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How to solve for solutions to the diophantine 5b^2*c^2 = 4a^2(b^2+c^2)

  1. Dec 3, 2012 #1
    I am trying to find a way to generate solutions to 5b^2*c^2 = 4a^2(b^2+c^2)

    Can anyone offer some insight?

    I know that (b^2+c^2) is the part that is divisible by 5
     
  2. jcsd
  3. Dec 3, 2012 #2

    haruspex

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    The interesting cases will be when HCF(a,b,c)=1, so assume that. First, consider what cases that leaves where some two of the three have a common factor.
    When no two have a common factor, look for an interesting factorisation. With squares, that's typically going to be of the form (x-y)(x+y). Hint: try writing the 5 as 4+1.
     
  4. Dec 3, 2012 #3
    I am sorry, I don't really understand what you mean here at all. Can you provide an example?
     
  5. Dec 4, 2012 #4

    haruspex

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    I'm suggesting breaking the problem into 3 cases:
    1. a, b, c have a common factor. This is trivially reducible to the other cases by factoring it out.
    2. Some pair of a, b, c have a common factor. E.g. consider p divides a m times and b n times. Then you can show either p = 2 or m=n, and maybe deduce some more consequences.
    3. No two have a common factor.
    Rewrite the equation as 4b2(c2-a2) = c2(4a2-b2) then factorise. What can the prime factors of c divide on the LHS?
     
  6. Dec 4, 2012 #5
    I know you are trying to help but I sincerely have absolutely no idea what that aims to solve?

    It seems like you're advocating some form of just iterating through all a,b,c in order to get all the cases and break them out into these three categories, but I'm not sure how this is any better than brute force.

    1. You're saying gcd(a,b,c)>1 here?
    2. either gcd(a,b)>1, gcd(a,c)>1, or gcd(b,c)>1?
    3. gcd(a,b,c)=1

    How does rewriting the equation that way help, and factorize which part?

    If it helps any, I am only looking for cases for which gcd(a,b,c)=1 and a<b<c<2*a
     
  7. Dec 4, 2012 #6
    for example

    [209, 247, 286],
    [341, 374, 527],
    [779, 950, 1025],
    [1711, 2146, 2183]
    ... etc
     
  8. Dec 4, 2012 #7
    Lastly, I just tried looking at the prime factors of c with respect to the lhs and nothing unusual cropped up (for instance using the cases I just posted). It's not like the primes all exclusively divide 4b^2 or (c^2-a^2) if that's what you're saying.

    It can divide 4*b*b*a*a though; not sure if this matters
     
  9. Dec 4, 2012 #8

    haruspex

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    No, not brute force. I'm suggesting that in each case you can make interesting but different logical deductions which might eventually allow you to characterise all solutions.
    Take e.g. gcd(a,c)=1 and gcd(b,c) = 1. Then c = 1 or 2. Indeed, if gcd(a,b)=1 then b|c.
     
  10. Dec 4, 2012 #9
    Why?
     
  11. Dec 4, 2012 #10
    I don't know why Dickfore; it just is

    haruspex: Yes but I am after the gcd(a,b,c) = 1 cases which don't seem to have those same unique attributes
     
  12. Dec 4, 2012 #11

    micromass

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    That doesn't really make any sense. If you don't know why, then why did you add the condition?? Is it given in the problem statement that 5 divides [itex]a^2+b^2[/itex]?? Or...
     
  13. Dec 4, 2012 #12

    haruspex

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    Well, that is very easy to prove.
    Yes, I understand that. The cases I elaborated upon, like (a,b)=1, necessarily satisfy (a,b,c)=1. Do you mean you are most interested in the cases where (a,b,c)=1 but some pair does have a common factor? Or perhaps where each pair has a common factor?
     
  14. Dec 4, 2012 #13
    Doesn't (b^2 + c^2) or a^2 have to be divisible by 5 as the left side of the equation has 5 as a factor. Also b or c must be even as 5(B^2)(C^2) is equal to 4 times a number.
     
  15. Dec 4, 2012 #14
    Apparently, it isn't for the OP. If one is incapable of verifying this statement, then the analysis you suggested is beyond their comprehension.
     
  16. Dec 4, 2012 #15
    So, what if we assume that [itex]b^2 + c^2[/itex] is not divisible by 5?
     
  17. Dec 4, 2012 #16
    I've since figured out my own problem, no need to discuss this any further
     
  18. Dec 5, 2012 #17
    Hi SeventhSigma,

    I am always fascinated by solutions to Diophantine equations.
    I am glad that you figured it out, but we haven't learned anything
    from this discussion. Can you please share your solution with us?

    Thanks,

    LittleNewton
     
  19. Dec 5, 2012 #18
    Well it would be helpful to us to give us some insight as to what you discovered or found. As to the proof that (b^2 + c^2) must be divisible by 5 Dickfore was trying to get you to think like a mathematican. If 5 doesn't divide (b^2 + c^2) then it must divide a^2. But if 5 divides a^2 and not b^2 + c^2 then the right hand side must be divisible by an even power of 5. Is ths so with the left hand side? Don't let this forum get you down, it is very helpful for those willing to think for themselves. It would be nice for you to return the favor.
    PS, Although you may have been irritated by one or more of us, please be advised that none of the posts in this thread appear to have been meant to belittle you. It just that its sometimes hard to choose from simply spoon feeding detailed information to someone and just giving that person enough information so that he or she will have the ability to effectively solve a problem. I know that it is human nature for each of us to sometimes have a mental block at times and at that time a little more information to wake us up and help us think straight may be more helpful in the long run than just spoon fed detailed information.
     
    Last edited: Dec 5, 2012
  20. Dec 5, 2012 #19
    So, could you post your solution here for others to use?
     
  21. Dec 5, 2012 #20
    I made the reply "I don't know, it just is" when I was really tired and cranky. I did know why and what I meant by that was "I don't know, just look at it dude, it's trivial to prove."

    And I feel little incentive to share my solution since I feel like I did not get any help from this thread at all whatsoever, but instead unwarranted belittling.
     
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