(d-2)(d+2)^2X^4 + (4d-8)Y^4 + Z^4 = (4d+8)W^4 Existence of solution for diophantine equation ax^4 + by^4 + cz^4 + dw^4 = 0 are known if abcd is square number. So, we are curious about whether above equation has a solution or not if abcd is not square number. In particular, when does this equation have infinitely many integer solutions? Parametric solutions of x^4 + ay^4 = z^4 + bt^4 are given below. x^4 + ay^4 = z^4 + bt^4 We show diophantine equation (d-2)(d+2)^2X^4 + (4d-8)Y^4 + Z^4 = (4d+8)W^4 has infinitely many integer solutions. d is arbitrary. (d-2)(d+2)^2X^4 + (4d-8)Y^4 + Z^4 = (4d+8)W^4.................................................(1) We use an identity p((t+1))^4+q(t)^4 = r(t^2+at+a)^2+s(ct^2+dt+a)^2,..........................(2) with (a,c,p,q,r)=(1+1/2d, 1/2d, 1/4(-4+d^2)s, s(-2+d)/(2+d), -4s/(2+d)). So, we look for the integer solutions {Z^2 = t^2+(1+1/2d)t+1+1/2d, W^2 = 1/2dt^2+dt+1+1/2d}...(3) By parameterizing the second equation and substituting the result to first equation, then we obtain quartic equation below. u^2 = 16k^4+(-32+32d)k^2+(-16d^2+32d)k-8d^2+2d^3+16+8d........................................(4) This quartic equation is birationally equivalent to an elliptic curve below. Y^2+(-128d^2+256d)Y = X^3+(-32+32d)X^2+(512d^2-128d^3-1024-512d)X-32768d^2+20480d^3+32768-16384d-4096d^4 The corresponding point is P(X,Y)=( 32-32d, 128d^2-256d ). Hence we get 2P(X,Y)=( d^2+12d+68, -404d-360+34d^2+d^3 ). This point P is of infinite order, and the multiples nP, n = 2, 3, ...give infinitely many points. This quartic equation has infinitely many parametric solutions below. n=2: X = 16d+96 Y = -d^2+36d+60 Z = 2(3d^2-28d-84) W = d^2+44d+36 n=3: X = 16(d^2+44d+36)(d+6)(3d^2-28d-84) Y = (d^2-36d-60)(d^4-200d^3-616d^2-3360d-5616) Z = 2(316224+426816d+215472d^2+25696d^3-7988d^4-396d^5+5d^6) W = 46656+355392d+262128d^2+39392d^3-2276d^4+388d^5+d^6 . . etc. Hence we can obtain infinitely many integer solutions for equation (1). Example: d=1: -9X^4-4Y^4+Z^4=12W^4 d=3: 25X^4+4Y^4+Z^4=20W^4

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