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^7 + az^3 = y^7 + aw^3 has infinitely many parametric solutions.
     
     
2.Theorem
     
If a=p^18+q^3p^15+q^6p^12+q^9p^9+q^12p^6+q^15p^3+q^18, then
equation x^7 + az^3 = y^7 + aw^3 has infinitely many parametric solutions.
p,q are arbitrary.
 
Proof.

x^7 + az^3 = y^7 + aw^3.................................................................................................(1)

Substitute x=p^3, y=q^3, z=mt+q, w=t+p to equation (1), we obtain

(am^3-a)t^3+(-3ap+3aqm^2)t^2+(-3ap^2+3aq^2m)t+p^21+aq^3-ap^3-q^21=0.....................................................(2)

Let a = p^18+q^3p^15+q^6p^12+q^9p^9+q^12p^6+q^15p^3+q^18, then equation (2) becomes equation (3).

((p^18+q^3p^15+q^6p^12+q^9p^9+q^12p^6+q^15p^3+q^18)m^3-p^18-q^3p^15-q^6p^12-q^9p^9-q^12p^6-q^15p^3-q^18)t^2
+(-3(p^18+q^3p^15+q^6p^12+q^9p^9+q^12p^6+q^15p^3+q^18)p+3(p^18+q^3p^15+q^6p^12+q^9p^9+q^12p^6+q^15p^3+q^18)qm^2)t
-3(p^18+q^3p^15+q^6p^12+q^9p^9+q^12p^6+q^15p^3+q^18)p^2+3(p^18+q^3p^15+q^6p^12+q^9p^9+q^12p^6+q^15p^3+q^18)q^2m=0.......(3)

In order to have rational solutions for t, we must have rational solutions of equation (4).
Let U=m.
V^2 = -3q^2U^4+12p^2U^3-18pqU^2+12q^2U-3p^2.............................................................................(4)

Since quartic equation (4) has a rational solution Q(U,V)=(p^2/q^2,3p(p-q)(p^2+pq+q^2)/(q^3)),
this quartic equation (4) is birationally equivalent to an elliptic curve below.

Y^2-4(q^3-2p^3)YX/(pq) = X^3-2(2q^6+q^3p^3-p^6)X^2/(q^2p^2)+108p^2(p^6-2q^3p^3+q^6)X/(q^4)-216(3q^3p^9-q^6p^6-3q^9p^3-p^12+2q^12)/(q^6)
Transformation is given, 
U = (24q^9-12p^3q^6-6p^2Xq^5-24p^6q^3+p^3Yq^3+6p^5Xq^2+12p^9)/(pYq^5)
V = -3(216p^19+100q^2p^15X+16q^3p^13Y-6q^4p^11X^2-300q^5p^12X-40q^6p^10Y
    -q^6p^7X^3+12q^7p^8X^2+332q^8p^9X+q^9p^4X^3+6q^10p^5X^2-164q^11p^6X
    +40q^12p^4Y+432p^4q^15-1080p^7q^12+432p^10q^9+864p^13q^6-864p^16q^3-12q^13X^2p^2-16q^15Yp+32q^17X)
    /(q^9p^3Y^2)
X = (6q^3p^4V-6p^8-6q^6pV+6q^6p^2+12q^8U-36q^5p^3U+24q^2p^6U)/(q^6U^2-2q^4Up^2+q^2p^4)
Y = (24q^13U^2-120q^11p^2U-12q^10p^3U^2+36q^9p^3V-12p^4q^9+312q^8p^5U-24q^7p^6U^2+24q^6p^7
    -72q^6p^6V-312q^5p^8U+12q^4p^9U^2+36q^3p^9V+12p^10q^3+120q^2p^11U-24p^13)
    /(q^9pU^3-3q^7p^3U^2+3q^5p^5U-p^7q^3)    
    
The point corresponding to point Q is P(X,Y)=(2(2q^6+q^3p^3-p^6)/(q^2p^2), 8(2q^9-3p^3q^6-3p^6q^3+2p^9)/(q^3p^3)).

This point P is of infinite order, and the multiples nP, n = 2, 3, ...give infinitely many points.

Hence we can obtain infinitely many parametric solutions for equation (1).

Case : n=1
Q(U,V)=(p^2/q^2,3p(p-q)(p^2+pq+q^2)/(q^3))
(x,y,z,w)=( p^3(p+q)^3(p^2-pq+q^2)^3,
            q^3(p+q)^3(p^2-pq+q^2)^3,
            -(p+q)^6(p^2-pq+q^2)^6q(2p^3-q^3),
            (p+q)^6(p^2-pq+q^2)^6p(-2q^3+p^3) )

Case : n=2
(x,y,z,w)=( p^3(q+p)^3(q^2-pq+p^2)^3(q^4-2q^3p-2qp^3+p^4)^3(q^8+2q^7p+4q^6p^2-4q^5p^3-5p^4q^4-4p^5q^3+4q^2p^6+2qp^7+p^8)^3,
            q^3(q+p)^3(q^2-pq+p^2)^3(q^4-2q^3p-2qp^3+p^4)^3(q^8+2q^7p+4q^6p^2-4q^5p^3-5p^4q^4-4p^5q^3+4q^2p^6+2qp^7+p^8)^3,
            (q+p)^6(q^2-pq+p^2)^6(q^4-2q^3p-2qp^3+p^4)^6(q^8+2q^7p+4q^6p^2-4q^5p^3-5p^4q^4-4p^5q^3+4q^2p^6+2qp^7+p^8)^6q(-q^3+2p^3)(-q^12-10q^9p^3+12q^6p^6-4q^3p^9+2p^12),
            (q+p)^6(q^2-pq+p^2)^6(q^4-2q^3p-2qp^3+p^4)^6(q^8+2q^7p+4q^6p^2-4q^5p^3-5p^4q^4-4p^5q^3+4q^2p^6+2qp^7+p^8)^6p(-2q^3+p^3)(-2q^12+4q^9p^3-12q^6p^6+10q^3p^9+p^12) )
           


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