Finding formula that solves $w^4+x^4=y^4+z^4$ over the integers. Several formulae that solve the diophantine equation
$$w^4 + x^4 = y^4+z^4 \tag{1}$$
are presented in this collection. The simplest one bases on
$$f_1 = a^7 + a^5 - 2 a^3 + 3 a^2 + a \tag{2}$$
and sets$\def\hf{{}^h\!f}$
$$ (w, x, y, z) = (\hf_1,\ \hf_1(b,-a),\ \hf_1(a,-b),\ \hf_1(b,a)) $$
where $\hf(a,b) = f(a/b)\cdot b^{\deg f}$ denotes the homogenized version of $f$.
Question: How does one find such formulae?
The collection refers to Hardy & Wright, and MathWorld mentions that parametric solutions to (1) were already known to Euler in 1802, but without mentioning which one.
Note: $f_1$ appears to be — in some sense — the simplest of such functions:  It produces (158, 59, 134, 133) from (1,2), the smallest solution of (1). All solutions that I checked using Sage were non-trivial except for the case $a=b=1$, and with $\gcd(a,b)=1$ all solutions satisfied $\gcd(w,x,y,z)\in\{1,2\}$.  The last two properties also showed up for the next 2 more complicted functions from the collection that I tried like
$$f_2 = -a^{13} + a^{12} + a^{11} + 5 a^{10} + 6 a^{9} - 12 a^{8} - 4 a^{7} + 7 a^{6} - 3 a^{5} - 3 a^{4} + 4 a^{3} + 2 a^{2} - a + 1
$$
 A: $$w^4 + x^4 = y^4 + z^4 \tag{1}$$
Let $w=mt+a, x=t-b, y=t+a, z=mt+b,$ then we get
$(-4a-4b-4bm^3+4am^3)t^3+(-6b^2m^2-6a^2+6b^2+6a^2m^2)t^2+(-4b^3m+4a^3m-4a^3-4b^3)t=0$
Hence let $$m = \frac{a^3+b^3}{-b^3+a^3}$$ then we get $$t = \frac{3(-b^3+a^3)b^2a^2}{a^6-2b^2a^4-2b^4a^2+b^6}$$
Thus we get a parametric solution as follows.
$w = a(a^6+b^2a^4-2b^4a^2+3b^5a+b^6)$
$x = b(a^6-3ba^5-2b^2a^4+b^4a^2+b^6)$
$y = a(a^6+a^4b^2-2a^2b^4-3ab^5+b^6)$
$z = b(a^6+3ba^5-2b^2a^4+b^4a^2+b^6)$
Let $b=1$, we get
$w = a^7+a^5-2a^3+3a^2+a$
$x = a^6-3a^5-2a^4+a^2+1$
$y = a^7+a^5-2a^3-3a^2+a$
$z = a^6+3a^5-2a^4+a^2+1$
This solution is not a complete solution.
Several solutions are found on A^4+B^4=C^4+D^4
A: The point here is to note that the equation
$$x^4 + w^4 - y^4 - z^4 = 0$$
cuts out a surface $X \subset \mathbb{P}^3$. Indeed, all (primitive, i.e., $\operatorname{gcd}(x,y,z,w) = 1$) integral solutions are in bijective correspondence with rational points on the surface $X$.
Now, one might hope for a solution with $2$ parameters (i.e., a birational map $\mathbb{P}^2 \to X$ defined over $\mathbb{Q}$) however, this is not the case. One may see this since $X$ is a nonsingular quartic surface, it is a $K3$ surface - so is not birational to $\mathbb{P}^2$.
Instead, the best one can hope for in terms of parametric solutions is to find a curve $C \subset X$ lying on $X$ which has genus $0$. Since in this case we may parametrise $C$ and find a $1$-parameter family of solutions to our original equation.
There are some obvious curves that you already know about. For example the curve given by $x=y$ and $w=z$ is pretty stupid, but you do get the parametrisation $(w,x,y,z) = (1, a, a, 1)$, which is a perfectly legitimate option, but I presume that you are calling these 'trivial' solutions.
I'm not going to regurgitate @Tomita's answer, which shows how to construct the particular genus $0$ curve you're interested in which lies on $X$.
In general it is an interesting question to find genus $0$ curves on a $K3$ surface. If $X$ is elliptic $K3$ then one can find a fibration and use the theory of elliptic curves over function fields, otherwise I am not sure of any general theory which allows one to do this.
