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Math Help - Integer solutions to a^2+b^2=c^3

  1. #1
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    Integer solutions to a^2+b^2=c^3

    a^2+b^2=c^3, a,b,c are positive integers, and b>=a.

    Just by inspection, a=b=c=2 is one set of solution.

    Also, a=26, b=18, c=10; a=1358, b=594, c=130; are two other sets of solutions.

    If you multiply a and b by n^3 and c by n^2 for any positive integers n, you can easily get related sets of solutions.

    So my the question is:
    - Do other solutions exists?
    - Are there any solutions, such that a,b,c are pairwise coprime, exists?
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  2. #2
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    One solution set is:

    (a,b,c) = (|x^3 - 3xy^2|,|3x^2y-y^3|, x^2+y^2) (and removing those where b<a)

    Nor is it completely unmotivated. Consider the Gaussian integers:

    a^2+b^2 = c^3 \Rightarrow (a+ib)(a-ib) = c^3

    Suppose a+ib = (x+iy)^3 = x^3 - 3xy^2 + i(3x^2y-y^3)
    Last edited by SimonM; June 11th 2009 at 08:04 AM.
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  3. #3
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    Re: Integer solutions to a^2+b^2=c^3

    And of course I wonder how she looks like a formula describing their solutions. For the special case when X^2+Y^2=Z^3

    You can get a basic formula. Has the solutions:

    X=2k^6+8tk^5+2(7t^2+8qt-9q^2)k^4+16(t^3+2qt^2-tq^2-2q^3)k^3+ 2(7t^4+12qt^3+6q^2t^2-28tq^3-9q^4)k^2+8(t^5+2qt^4-2q^3t^2-5tq^4)k+

    +2(q^6-4tq^5-5q^4t^2-5q^2t^4+4qt^5+t^6)

    .................................................. .................................................. .............................................

    Y=2k^6+4(3q+t)k^5+2(9q^2+16qt+t^2)k^4+32qt(2q+t)k^  3+ 2(-9q^4+20tq^3+30q^2t^2+12qt^3-t^4)k^2+

    +4(-3q^5-tq^4+10q^3t^2+6q^2t^3+5qt^4-t^5)k- 2(q^6+4tq^5-5q^4t^2-5q^2t^4-4qt^5+t^6)

    .................................................. .................................................. .............................................

    Z=2k^4+4(q+t)k^3+4(q+t)^2k^2+4(q^3+tq^2+qt^2+t^3)k  +2(q^2+t^2)^2

    q,t,k - What are some integers any sign. After substituting the numbers and get a result it will be necessary to divide by the greatest common divisor. This is to obtain the primitive solutions.

    In the equation: X^2+Y^2=qZ^3 If the ratio is such that the root of an integer: c=\sqrt{q-1} Then the solution is:

    X=-2(c+1)p^6+4(2c(q-2)-3q)p^5s+2(c(5q^2-2q-8)-q^2-22q+8)p^4s^2 +8q(5q^2-14q+4)p^3s^3+2(c(5q^2-2q-8)+q^2+22q-8)q(q-2)p^2s^4-

    -4(2c(q-2)+3q)q^2(q-2)^2ps^5-2(c-1)q^3(q-2)^3s^6

    .................................................. .................................................. .................................................. .....

    Y=2(c-1)p^6+4(2qc+q-4)p^5s+2(c(-5q^2+18q-8)+15q^2-22q-8)p^4s^2 +8q(q^2+6q-12)p^3s^3-2(c(5q^2-18q+8)+15q^2-22q-8)q(q-2)p^2s^4-

    -4(2qc-q+4)q^2(q-2)^2ps^5+2(c+1)q^3(q-2)^3s^6

    .................................................. .................................................. .................................................. ......

    Z=2p^4+8p^3s+4(q^2-2q+4)p^2s^2-8q(q-2)ps^3+2q^2(q-2)^2s^4

    And more.

    X=-2(c-1)p^6-4(2c(q-2)+3q)p^5s+2(c(5q^2-2q-8)+q^2+22q-8)p^4s^2 +8q(5q^2-14q+4)p^3s^3+2(c(5q^2-2q-8)-q^2-22q+8)q(q-2)p^2s^4+

    +4(2c(q-2)-3q)q^2(q-2)^2ps^5-2(c+1)q^3(q-2)^3s^6

    .................................................. .................................................. .................................................. .......

    Y=2(c+1)p^6-4(2qc-q+4)p^5s+2(c(-5q^2+18q-8)-15q^2+22q+8)p^4s^2 +8q(q^2+6q-12)p^3s^3+2(c(-5q^2+18q-8)+15q^2-22q-8)q(q-2)p^2s^4+

    +4(2qc+q-4)q^2(q-2)^2ps^5+2(c-1)q^3(q-2)^3s^6

    .................................................. .................................................. .................................................. .......

    Z=2p^4-8p^3s+4(q^2-2q+4)p^2s^2+8q(q-2)ps^3+2q^2(q-2)^2s^4
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  4. #4
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    Re: Integer solutions to a^2+b^2=c^3

    Quote Originally Posted by linshi View Post
    a^2+b^2=c^3, a,b,c are positive integers, and b>=a.

    Just by inspection, a=b=c=2 is one set of solution.

    Also, a=26, b=18, c=10; a=1358, b=594, c=130; are two other sets of solutions.

    If you multiply a and b by n^3 and c by n^2 for any positive integers n, you can easily get related sets of solutions.

    So my the question is:
    - Do other solutions exists?
    - Are there any solutions, such that a,b,c are pairwise coprime, exists?
    A pairwise coprime solution

    2^2+11^2=5^3
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