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^2 = y^9 + aw^2 has a parametric solution.
     
     
2.Theorem
     
Diophantine equation x^9 + az^2 = y^9 + aw^2 has a parametric solution.

(x,y,z,w)=( 4pt^2a^2, 4qt^2a^2, 256(t^2q^9-t^2p^9+a)t^8a^8, -256t^8a^8(-t^2q^9+t^2p^9+a) )
a,p,q,t are arbitrary.
 
Proof.

x^9 + az^2 = y^9 + aw^2....................................................(1)

Substitute x=p, y=q to equation (1), we obtain

a(z+w)(z-w) = q^9-p^9......................................................(2)

Hence we obatin (z,w) =( 1/2(t^2q^9-t^2p^9+a)/(ta), -1/2(-t^2q^9+t^2p^9+a)/(ta) ).

Finally, we obtain a parametric solution below.

(x,y,z,w)=( 4pt^2a^2, 4qt^2a^2, 256(t^2q^9-t^2p^9+a)t^8a^8, -256t^8a^8(-t^2q^9+t^2p^9+a) )

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