-dm^2(d+2)X^4 + (-2m^2+d+2)m^2(d-2m^2)Y^4 = d(d-2m^2)m^4Z^4 + (d+2)(2m^2-d-2)W^4 Existence of solution for diophantine equation ax^4 + by^4 + cz^4 + dw^4 = 0 are known if abcd is square number. Furthermore, condition of a + b + c + d = 0 or having one known solution is necessary to have a rational solution. We show some parametric solutions of ax^4 + by^4 + cz^4 + dw^4 = 0 with abcd is square number. General solutions of ax^4 + by^4 + cz^4 + dw^4 = 0 are given below. ax^4 + by^4 + cz^4 + dw^4 = 0 ax^4 + by^4 + cz^4 + dw^4 = 0 Part 2 Parametric solutions of x^4 + hy^4 = z^4 + ht^4 are given below. x^4 + hy^4 = z^4 + ht^4 Many numeric solutions of x^4 + hy^4 = z^4 + ht^4 are given below. x^4 + hy^4 = z^4 + ht^4 We show diophantine equation -dm^2(d+2)X^4 + (-2m^2+d+2)m^2(d-2m^2)Y^4 = d(d-2m^2)m^4Z^4 + (d+2)(2m^2-d-2)W^4 has infinitely many integer solutions. d,m are arbitrary. -dm^2(d+2)X^4 + (-2m^2+d+2)m^2(d-2m^2)Y^4 = d(d-2m^2)m^4Z^4 + (d+2)(2m^2-d-2)W^4............................(1) We use an identity p(t+1)^4+q(t)^4=r(t^2+at+b)^2+pq/r(ct^2+dt+e)^2,.........................................(2) with (a,b,e,p,q)=((d+2)/c, 1/2(d+2)/c, 1/2d, r(d+2)/(c(-d+2c)), -(2c-d-2)r/(dc)). So, we look for the integer solutions {Z^2 = t^2+(d+2)t/c+1/2(d+2)/c, W^2 = ct^2+dt+1/2d}...................(3) By parameterizing the first equation and substituting the result to second equation, then we obtain quartic equation below. Let c=m^2, then u^2 = m^4k^4+(-4m^2d+4m^4)k^3+(4d^2-8m^2d+6m^4-8m^2+8d)k^2+(-16m^2+4m^4-4m^2d)k+16+m^4-8m^2+8d..............(4) This quartic equation is birationally equivalent to an elliptic curve below. Y^2+(-4d+4m^2)YX+(-32m^4+8m^6-8m^4d)Y = X^3+(2m^4-8m^2+8d)X^2+(-64m^4-4m^8+32m^6-32m^4d)X-384m^8-8m^12+96m^10-96m^8d+512m^6+512m^6d-512m^4d-256m^4d^2 It has a point P(X,Y)=( -2m^4+8m^2-8d, 64m^2d-32d^2 ). Hence we get 2P(X,Y)=( 4+2m^4, 8m^4+8+16d-16m^2 ). 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 = 3m^4+(-2-2d)m^2-1 Y = m^4+(-2-2d)m^2+1 Z = m^4+2m^2-2d-3 W = m(m^4-2m^2+2d+1) n=3: X = 5m^12+(-10-10d)m^10+(8d-9+4d^2)m^8+(36+36d)m^6+(-28d^2-56d-29)m^4+(24d^2+8d^3+6+22d)m^2+1 Y = m^12+(-6-6d)m^10+(8d+4d^2+15)m^8+(-20-20d)m^6+(40d+20d^2+15)m^4+(-24d^2-8d^3-6-22d)m^2+1 Z = m^12+6m^10+(-22d-29)m^8+(56d+36+24d^2)m^6+(-9-8d^3-28d^2-36d)m^4+(-8d-10)m^2+4d^2+5+10d W = m^13-6m^11+(22d+15)m^9+(-40d-20-24d^2)m^7+(15+8d^3+20d^2+20d)m^5+(-8d-6)m^3+(4d^2+1+6d)m . . etc. Hence we can obtain infinitely many integer solutions for equation (1). Example: (d,m)=(2,2): -2X^4 + 6Y^4 = -12Z^4 + W^4: [X,Y,Z,W]=[23, 7, 17, 26],[647, 4183, 2993, 7274],...etc.

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