A. Bremner, AJAI CHOUDHRY and M. Ulas([1]) gave a solution of x^6 + y^6 - 36z^6 + 2w^3 =0 as a special case of PX^6+ QY^6+ Sm^2Z^6+ RW^3= 0. We show five solutions of ax^8+ by^8+ cz^4+ dw^4= 0 by their method.1.IntroductionWe consider the identity a^2T^8+b^2+2abT^4 = (aT^4+b)^2........................(1) Let aT^4+b = mz^2..............................................................(2) If we can find the rational solution (T,z) of (2) for some m, we obtain the solution of (3). a^2X^8+b^2Y^8-m^2Z^4+2abW^4=0..................................................(3) Hence if equation (2) has infinitely many rational solutions, equation (3) has infinitely many integer solutions.2.Method3.ResultsCase 1: 4x^8+y^8-9z^4+4w^4 = 0By using (a,b,m)=(2,1,3), equation (2) has rational solution (T,z)=(-1,-1). We obtain quartic elliptic curve and Weierstrass form, V^2 = 6U^4 +3. Y^2 = X^3 - 72X. This rank is 1 and generator: P=(9, -9). Thus, we can obtain infinitely many integer solutions. P: (x,y,z,w)=(11, 23, 321, 253) 2P: (x,y,z,w)=(6227, 2375, 31827137, 14789125)Case 2: 9x^8+y^8-z^4+6w^4 = 0By using (a,b,m)=(3,1,1), equation (2) has rational solution (T,z)=(1,2). We obtain quartic elliptic curve and Weierstrass form, V^2 = 3U^4 + 1. Y^2 = X^3 - 12X. This rank is 1 and generator: P=(-2, -4). Thus, we can obtain infinitely many integer solutions. P: (x,y,z,w)=(1, 1, 2, 1) 2P: (x,y,z,w)=(2, 1, 7, 2) 3P: (x,y,z,w)=(3, 11, 122, 33)Case 3: 16x^8+y^8+8w^4-25z^4 = 0By using (a,b,m)=(4,1,5), equation (2) has rational solution (T,z)=(1,1). We obtain quartic elliptic curve and Weierstrass form, V^2 = 20U^4 + 5. Y^2 = X^3 - 25X. This rank is 1 and generator: P=(-4, -6). Thus, we can obtain infinitely many integer solutions. P: (x,y,z,w)=(11, 71, 2257, 781) 2P: (x,y,z,w)=(99731, 35689, 8914433905, 3559299659)Case 4: x^8+y^8-25z^4-2w^4 = 0By using (a,b,m)=(1,-1,5), equation (2) has rational solution (T,z)=(3,4). We obtain quartic elliptic curve and Weierstrass form, V^2 = 5U^4 - 5. Y^2 = X^3 - 100X. This rank is 1 and generator: P=(5, 25). Thus, we can obtain infinitely many integer solutions. P: (x,y,z,w)=(3, 1, 4, 3) 2P: (x,y,z,w)=(49, 31, 984, 1519)Case 5: 4x^8+y^8-z^4-4w^4 = 0By using (a,b,m)=(2,-1,1), equation (2) has rational solution (T,z)=(1,1). We obtain quartic elliptic curve and Weierstrass form, V^2 = 2U^4 - 1. Y^2 = X^3 - 8X. This rank is 1 and generator: P=(1, -3). Thus, we can obtain infinitely many integer solutions. P: (x,y,z,w)=(13, 1, 239, 13) 2P: (x,y,z,w)=(1525, 1343, 2750257, 2048075)[1]. A. Bremner, AJAI CHOUDHRY and M. Ulas, CONSTRUCTIONS OF DIAGONAL QUARTIC AND SEXTIC SURFACES WITH INFINITELY MANY RATIONAL POINTS, arxiv.org/pdf/1402.4583, 20144.Reference

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