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^6 + az^4 = y^6 + aw^4 has a parametric solution. BTW, when a=p^8+q^4p^4+q^8, equation x^6 + az^4 = y^6 + aw^4 has a obvious solution (x,y,z,w)=(p^2,q^2,q,p). Similarly, when a=p^6+q^6, equation x^6 + az^6 = y^6 + aw^6 has a obvious solution (x,y,z,w)=(p^2,q^2,q,p). Furthermore, when a=p^5+q^5, equation x^5 + az^5 = y^5 + aw^5 has a obvious solution (x,y,z,w)=(p^2,q^2,q,p).1.IntroductionDiophantine equation x^6 + az^4 = y^6 + aw^4 has a parametric solution. (x,y,z,w)=(2s(3+10s^2)(10s^2+6s-3), 2s(3+10s^2)(10s^2-6s-3), 2s(-9+100s^4-36s^2)(3+10s^2), 2s(-9+100s^4+36s^2)(3+10s^2)) a=(3+10s^2)s, s is arbitrary. Proof. x^6 + az^4 = y^6 + aw^4....................................................(1) Substitute x=-t+p, y=t+p, z=t+q, w=-t+q to equation (1), we obtain -12pt^5+(-40p^3-8*aq)t^3+(-12p^5-8aq^3)t=0.................................(2) Let a = -3/2p^5/(q^3), then we obtain -12pt^2+(-40p^3-8aq)=0...............(3) In order to have rational solutions for t, we must have rational solutions of equation (4). Let p=nq, then v^2 = 3q^2(-10+3n^2)........................................(4) Equation -10+3n^2 = 3r^2 has a solution (n,r)=(-1/6(3+10s^2)/s, -1/6(-3+10s^2)/s). Let q=72s^5 and a=(3+10s^2)s, we obtain a parametric solution below. (x,y,z,w)=(2s(3+10s^2)(10s^2+6s-3), 2s(3+10s^2)(10s^2-6s-3), 2s(-9+100s^4-36s^2)(3+10s^2), 2s(-9+100s^4+36s^2)(3+10s^2)) Q.E.D.2.Theorem[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:4,20153.Reference

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