4X^4 + Y^4 + Z^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 4X^4 + Y^4 + Z^4 = W^4 has infinitely many integer solutions.
This equation is related to X^4 + 4Y^4 + 4Z^4 = 4W^4.

This equation has a parametric solution.
(p^4+2q^4)^4 = (p^4-2q^4)^4+(2p^3q)^4+4(2pq^3)^4 (R.D.Carmichael).


4X^4 + Y^4 + Z^4 = W^4......................................................(1)
We use an identity 4(t+1)^4+(t)^4+( 2t^2+2t)^2 = (2+4t+3t^2)^2..............(2)

So, we look for the integer solutions {Z^2 = 2t^2+2t, W^2 = 2+4t+3t^2}......(3)

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

u^2 = k^4+8.................................................................(4)

This quartic equation is birationally equivalent to an elliptic curve below.
y^2 = x^3 - 2x.
Rank is 1 and generator is [-1,-1].

Hence we can obtain infinitely many integer solutions for equation (1).

Numerical example:

[X,Y,Z,W] = [72, 49, 84, 113], [338, 57121, 6214, 57123]