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Thread: diophintine equation perhaps?

  1. #1
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    diophintine equation perhaps?

    Are there integers m and n so that $\displaystyle m^{2} + (m+1)^{2} = n^{4} + (n + 1)^{4}$?
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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 $\displaystyle m=-1\,,\,n=0\,,\,\,m=0\,,\,n=-1$ ...

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

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

    Substitute them into the original equation :

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

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

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

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

    Back substitution ,

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

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

    $\displaystyle (2m+1)^2 + 3 = 4 (n^2+n+1)^2 $

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

    we obtain $\displaystyle x^2 - y^2 = 3 $

    $\displaystyle |x| = 2 , |y|=1 $



    If

    $\displaystyle n^2+n+1 = 1 $
    $\displaystyle n= 0 $ or $\displaystyle n=-1$

    $\displaystyle 2m+1 = 1$ or $\displaystyle 2m+1 = -1 $
    $\displaystyle m = 0 $ or $\displaystyle n=-1$

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