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