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.

If we allow obvious solutions, we can obtain the parametric solution of x^7+az^n=y^7+aw^n for any n.
We show that x^7 + az^n = y^7 + aw^n, n=2,3,4,5,6,7 has the parametric solutions.
     
     
2.Theorem
     
Diophantine equation x^7 + az^n = y^7 + aw^n, n=2,3,4,5,6,7 has the parametric solutions.

x^7 + az^2 = y^7 + aw^2:

a=q^7-p^7
(x,y,z,w)=(4pt^2a^2, 4qt^2a^2, 64(t^2q^7-t^2p^7+a)t^6a^6, 64t^6a^6(-t^2q^7+t^2p^7+a))
a,p,q,t are arbitrary.

a=p^12+q^2p^10+q^4p^8+q^6p^6+q^8p^4+q^10p^2+q^12
(x,y,z,w)=(p^2, q^2, q, p)
p,q are arbitrary.

x^7 + az^3 = y^7 + aw^3:
a=p^18+q^3p^15+q^6p^12+q^9p^9+q^12p^6+q^15p^3+q^18
(x,y,z,w)=(p^3, q^3, q, p).
p,q are arbitrary.

x^7 + az^4 = y^7 + aw^4:
a=p^24+q^4p^20+q^8p^16+q^12p^12+q^16p^8+q^20p^4+q^24
(x,y,z,w)=(p^4, q^4, q, p).
p,q are arbitrary.

x^7 + az^5 = y^7 + aw^5:
a=p^30+q^5p^25+q^10p^20+q^15p^15+q^20p^10+q^25p^5+q^30
(x,y,z,w)=(p^5, q^5, q, p).
p,q are arbitrary.

x^7 + az^6 = y^7 + aw^6:
a=p^36+q^6p^30+q^12p^24+q^18p^18+q^24p^12+q^30p^6+q^36
(x,y,z,w)=(p^6, q^6, q, p).
p,q are arbitrary.

(x,y,z,w)=(-pa(-r^6+s^6)(-p^7+q^7)^5, -qa(-r^6+s^6)(-p^7+q^7)^5, ra(-r^6+s^6)(-p^7+q^7)^6, sa(-r^6+s^6)(-p^7+q^7)^6)
a,p,q,r,s are arbitrary.

x^7 + az^7 = y^7 + aw^7:
a=p^7+q^7
(x,y,z,w)=(p^2, q^2, q, p).
p,q are arbitrary.

 
Proof.

x^7 + az^2 = y^7 + aw^2

Substitute x=p, y=q to above equation, we obtain
a(z+w)(z-w) = q^7-p^7
Hence we obatin (z,w) =(1/2*(t^2q^7-t^2p^7+a)/(ta),  -1/2*(-t^2q^7+t^2p^7+a)/(ta)).
Finally, we obtain a parametric solution below.
(x,y,z,w)=(4pt^2a^2, 4qt^2a^2, 64(t^2q^7-t^2p^7+a)t^6a^6, 64t^6a^6(-t^2q^7+t^2p^7+a))
a,p,q,t are arbitrary.

Another solution:
a=p^12+q^2p^10+q^4p^8+q^6p^6+q^8p^4+q^10p^2+q^12
(x,y,z,w)=(p^2, q^2, q, p)
p,q are arbitrary.

x^7 + az^3 = y^7 + aw^3
Substitute x=p^3, y=q^3 to above equation, we obtain
a=p^18+q^3p^15+q^6p^12+q^9p^9+q^12p^6+q^15p^3+q^18
(x,y,z,w)=(p^3, q^3, q, p).
p,q are arbitrary.

x^7 + az^4 = y^7 + aw^4
Substitute x=p^4, y=q^4 to above equation, we obtain
a=p^24+q^4p^20+q^8p^16+q^12p^12+q^16p^8+q^20p^4+q^24
(x,y,z,w)=(p^4, q^4, q, p).
p,q are arbitrary.

x^7 + az^5 = y^7 + aw^5
Substitute x=p^5, y=q^5 to above equation, we obtain
a=p^30+q^5p^25+q^10p^20+q^15p^15+q^20p^10+q^25p^5+q^30
(x,y,z,w)=(p^5, q^5, q, p).
p,q are arbitrary.

x^7 + az^6 = y^7 + aw^6
Substitute x=p^6, y=q^6 to above equation, we obtain
a=p^36+q^6p^30+q^12p^24+q^18p^18+q^24p^12+q^30p^6+q^36
(x,y,z,w)=(p^6, q^6, q, p).
p,q are arbitrary.

Another solution:
(x,y,z,w)=(-pa(-r^6+s^6)(-p^7+q^7)^5, -qa(-r^6+s^6)(-p^7+q^7)^5, ra(-r^6+s^6)(-p^7+q^7)^6, sa(-r^6+s^6)(-p^7+q^7)^6)
a,p,q,r,s are arbitrary.

x^7 + az^7 = y^7 + aw^7
Substitute x=p^2, y=q^2 to above equation, we obtain
a=p^7+q^7
(x,y,z,w)=(p^2, q^2, q, p).
p,q are arbitrary.



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






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