1.Introduction

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

V. Bouniakowsky: 

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


E. Catalan:

(a^4+2ab^3)^3 + (b^4+2a^3b)^3 + (3a^2b^2)^3 = (a^6+7a^3b^3+b^6)^2


A. Gerardin:

(9a^4+8ab^3)^3 + (4ab^3)^3 + (4b^4)^3 = (27a^6+36a^3b^3+8b^6)^2



I add four different parametric solutions of x^3+y^3+z^3 = t^2.

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



2.Result
      
 

 1. ( (s^2+3t^2)^2 )^3 + ( -(s^2-3t^2)^2 )^3 + (-12s^2t^2)^3 = (-6s^5t+54st^5)^2

 

    (x+a)^3-x^3-a^3 = 3ax(a+x)................................................(1)

　  Set a=3(2st)^2 and x=(s^2-3t^2)^2 then 3ax(a+x) becomes to square.


 2. ( 18t^6-1 )^3 + ( 9t^6+1)^3 + (-9t^4)^3 = ( 9t^3(9t^6-1) )^2


    (x+a)^3+(-x+b)^3+(-1)^3 = (3a+3b)x^2+(3a^2-3b^2)x+a^3+b^3-1...............(2)

    Set a=t^2 and b=2t^2 then (2) becomes to 9t^2x^2-9t^4x+9t^6-1.

    Hence,set x=1/9(9t^6-1)/(t^4) then we obtain above identity.
 

 3. ( 12s^2t^3+2s^2 )^3 + ( -12s^2t^3+2s^2 )^3 + (-12s^2t^2)^3 = ( 4s^3 )^2


   (x+a)^3-(x-a)^3 = 6ax^2+2a^3...............................................(3)

   Set x=6at^3 and a=2s^2 then (3) becomes to 1728s^6t^6+16s^6.


 4. ( 1+3s^3 )^3 + ( 2-3s^3 )^3 + (-3s)^3 = ( -3+9s^3 )^2


   (x+a)^3+(-x+b)^3 = (3a+3b)x^2+(3a^2-3b^2)x+a^3+b^3.........................(4)

   Set b = -a+3,then (4) becomes to (5).

   (3x)^2+(3a^2-3b^2)x+a^3+b^3................................................(5)

   Let (3a^2-3b^2)x+a^3+b^3=27s^3.

   Hence,x=-(a^2-3a+3-3s^3)/(2a-3)............................................(6)

   Substitute a=2 to (6),then x=-1+3s^3.

   Substitute x=-1+3s^3 to (4),then we obtain above identity.

   This identity is very similar to Bouniakowsky's one!