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Math Help - diophintine equation perhaps?

  1. #1
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    diophintine equation perhaps?

    Are there integers m and n so that m^{2} + (m+1)^{2} = n^{4} + (n + 1)^{4}?
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  2. #2
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    m=n=0
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  3. #3
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    Quote Originally Posted by FancyMouse View Post
    m=n=0

    And m=-1\,,\,n=0\,,\,\,m=0\,,\,n=-1 ...

    Tonio
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  4. #4
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     m^2 + (m+1)^2 = n^4 + (n+1)^4

    Let  m= a- \frac{1}{2} ,  n = b-\frac{1}{2}

    Substitute them into the original equation :

     (a-\frac{1}{2})^2 + (a+\frac{1}{2})^2 = (b-\frac{1}{2})^4 + (b+\frac{1}{2})^4

     2a^2 + 2(\frac{1}{2})^2 = 2b^4 + 12b^2(\frac{1}{2})^2 + 2(\frac{1}{2})^4

     a^2 + (\frac{1}{2})^2 = b^4 + \frac{3}{2} b^2 + \frac{1}{16}

     a^2 + (\frac{1}{2})^2 = (b^2 + \frac{3}{4} )^2 + \frac{1}{16} - \frac{9}{16}

    Back substitution ,

     (m+\frac{1}{2})^2 + ( \frac{1}{4} + \frac{1}{2}) = [ (n+\frac{1}{2})^2 + \frac{3}{4} ]^2

     (2m+1)^2 + 3 = 4 ( n^2 + n + \frac{1}{4} + \frac{3}{4} )^2

     (2m+1)^2 + 3  = 4 (n^2+n+1)^2

    Put , owing to brevity  x = 2n^2 + 2n + 2 , ~y = 2m+1

    we obtain  x^2 - y^2 = 3

     |x| = 2 , |y|=1



    If

     n^2+n+1 = 1
     n= 0 or  n=-1

     2m+1 = 1 or  2m+1 = -1
     m = 0 or  n=-1

    Note that  n^2 + n + 1 = -1 doesn't have real solution so we conclude that  m,n are either  0 or  -1 .
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