d^2a^2X^4 + (-d+2a)^2a^2Y^4 + 2d(-d+2a)Z^4 = 4a^2W^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^2a^2X^4 + (-d+2a)^2a^2Y^4 + 2d(-d+2a)Z^4 = 4a^2W^4 has infinitely many integer solutions. a,d are arbitrary. d^2a^2X^4 + (-d+2a)^2a^2Y^4 + 2d(-d+2a)Z^4 = 4a^2W^4............................(1) We use an identity p(t+1)^4+q(t)^4+r(at^2+at)^2=s(at^2+dt+e)^2,.................(2) with (p,q,r,s,e)=(1/4sd^2, 1/4s(2a-d)^2, 1/2sd(2a-d)/(a^2), s, 1/2d). So, we look for the integer solutions {Z^2 = at^2+at, W^2 = at^2+dt+1/2d}.......(3) By parameterizing the second equation and substituting the result to first equation, then we obtain quartic equation below. u^2 = (4a-2d)k^4+2da^2..........................................................(4) This quartic equation has infinitely many rational solutions for (a,d) with (a,d)<=5 below. (a,d)= (1,3),(1,4),(1,5),(3,2),(3,4),(4,1),(4,2),(4,3),(4,4),(4,5),(5,2). Hence we can obtain infinitely many integer solutions for equation (1). Example: (a,d)=(3,2): 9X^4 + 36Y^4 + 4Z^4 = 9W^4 : see details 9X^4 + 36Y^4 + 4Z^4 = 9W^4 (a,d)=(4,2): X^4 + 9Y^4 + 6Z^4 = W^4 : see details X^4 + 9Y^4 + 6Z^4 = W^4 (a,d)=(5,2): 25X^4 + 25Y^4 + 8Z^4 = 25W^4 : see details 25X^4 + 25Y^4 + 8Z^4 = 25W^4

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