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Math Help - Cube numbers

  1. #1
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    Cube numbers

    Show that if the difference of two consecutive cube numbers is a square number ( (k+1)^3-k^3=n^2) then there exists two integers a and b for which n=a^2+b^2.
    Last edited by james_bond; March 22nd 2010 at 09:15 AM.
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  2. #2
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    Quote Originally Posted by james_bond View Post
    Show that if the difference of two consecutive cube numbers is a square number ( (k+1)^3-k^3=n^2) then there exists two integers a and b for which n=a^2+b^2.
    This looks like an interesting problem!

    If (k+1)^3-k^3=n^2 then 3k^2 + 3k -(n^2-1) = 0. For that quadratic equation in k to have an integer solution, the discriminant 9+12(n^2-1) = 3(4n^2-1) must be a square. So let 4n^2 - 1 = 3m^2, and hence (2n)^2 - 3m^2 = 1. The solution to that Pell's equation comes from convergents to the continued fraction expansion of \sqrt3. The j'th solution (k_j,n_j) is given by 2n_j+(2k_j+1)\sqrt3 = (2+\sqrt3)^{2j+1}.

    The solutions n_j turn out to be sums of consecutive squares, satisfying the equation n_j = p_j^2 + (p_j+1)^2, where p_j = \tfrac12(\sqrt{2n_j-1}-1). That raises two questions. (1) Why should 2n_j-1 be a square? (2) Why is that formula true? I can't answer either of those questions.

    The first few values of k, n and p are given in this table (entries taken from the Sloane sequences A001921, A001570 and A001571).

    \begin{array}{r|r|r}k&n&p\\ \hline 0&1&0\\ 7&13&2\\ 104&181&9\\ 1455&2521&35\\ 20272&35113&132\\ 282359&489061&494\\ 3932760&6811741&1845\end{array}
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  3. #3
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by Opalg View Post
    The j'th solution (k_j,n_j) is given by 2n_j+(2k_j+1)\sqrt3 = (2+\sqrt3)^{2j+1}.

    The solutions n_j turn out to be sums of consecutive squares, satisfying the equation n_j = p_j^2 + (p_j+1)^2, where p_j = \tfrac12(\sqrt{2n_j-1}-1). That raises two questions. (1) Why should 2n_j-1 be a square? (2) Why is that formula true? I can't answer either of those questions.
    1.)  2n_j-1 square implies  n_j = 2t^2+2t+1 for some  t (I leave out my work ).

    We also know  n_j = \frac{(2+\sqrt{3})^{2j+1}-(2k_j+1)\sqrt{3}}{2} .

    Maybe it's possible to use the latter formula to derive the desired relation... I save this for another day.


    2.) Does n_j = p_j^2 + (p_j+1)^2, where p_j = \tfrac{1}{2}(\sqrt{2n_j-1}-1)?

    This can be shown true by simple algebra.
    Last edited by chiph588@; March 27th 2010 at 09:26 PM.
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