Parametrizing the solutions to a diophantine equation of degree four Good evening,
Consider $x^4+y^4+z^4=2t^4$ where x,y,z,t integer.
Is it known how to find all parametrisation of this equation ?
If you have any parametrisation or reference of this equation, please post it
Thank you.

Ramanujan gave two parametrizations: if $a+b+c=0$ then
$$a^4(b-c)^4+ b^4(c-a)^4+ c^4(a-b)^4= 2(ab+bc+ca)^4$$
and
$$(a^3+2abc)^4(b-c)^4+(b^3+2abc)^4(c-a)^4+(c^3+2abc)^4(a-b)^4=2(ab+ac+bc)^8$$
(listed in Mathworld, equations 144 and 146). See also Bhargava, S. (1992) On a family of Ramanujan's formulas for sums of fourth powers. Ganita, 43 (1-2). pp. 63-67. [abstract] 
[summary]
The F. Ferrari Identity (1909) gives
$$(a^2+2ac-2bc-b^2)^4+ (b^2-2ba-2ca-c^2)^4+ (c^2+2cb+2ab-a^2)^4 = 2(a^2+b^2+c^2-ab+bc+ca)^4$$
There is also K. Ford's Theorem: if
$S_j=\sum_{i=j\,(\text{mod}\,3)}(-1)^i\binom ki a^{k-i}b^i$  then
$$(S_0-S_1)^4+(S_1-S_2)^4+(S_2-S_0)^4=2(a^2+ab+b^2)^{2k}$$
For $k=2$ and $k=4$, the Ford parametrizations are the same as the two Ramanujan ones. For $k=2$, this is simple, and for $k=4$, $S_0 = a^4-4 a b^3$, $S_1 = b^4-4 a^3 b$, $S_2 = 6 a^2 b^2$, and
$$\begin{align}
S_0 - S_1 = -(c^3+2abc)(a-b)\\
S_1 - S_2 = -(b^3+2abc)(c-a)\\
S_2 - S_0 = -(a^3+2abc)(b-c)\\
\end{align}$$
where $c=-a-b$.
Edit: A complete discussion of these identities, by S. Ramanujan, F. Ferrari and Kevin Ford, plus further identities by S. Bhargava, may be found on pages 96 to 101 of Bruce Berndt's Ramanujan's Notebooks, Part IV (1994), at e.g. http://www.plouffe.fr/simon/math/Ramanujan%27s%20Notebooks%20IV.pdf.
