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^8 + az^2 = y^8 + aw^2 has a parametric solution.
     
     
2.Theorem
     
Diophantine equation x^8 + az^2 = y^8 + aw^2 has a parametric solution.

(x,y,z,w)=( 2pta, 2qta, 8(t^2q^8-t^2p^8+a)t^3a^3, -8t^3a^3(-t^2q^8+t^2p^8+a) )
a,p,q,t are arbitrary.
 
Proof.

x^8 + az^2 = y^8 + aw^2....................................................(1)

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

a(z+w)(z-w) = q^8-p^8......................................................(2)

Hence we obatin (z,w) =( 1/2(t^2q^8-t^2p^8+a)/(ta), -1/2(-t^2q^8+t^2p^8+a)/(ta)  ).

Finally, we obtain a parametric solution below.

(x,y,z,w)=( 2pta, 2qta, 8(t^2q^8-t^2p^8+a)t^3a^3, -8t^3a^3(-t^2q^8+t^2p^8+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