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Pythagorean tripple

  1. Jan 24, 2010 #1
    The problem is 3^x + 4^x = 5^x. I recognize that this is a pythagorean triple but I am curious as to how you could solve this without just knowing that it is a pythagorean triple.
     
  2. jcsd
  3. Jan 24, 2010 #2

    Char. Limit

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    FLT states that the only way all of those solutions could be integers is if x=2.
     
  4. Jan 25, 2010 #3
    I am sure Fermats Theorem says what you say it says, however I am not sure that answer my question. I am trying to see if there is a way to solve the equation I have posted without just knowing the answer is 2.
     
  5. Jan 25, 2010 #4

    D H

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    Solve what equation? 3^2+4^2 = 5^2 is obviously true. Are you looking for a general way to generate a three positive integers (a,b,c) such that a^2+b^2=c^2?
     
  6. Jan 25, 2010 #5
    I guess I am looking for an algebraic way to solve 3^x + 4^x = 5^x for x. (sorry if I used the term algebraic incorrectly). So what I mean is take the natural log of both sides, then do this etc. Or perhaps integrate both sides, then do something else etc.

    Is it possible?
     
  7. Jan 25, 2010 #6
    No. It's called Fermat's last theorem.

    Fermat's Last Theorem states that no three positive integers a, b, and c can satisfy the equation [tex]a^x + b^x = c^x[/tex] for any integer value of x greater than two.

    So, assuming x is an integer, the only possible answers are 0, 1, and 2. Obviously, x = 2 works and the rest fail.
     
  8. Jan 25, 2010 #7
    Fermat-Wiles theorem. Give the guy credit who proved it along with the one who conjectured it.
     
  9. Jan 25, 2010 #8
    Ok. Fine. Say I didn't use 3,4,and 5 as a, b, and c. HOw do we know there are no non integer solutions.

    I'm making this up off the top of my head, so say we had 4^x + 10^x = 13.4534543^x How would you solve for x?

    0, 1 and 2 don't work obviously, if there were any other solutions for x, how would we solve it, if at all....
     
  10. Jan 25, 2010 #9

    Char. Limit

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    That is possibly a very good question...

    Could we take the log of it?Maybe. Would the answer then make sense?... Maybe. The question
    is... do we need an integer solution of x?
     
  11. Jan 25, 2010 #10
    [tex]
    \begin{array}{1}
    f(x) = 4^x + 10^x - 13.4534543^x \\
    f(1) > 0 \\
    f(2) < 0 \\
    [/tex]
    So by the intermediate value theorem, there should exist an [tex]x_0 \in [1,2] [/tex]such that
    [tex]
    f (x_0) = 0
    [/tex]
    So there is a solution, I just don't think it can be solved by standard methods. These kind of things can usually only be solved in special cases or through approximations.
     
  12. Jan 25, 2010 #11
    No I don't require integer solutions.

    L'hopital -- Thanks, I think I am finally getting the answer I was looking for. I think starting with such a well know triple threw people off.
     
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