1.Introduction

By Tito Piezas[1],it seems that x^3+y^3+z^3 = t^2 has a follwing parametric solution.

V. Bouniakowsky: 

(n^3+1)^3 + (-n^3+2)^3 + (3n)^3 = (3(n^3+1))^2

Generalize the above identity,I found a parametric solution of x^3+y^3+z^3 = t^n, n>=2.

[1].Tito Piezas:http://sites.google.com/site/tpiezas/011



2.Theorem
      
 x,y,z,t,m,n:integer

 condition: 
     3m+1=0 mod n.
     n>=2.
     

There is a parametric solution of x^3+y^3+z^3 = t^n,

(6*2^m*y^n*x^3 + 2^m*y^n)^3 + (-6*2^m*y^n*x^3 + 2^m*y^n)^3 + (-6*2^m*y^n*x^2)^3 

= {2^((3m+1)/n)*y^3}^n.
     
Proof.

(t+a)^3-(t-a)^3 = 6*a*t^2+2*a^3...........................................................(1)

Set t=6*a*x^3 and a=2^m*y^n,then 

((6*2^m*y^n*x^3+2^m*y^n)^3-(6*2^m*y^n*x^3-2^m*y^n)^3-(6*2^m*y^n*x^2)^3 = 2^(3m+1)*y^3n.....(2)

If 3m+1=0 mod n,then right hand side of (2) becomes to {2^((3m+1)/n)*y^3}^n.

 
Q.E.D.
　

 


3.Example

I show only the case of n=2 ,4,and 5.


Case. n=2

      (12*y^2*x^3 + 2*y^2)^3 + (-12*y^2*x^3 + 2*y^2)^3 + (-12*y^2*x^2)^3 = (4*y^3)^2

Case. n=4

      (12*y^4*x^3 + 2*y^4)^3 + (-12*y^4*x^3 + 2*y^4)^3 + (-12*y^4*x^2)^3 = (2*y^3)^4

Case. n=5

      (48*y^5*x^3 + 8*y^5)^3 + (-48*y^5*x^3 + 8*y^5)^3 + (-48*y^5*x^2)^3 = (4*y^3)^5