The background to the question:
$$a^4+b^4=c^4+d^4 \label{1}\tag 1 $$
Tito Piezas & others have recently given some parametric solutions on Math stack exchange & Math overflow. In math literature there are parametric solutions given in Dickson’s book (vol 2), which includes solutions by Euler & others. Solutions are shown by Zajta (In a AMS journal paper). For degree two we have general solution for $a^2+b^2=c^2+d^2.$ For degree three $a^3+b^3=c^3+d^3$ two general solution has been given. One by Noam Elkies & second by Ajai Choudhry. Even though this problem of the quartic equation (# \ref{1} above) has been around since the time of Euler (for over 200) years a general solution has been evasive. Eight relevant Links are shown below:
Noam Elkies " Complete cubic parametrization of the Fermat cubic surface ".
Tito Piezas III " On Euler's elliptic curve for $A^4+B^4 = C^4+D^4$? ".
Tito Piezas III " Finding rational points on the elliptic curve for $A^4+B^4 = C^4+D^4$? ".
emacs drives me nuts "Finding formula that solves $w^4+x^4=y^4+z^4$ over the integers."
Edward Brisse "Identities Of Equal Sums Of Like Power"
Wolfram MathWorld "Diophantine Equation 4th Powers"
Aurel J. Zajta "Solutions of the Diophantine equation $A^4+B^4 = C^4+D^4$", Mathematics of Computation 41, 635-659 (1983), MR717709, Zbl 0525.10011
[Ajai Choudhry paper for the general solution of the cubic (3-2-2) equation, link is below, select item # 14]
https://sites.google.com/view/ajaichoudhry/publications
Remark:
There are more than a couple of dozen parametric solutions available for equation \eqref{1}. There is a possibility that one of them could be a general solution or maybe not. There are two options. First is, someone needs to write an algorithm to see if the (1420 different) numerical solutions (link as shown above & given at Jaroslaw Wrobelewski website -uni.wroc) are satisfied by one of the published parametric solution. The other option is that someone needs to give a general solution along with a proof. Any response to the above will be appreciated.