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Math Help - x^n+y^n=z^(n+1)

  1. #1
    MHF Contributor Also sprach Zarathustra's Avatar
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    x^n+y^n=z^(n+1)

    How many solution in integers have to the following equation:
    ( n\in \mathbb{N})

    x^n+y^n=z^{n+1}
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    Quote Originally Posted by Also sprach Zarathustra View Post
    How many solution in integers have to the following equation:
    ( n\in \mathbb{N})

    x^n+y^n=z^{n+1}
    Obviously loads of solutions if n=1. Also if n=2, for example 11^2+2^2 = 125 = 5^3. In fact, let z be any number with the property that each of its prime factors of the form 4k+3 occurs to an even power. Then z^3 is the sum of two squares.

    Another solution, valid for all n, is that 2^n+2^n=2^{n+1}.
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    MHF Contributor Also sprach Zarathustra's Avatar
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    And what about n>3?
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    Quote Originally Posted by Also sprach Zarathustra View Post
    And what about n>3?
    Apart from that solution x = y = z = 2, I have no idea.
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  5. #5
    MHF Contributor undefined's Avatar
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    For n=3, we have some solutions x=y=2^1, 2^5, 2^9, 2^13, ...

    For n=4, we have some solutions x=y=2^1, 2^6, 2^11, 2^16, ...

    etc.

    Edit: More solutions for n=3 given by (x,y)

    9,18
    28,84
    65,260
    70,105
    126,630
    144,288
    162,162
    266,665
    273,364
    456,760
    469,603
    756,945
    793,854

    n = 4

    17,34
    82,246
    194,291
    486,486

    PARI/GP

    Code:
    f(x,y,z)=for(i=1,x,for(j=i,y,a=i^z+j^z;b=round(a^(1/(z+1)));if(b^(z+1)==a,print(i," ",j))))
    Then for example

    Code:
    f(1000,1000,3)
    
    f(1000,1000,4)
    Last edited by undefined; July 17th 2010 at 03:14 PM.
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  6. #6
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    Quote Originally Posted by Opalg View Post
    Obviously loads of solutions if n=1. Also if n=2, for example 11^2+2^2 = 125 = 5^3. In fact, let z be any number with the property that each of its prime factors of the form 4k+3 occurs to an even power. Then z^3 is the sum of two squares.

    Another solution, valid for all n, is that 2^n+2^n=2^{n+1}.
    Yes, then the general solution for  n = 2 is given by :


     (x,y,z) = ( p^3 - 3pq^2 ,  3p^2 q - q^3  , p^2 + q^2 )
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  7. #7
    MHF Contributor Also sprach Zarathustra's Avatar
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    Quote Originally Posted by simplependulum View Post
    Yes, then the general solution for  n = 2 is given by :


     (x,y,z) = ( p^3 - 3pq^2 ,  3p^2 q - q^3  , p^2 + q^2 )
    You are very close to the solution...
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