I. Question
We have a quartic in $v$ to be made a square,
$D^2 = 4(-6 - 2u + u^2)(2 - 2u + u^2) - 8(-2 - 4u + u^2)(2 - 2u + u^2)v - 16u(4 - 3u + u^2)v^2 - 4(4 - 12u + 4u^2 - 2u^3 + u^4)v^3 + (4 - 8u - 4u^3 + u^4)v^4$
If there is rational $(u,v)$ such that $D$ is also rational, then the quartic is birationally equivalent to an elliptic curve. And if we consider only $u$ of small height (numerator and denominator with absolute value less than $1000$), then only $16 $ $u$ are known.
Q: How do we find more $u$ of small height? (The context of this curve is given below.)
II. Background
Noam Elkies found the first counterexample to Euler's conjecture that $x^4+y^4+z^4 = 1$ has no rational point $xyz \neq0$. We provide a slightly different method that produces pairs of solutions. Let,
$$x^4+y^4+z^4 = 1\tag1$$ $$\frac{(x-y)^2-z^2-1}{x^2-xy+y^2+(x-y)}=u\tag2$$ $$\frac{(y-z)^2-x^2-1}{y^2-yz+z^2+(y-z)}=v\tag3$$ $$\frac{(z-x)^2-y^2-1}{z^2-zx+x^2+(z-x)}=w\tag4$$
where the $(u,v,w)$ have the simple relationship,
$$2(u+v+w)-uvw-4=0$$
Use the first three equations to solve for the three unknowns $(x,y,z)$ to get, \begin{align} x &= \frac{P_1\pm(u^2-2u+2)(v^2-2v)\sqrt{D^2}}{P_0+D^2}\\ y &= \frac{-(P_1+P_2+P_3)\pm 2(u+v-2)(u+v-uv)\sqrt{D^2}}{P_0+D^2 \quad}\\ z &= \frac{P_3\pm(v^2-2v+2)(u^2-2u)\sqrt{D^2}}{P_0+D^2} \end{align}
with $D$ as defined in the first section. The $P_k$ are,
\begin{align} P_0 &= (2 + u^2)(2 + v^2)(12 - 8u + 2u^2 - 8v + 2v^2 + u^2v^2)\\ P_1 &= (-4 + 4u + 2v - u^2v)(8u - 4u^2 + 4v - 8 u v + 2u^2v - 4v^2 + 2v^3 + u^2v^3)\\ P_2 &= 2(- 4 + 4u + 2v - u^2v)(4 - 2u - 4v + u v^2)(-u + v)\\ P_3 &= (4 - 2u - 4v + u v^2)(4u - 4u^2 + 2u^3 + 8v - 8 u v - 4v^2 + 2 u v^2 + u^3v^2)\end{align}
For example, let $u = -\frac{9}{20}$, then,
$$D^2 = \frac{-9724476 - 9928v + 6396480v^2 - 6677284v^3 + 1280881v^4}{20^4}$$
One solution is $v = -\frac{1041}{320}$, or $v = \frac{1000}{47}$. Using the positive and negative cases of $\pm \sqrt{D^2}$, either $v$ yields,
$$\left(\frac{414560}{422481}\right)^4 + \left(\frac{95800}{422481}\right)^4 + \left(\frac{217519}{422481}\right)^4 = 1$$
$$\left(\frac{632671960}{1679142729}\right)^4 + \left(\frac{1670617271}{1679142729}\right)^4 + \left(\frac{50237800}{1679142729}\right)^4 = 1$$
From this initial point $v$, one can then find infinitely more $v_k$.
III. The known $u$
The $16$ known "small" $u$ (and their initial $v_1$) are,
\begin{array}{|c|c|c|c|c|c|c|} \hline \text{#} & u & v & &\text{#} & u & v\\ \hline 1 & -\dfrac{9}{20} & \; -\dfrac{1041}{320} & & 9 & -\dfrac{41}{36} & -\dfrac{4061}{16308}\\ \hline 2 & -\dfrac{29}{12} & \;\dfrac{1865}{132} & & \color{blue}{10} & -\dfrac{5}{44} & \dfrac{57878913}{12642040}\\ \hline 3 & -\dfrac{93}{80} & -\dfrac{400}{37} & & \color{blue}{11} & +\dfrac{233}{60} & \;\dfrac{7584}{54605}\\ \hline \color{red}4 & -\dfrac{400}{37} & -\dfrac{93}{80} & & \color{blue}{12} & -\dfrac{56}{165} & -\dfrac{383021}{380940}\\ \hline 5 & -\dfrac{136}{133} & +\dfrac{201}{4} & & \color{blue}{13} & -\dfrac{125}{92} & -\dfrac{936}{5281}\\ \hline \color{red}6 & +\dfrac{201}{4} & -\dfrac{136}{133} & & 14 & -\dfrac{361}{540} & +\dfrac{1861}{240}\\ \hline 7 & -\dfrac{5}{8} & -\dfrac{477}{692} & & 15 & -\dfrac{817}{660} & -\dfrac{1581}{1520}\\ \hline \color{red}8 & -\dfrac{477}{692} & -\dfrac{5}{8} & & 16 & -\dfrac{865}{592} & -\dfrac{14177}{20156}\\ \hline \end{array}
Can you find more $u$ of small height that will yield new primitive solutions to $a^4+b^4+c^4 = d^4$ with $d<10^{27}$ different from the 30 known $d$? (Update: Now 93 $d$ as of Feb 21, 2024 in this table.)
Note 1: Since $(u,v_k)$ and $(v_k,u)$ both solve the quartic, then there are in fact infinitely many $u$ of increasingly large height. Likewise, the $w$ of $2(u + v + w) - u v w - 4 = 0$ also solves the quartic.
Note 2: The first ten (except the red ones) were considered by Tomita here and here. The blue ones were found by Andrew Bremner in this post. It may be that #10 has a smaller v. (Update: No smaller v. See this MO post.)
IV. Update
Several new $u$ found in the answers below can also yield small $d<10^{14}.\,$ S. Tomita found 6 while D. Fulea found 3 with Fulea's $u =\frac{553}{80}$ yielding $d\approx 5\times10^9$,
$$24743080^4 + 3971389576^4 + 4657804375^4 = 5179020201^4$$
the smallest known since 2008.
