1.Introduction

A.S. Janfada and  A. Abbaspour[1] showed that equation x^6+6z^3=y^6+6w^3 has infinite non-trivial primitive integer solutions.

We show that x^9 + az^8 = y^9 + aw^8 has a parametric solution.
     
     
2.Theorem
     
Diophantine equation x^9 + az^8 = y^9 + aw^8 has two parametric solutions.

(x,y,z,w)=(  -pa(r^8-s^8)(p^9-q^9)^7, -qa(r^8-s^8)(p^9-q^9)^7, -ra(r^8-s^8)(p^9-q^9)^8, -sa(r^8-s^8)(p^9-q^9)^8 )
a,p,q,r,s are arbitrary.

(x,y,z,w)=(  p(-r^8+s^8), q(-r^8+s^8), r(-r^8+s^8), s(-r^8+s^8) )
a = p^9-q^9
p,q,r,s are arbitrary.

 
Proof.

x^9 + az^8 = y^9 + aw^8................................................(1)

Substitute x=pt, y=qt, z=rt, w=st to equation (1), we obtain

t(p^9-q^9) = -a(r^8-s^8)...............................................(2)

Hence we obatin t = -a(r^8-s^8)/(p^9-q^9).

Finally, we obtain a parametric solution below.

(x,y,z,w)=( -pa(r^8-s^8)(p^9-q^9)^7, -qa(r^8-s^8)(p^9-q^9)^7, -ra(r^8-s^8)(p^9-q^9)^8, -sa(r^8-s^8)(p^9-q^9)^8 )

Let a=p^9-q^9, we obtain another parametric solution.
(x,y,z,w)=(  p(-r^8+s^8), q(-r^8+s^8), r(-r^8+s^8), s(-r^8+s^8) )


Q.E.D.



3.Reference

[1].A.S. Janfada and  A. Abbaspour, On Diophantine equations X^6+ 6Z^3 = Y^6± 6W^3,
    International Journal of Pure and Applied Mathematics, VOL:105, NO:4,2015