pX^4 + pY^4 = 2Z^4 + 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 pX^4 + pY^4 = 2Z^4 + W^4 has infinitely many integer solutions for p=9,19,129,243,289,801 with p<1000.

pX^4 + pY^4 = 2Z^4 + W^4.....................................................................(1)

We use an identity p(t+1)^4+p(t)^4=2(t^2+at+b)^2+(ct^2+dt+e)^2,..............................(2)
with p = 3-2d+1/2d^2,  e = 1, c = -2+d, a = -d+3, b = 1-1/2d.

So, we look for the integer solutions {Z^2 = t^2+(-d+3)t+1-1/2d, W^2 = (-2+d)t^2+dt+1}.......(3)

By parameterizing the second equation and substituting the result to first equation, then we obtain quartic equation below.

u^2 =(-2d+4)k^4+(-24+8d)k^3+(32-4d)k^2+(-8d^2-48+24d)k+16+2d^3-4d^2..........................(4)

This quartic equation has infinitely many rational solutions for d=6,8,18,24,26,42,48,50 with d<=50.

(d,p)= (6,9),(8,19),(18,129),(24,243),(26,289),(42,801),(48,1059),(50,1153).

Hence we can obtain infinitely many integer solutions for equation (1) where p=9,19,129,243,289,801 with p<1000.

Example:

d=6: 9X^4+9Y^4 = 2Z^4+W^4    : see details 9X^4+9Y^4 = 2Z^4+W^4

d=8: 19X^4+19Y^4 = 2Z^4+W^4  : see details 19X^4+19Y^4 = 2Z^4+W^4