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Beale Conjecture Reduced to Practicality?

  1. Jun 6, 2013 #1
    What is the easiest way to explain the Beale Conjecture to someone who isn't math literate?

    BEAL'S CONJECTURE: If Ax + By = Cz, where A, B, C, x, y and z are positive integers and x, y and z are all greater than 2, then A, B and C must have a common prime factor.

    What exactly is Beale trying to extract or squeeze into his equation and what is it's practical application or outcome.If it could be proved,what is it's benefit to math or to anything?
  2. jcsd
  3. Jun 7, 2013 #2


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    Hey Skip Hawley and welcome to the forums.

    Basically in a nut-shell it means that you can simplify the equation and take a prime factor out of both sides. In other words:

    Ax + By = Cz is equivalent to:

    p(Dx + Ey) = p(Fz) for some integers D, E, and F where p is a prime number.

    In other words, if this theorem is true you will be able to do this factorization.
  4. Jun 7, 2013 #3


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    The conjecture as you state it doesn't look right to me.

    Take x=y=z=3, so you have: A+B=C, take for example A=1 and B=2, and C=3, so they don't have common prime factor, unless you include 1 as a prime number.

    Edit: it should be raised to the power...

  5. Jun 7, 2013 #4


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    If we assume that they don't have a common prime factor, which means that gcd(A,B,C)=1, then there exist integers w,v,u s.t: Aw+Bv+Cu=1, now I can plug this back to the equation, and check for some contradiction.

    Has someone tried this already?
  6. Jun 7, 2013 #5


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    I'm not sure adding more variables into the mix is going to make things easier!

    Furthermore, Beal's conjecture can be used to prove Fermat's Last Theorem by contradiction. I would be quite surprised if a proof of Beal's conjecture turned out to be easier than the proof of Fermat's Last Theorem. =)

    (As for counter-examples, it has been checked computationally that it holds for all variables up to 1000, so if a counter-example exists at least one of the numbers must be greater than 1000.)
  7. Jun 7, 2013 #6


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    Well, I recently heard that the proof of Wiles depends on non trivial axioms in set theory which involves large cardinals. And large cardinal are not that intuitive as peano axioms system. I mean the proof is proved in a stronger system, which means that there may be some obstacles to use the same proof in a system such as peano system, but I am just speculating here.
  8. Jun 7, 2013 #7
    Mathematical Physicist has already pointed out that you have misstated the conjecture, but in hopes of avoiding further confusion, the equation should be [tex]A^x + B^y = C^z[/tex].
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