From: kohmoto@z2.zzz.or.jp (y.kohmoto)
Subject: Identity for 4 cubes
Date: 22 Jan 01 03:06:24 GMT
Newsgroups: sci.math.numberthy
Summary: Parameterized solutions for sums of 3 cubes to be a cube

    I discoverd a sequence of identities for a diophantine equation :
A^3+B^3+C^3=D^3 .
          The iteration for the identity is as follows :
         x0=9*a^4; y0=-9*a^4+3*a*b^3; z0=-9*a^3*b+b^4; u0=b^4
         x1=9*a^4; y1=-9*a^4-3*a*b^3; z1=9*a^3*b+b^4;   u1=b^4
         for n=2
         xn=(432*a^6-2*b^6) * x_(n-1) - x_(n-2)*b^6 - 108 * a^4*b^6
         yn=(432*a^6-2*b^6) * y_(n-1) - y_(n-2) *b^6- 108 * a^4*b^6
         zn=(432*a^6-2*b^6) * z_(n-1) - z_(n-2)*b^6 + 216 * a^6*b^4 + 4*b^10
         un= b^(6*n-2)
         for 3<=n
         xn=(432*a^6-2*b^6) * x_(n-1) - x_(n-2)*b^12 - 108 * a^4*b^(6*n-6)
         yn=(432*a^6-2*b^6) * y_(n-1) - y_(n-2) *b^12- 108 * a^4*b^(6*n-6)
         zn=(432*a^6-2*b^6) * z_(n-1) - z_(n-2)*b^12 + (216 * a^6*b^4
+4*b^10)*b(6*n-12)
         un= b^(6*n-2)

         for all n
         xn^3+yn^3+zn^3=un^3

         ex.
         If n=4, the identity is as follows :
    (725594112*a^22-31912704*b^6*a^16+194400*b^12*a^10-279*b^18*a^4)^3+
(  -725594112*a^22-241864704*b^3*a^19-8398080*b^6*a^16+2799360*b^9*a
^13+85536*b^12*a^10-7776*b^15*a^7-153*b^18*a^4+3*b^21*a)^3+
(  725594112*b*a^21+120932352*b^4*a^18-8398080*b^7*a^15-839808*b^10*
a^12+23328*b^13*a^9+1296*b^16*a^6-9*b^19*a^3+b^22    )^3=b^66

         If b=1, then it becomes the same thing as the iteration on
"Identities Of Equal Sums Of Like Power" :
         http://euler.free.fr/identities.htm

    Is it known?