c and a^3 and b^3 are +/-1 mod 7 and mod 17 that a+b=0
     
  10. Jun 15, 2005 #9

    saltydog

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    Alright, I'd like to qualify Zurtx's statement if I may: Mathematica cannot find any value of a and b under 5000 which satisfy the equation. Frankly, if I had access to a faster PC I'd run it up to a million at least as well as optimize my algorithm. It kinds looks like it's related to Fermat's theorem. Is there a proof that there is no solution?

    Wait, he did say integers. Need to check some negative numbers too. Didn't say I was good in Number Theory, only that I'm interested in it. :smile:
     
    Last edited: Jun 15, 2005
  11. Jun 15, 2005 #10

    Zurtex

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    Actually I just used the FindInstance function for c > 0 and then for c < 0.
     
  12. Jun 15, 2005 #11

    saltydog

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    FindInstance? Suppose that's a 5.0 feature. I have 4.1.

    IndustriaL, where you got this anyway? I'd like to see some closure in the matter. You know Richard Guy has a nice book out, "Unsolved Problems in Number Theory". Don't suppose anything like this is in there?
     
  13. Jun 15, 2005 #12

    matt grime

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    i suspect if you have a basic knowledge of cubic number fields then the answer is known - we can do it for quadratics in quadratic number fields.
     
  14. Jun 15, 2005 #13

    saltydog

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    Very well Matt. I'll leave it there.
     
  15. Jun 15, 2005 #14

    matt grime

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    i say that but i am only guessing from the results about when we can solve x^2-dy^2=something; i don't have any knowledge of this at all.
     
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