(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