1.Introduction

By x^3+2y^3+3z^3=n with the Mathoverflow, it seems that x^3+2y^3+3z^3=36  has no intger solution.

If we accept the rational number solution, we can easily obtain the solution x^3+2y^3+3z^3=n for any n.

We show a parametric solution of ax^3 + by^3 + cz^3 = n for any n.


2.Theorem
      
 a,b,p,q,n: arbitrary

 condition: 
 a+b=c.
     
There is a parametric solution of ax^3 + by^3 + cz^3 = n,

      x = 1/3((2ab^2+a^2b)p^3+(-3a^2b-6ab^2)qp^2+(6ab^2+3a^2b)q^2p+(-a^2b-2ab^2)q^3+a^2n+nb^2+2anb)/((a+b)ab(p-q)^2)

      y = 1/3((-2a^2b-ab^2)p^3+(6a^2b+3ab^2)qp^2+(-3ab^2-6a^2b)q^2p+(2a^2b+ab^2)q^3+a^2n+nb^2+2anb)/((a+b)ab(p-q)^2)

      z = 1/3((ab^2-a^2b)p^3+(3a^2b-3ab^2)qp^2+(3ab^2-3a^2b)q^2p+(-ab^2+a^2b)q^3-a^2n-nb^2-2anb)/((a+b)ab(p-q)^2).


     
Proof.

ax^3 + by^3 + cz^3 = n.........................................................(1)

Let c = a+b.

Set x=t+p, y=t+q, z=-t+r.......................................................(2)

(3br+3ar+3bq+3ap)t^2+(3ap^2-3br^2-3ar^2+3bq^2)t+ar^3+bq^3+ap^3+br^3-n..........(3)

Set r = -(bq+ap)/(a+b), then we obtain t = -1/3(2a^2bp^3+a^2bq^3-a^2n-3bqa^2p^2+2ab^2q^3-3b^2q^2ap-2anb+ap^3b^2-nb^2)/((a+b)ab(p^2-2qp+q^2)).

Substitute r and t to (2), and obtain a parametric solution.

 
Q.E.D.
　

3.Example


Case. x^3 + 2y^3 + 3z^3 = 36

x = 1/9(5p^3-5q^3+162-15qp^2+15q^2p)/((p-q)^2)

y = -2/9(2p^3-2q^3-81-6qp^2+6q^2p)/((p-q)^2)

z = 1/9(p^3-q^3-162-3qp^2+3q^2p)/((p-q)^2)

p,q: arbitrary