Parametrizations of $ x^4+y^4+z^4=9t^2 $ integer solutions I would like to derive all the parametrizations for the nontrivial solutions of this Diophantine equation:
$ x^4+y^4+z^4=9t^2 $
I already know that with the Fauquembergue's parametrization I can find infinite solutions (essentially multiplying by 3 the terms in the parametrization, or in the Pythagorean triple), like this one:
$ 60^4 + 45^4 + 36^4 = 9(1443)^2 $
$ 108^4+135^4+180^4 = 9(12987)^2 $
But I even know that there are more solutions (I found the last two from $ x^4+y^4+z^4=t^4 $):
$(155, 260, 296, 37747)$
$(95800, 217519, 414560, 59496731787)$
$ (2682440, 15365639, 18796760, 141668657747643) $
and they don't seem to come out from the result I found on here (none of the fourth powers is divisible by 3).
So which kind of parametrization will give those solutions?
 A: The Diophantine equation,
$$x^4+y^4+z^4 = nt^2$$
has quite an interesting history, being studied by Fauquembergue, Proth, Ramanujan, etc.
I. Case n = 1
As alluded to by the OP, Fauquembergue found that if $a^2+b^2 = c^2$, then,
$$(ab)^4+(ac)^4+(bc)^4 = (a^4+a^2b^2+b^4)^2$$
So using the simplest Pythagorean triple $(3,4,5)$ and multiplying both sides by $3^4$, we get the first solution of the OP,
$$ 60^4 + 45^4 + 36^4 = 9(1443)^2 $$
II. Case n = 2
Even simpler, Proth found that if $a+b+c = 0$, then,
$$a^4+b^4+c^4 = 2(ab+ac+bc)^2$$
extended by Ramanujan to,
$$a^4(b-c)^4 + b^4(a-c)^4 + c^4(a-b)^4 = 2(ab+ac+bc)^4$$
and so on for all $2(ab+ac+bc)^k$ with even $k.$
III. Case n = 9
We can use Fauquembergue's identity and multiply both sides by $3^4$ to get $n=9$. But if we wish that both sides have no common factor, we can employ a particular case of the result by Demjanenko-Elkies,
$$(85v^2 + 484v - 313)^4 + (68v^2 - 586v + 10)^4 + (2u)^4 = 9t^2$$
where $t = 3(119v^2 - 68v + 121)^2$ and,
$$22030 + 28849v - 56158v^2 + 36941v^3 - 31790v^4 = u^2$$
This is an elliptic curve, with one initial rational point $v = -31/467$ and an infinite more. The first point gives,
$$2682440^4+15365639^4+18796760^4 = 9\times141668657747643^2$$
noticed by the OP.
IV. Remark
The small co-prime solution found by the OP,
$$155^4+260^4+296^4 = 9\times37747^2$$
can also be treated as the initial point of an elliptic curve, and it would be nice to know if it satisfies a simpler polynomial identity than the Demjanenko-Elkies.
