Susil Kumar Jena[1] showed that x1^4 + 4X2^4 = X3^8 + 4X4^8 has infinitely many integral solutions. We show that x1^4 + 2nX2^4 = X3^8 + n^2X4^8 has infinitely many integral solutions. Furthermore, we searched the rank of Y^2=X^3-nX with n<100 and give the parametric solution of x1^4 + 2nX2^4 = X3^8 + n^2X4^8.1.IntroductionDiophantine equation x1^4 + 2nX2^4 = X3^8 + n^2X4^8 has infinitely many integral solutions if the elliptic curve Y^2=X^3-nX has nonzero rank. n is integer. Proof. x1^4 + 2nX2^4 = X3^8 + n^2X4^8....................................(1) We consider an identity: (c^4-nd^4)^2+2n(cd)^4=c^8+n^2d^8.........(2) Let v^2=c^4-nd^4..................................................(3) Let X=(c/d)^2 and Y=cv/d^3, the equation (3) becomes to below elliptic curve. Y^2=X^3-nX........................................................(4) When the elliptic curve Y^2=X^3-nX has the rational point P(X,Y), rational point 2P(X) is always perfect square number. Thus we can obtain the integral solution (c,d) from the solution of X=(c/d)^2. Hence if the elliptic curve Y^2=X^3-nX has nonzero rank, then equation (1) infinitely many integral solutions. Q.E.D.2.TheoremThere are parametric solution of x1^4 + 2nX2^4 = X3^8 + n^2X4^8 with n=m^2+1. We show only the solution for the point 2P and 4P. [x1,x2,x3,x4] = [m^4-4m^2-4, 2m(2+m^2), 2+m^2, 2m] = [(256+1024m^2+256m^4-2816m^6-4000m^8-2112m^10-496m^12-80m^14+m^16), 4(m^8+24m^6+40m^4+32m^2+16)m(m^2+2)(m^4-4m^2-4), m^8+24m^6+40m^4+32m^2+16, 4m(m^2+2)(m^4-4m^2-4)] n rank [2, 1] [5, 1] [6, 1] [7, 1] [10, 1] [12, 1] [14, 1] [15, 1] [17, 2] [20, 1] [21, 1] [22, 1] [23, 1] [25, 1] [26, 1] [30, 1] [31, 1] [32, 1] [34, 1] [36, 1] [37, 1] [38, 1] [39, 1] [41, 1] [42, 1] [45, 1] [46, 1] [47, 1] [49, 1] [50, 1] [52, 1] [53, 1] [54, 1] [55, 1] [56, 2] [57, 1] [58, 1] [60, 1] [62, 1] [65, 2] [66, 1] [69, 1] [70, 1] [71, 1] [72, 1] [73, 1] [74, 1] [76, 1] [77, 2] [78, 1] [79, 1] [80, 1] [82, 3] [84, 1] [85, 1] [86, 1] [87, 1] [89, 1] [90, 2] [94, 1] [95, 1] [96, 1] [97, 2] [99, 1]3.Results[1]. Susil Kumar Jena, On X1^4 + 4X2^4 = X3^8 + 4X4^8 and Y1^4 = Y2^4 + Y3^4 + 4Y4^4, Communications in Mathematics, Vol. 23 (2015).4.Reference

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