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A^2 + b^2 = c^3

  1. Mar 8, 2006 #1
    a^2 + b^2 = c^3
    (a,b,c - integers, different from each other)
    Is there a solution to this?
     
    Last edited: Mar 8, 2006
  2. jcsd
  3. Mar 8, 2006 #2

    Zurtex

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    A trivial solution would be:

    a = -1
    b = 0
    c = 1
     
  4. Mar 8, 2006 #3
    Of course, for a,b,c you can have:
    -5, 10, 5
    -17, 68, 17
    ... and so on.
    But lets add a condition |a|<>|b|<>|c|, or just
    a, b, c - positive integers, different from each other.
    Is there a solution, any idea?
     
  5. Mar 8, 2006 #4

    NateTG

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    There's also
    a=x^3,b=0,c=x^2
     
  6. Mar 8, 2006 #5
    I wrote a program in C++ for this and got the following results for
    a,b,c upto 30

    11, 2, 5
    18, 26, 10

    Of course u can put - signs for a and b.
     
  7. Mar 9, 2006 #6
    For any prime congruent to 1 mod 4, it is the sum of two squares. For that matter so is the product of two such cases:

    (a^2+b^2)(c^2+d^2) = (ac=+/-bd)^2+(ad-/+bc)^2.

    This then can be carried to any degree. For example: 5=2^2+1^2 and fits the bill. We have:

    5^3=5^2+10^2, and better: 2^2+11^2.
     
  8. Mar 9, 2006 #7
    :rofl: :rofl: :rofl:
     
  9. Mar 9, 2006 #8
    Let there be a pythagorian triplet x^2 + y^2 = z^2. Now multiply both sides with z^4,
    x^2*z^4 + y^2*z^4 = (xz^2)^2 + yz^2)^2 = z^6 = (z^2)^3.
    Find individual pythagorian triplets and you will find corresponding solutions by this method.
     
  10. Mar 10, 2006 #9
    Let x^2 + y^2 = z. Now multiply both sides with z^2,
    x^2*z^2 + y^2*z^2 = xz^2 + yz^2 = z^3
     
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