Jacobi and Madden found infinitely many primitive solutions to,
$$a^4+b^4+c^4+d^4 = (a+b+c+d)^4$$
using an elliptic curve. We will use a different approach that, like the method for,
$$a^4+b^4+c^4 = d^4$$
discussed in this post, also yields pairs of solutions.
I. The system
Define,
$$x^4+y^4+z^4+1 = (x+y+z+1)^4\tag1$$ $$\frac{x^2+x+1}{(x+y+1)(x+z+1)}=u\tag2$$ $$\frac{y^2+y+1}{(y+z+1)(y+x+1)}=v\tag3$$ $$\frac{z^2+z+1}{(z+x+1)(z+y+1)}=w\tag4$$
The variable $w$ is dependent via a rather complicated expression on $(u,v)$ so $(4)$ is redundant. Use the first three equations to solve for the three unknowns $(x,y,z)$. After some algebra, we find they are roots of quadratics hence yields pairs of solutions. The discriminant of the quadratic is,
$$D^2 = -3(2 - u + u^2)^2 + 6(2 + u + 3u^2 - 3u^3 + u^4)v - 3(5 - 6u + 6u^2 - 2u^3 + u^4)v^2 + 6(1 - u)(1 - 2u - u^2)v^3 - 3(1 - u)^2v^4$$
II. Table of known u
If there is rational $(u,v)$ such that $D$ is also rational, then the quartic in $v$ is birationally equivalent to an elliptic curve. Furthermore, if $(u,v)$ is a solution, then $\big(\frac{u+1}{u-1},v\big)$ is also a solution and which slightly complicates things. As of 2015, there are only eight $u$ (Update: Now twelve as of 2024) of small height (numerator and denominator < $1000$) that are known, and not counting its partner $u'=\frac{u+1}{u-1}$. Namely,
\begin{array}{|c|c|c|c|c|} \hline \text{#} & \color{red}u & u'=\frac{u+1}{u-1} & \color{red}v & \text{Discoverer}\\ \hline 1 & \dfrac{511}{450} & \dfrac{961}{61} & \dfrac{1651}{126} & \text{Brudno}\\ \hline 2 & \dfrac{193}{18} & \dfrac{211}{175} & \dfrac{619}{450} & \text{Wroblewski}\\ \hline 3 & \dfrac{31}{6} & \dfrac{37}{25} & \dfrac{6619}{5550} & \text{Rouse} \\ \hline 4 & \dfrac{211}{150} & \dfrac{361}{61} & \dfrac{2041}{150} & \text{Tomita}\\ \hline 5 & \dfrac{157}{150} & \dfrac{307}{7} & \dfrac{8467}{150} & \text{Rouse}\\ \hline 6 & \dfrac{181}{150} & \dfrac{331}{31} & \dfrac{277567}{31675} & \text{Tomita}\\ \hline 7 & \dfrac{373}{150} & \dfrac{523}{223} & \dfrac{9785779}{952879} & \text{Tomita}\\ \hline 8 & \dfrac{121}{96} & \dfrac{217}{25} & \dfrac{6250987}{506400} & \text{Tomita}\\ \hline \end{array}
III. Example
From the table, choose #4 and let $u=\dfrac{211}{150}$ and $v_1 =\dfrac{2041}{150}$ where $u$ was also employed by Tomita here using a different method. Solving the system,
$$x^4+y^4+z^4+1 = (x+y+z+1)^4$$ $$\frac{x^2+x+1}{(x+y+1)(x+z+1)}=\frac{211}{150}$$ $$\frac{y^2+y+1}{(y+z+1)(y+x+1)}=\frac{2041}{150}$$
yields the pair of solutions,
$$ \left(- \frac{1984340}{107110}\right)^4 + \left( \frac{1022230}{107110}\right)^4 + \left( - \frac{1229559}{107110}\right)^4 + 1 = (x+y+z+1)^4$$
$$ \left(-\frac{3597130}{1953890}\right)^4 + \left(- \frac{561760}{1953890}\right)^4 + \left(- \frac{1493309}{1953890}\right)^4 + 1 = (x+y+z+1)^4$$
where $(x,y,z)$ are the first three addends. These are the 3rd and 5th smallest known solutions, also found by Tomita in the link above. From the initial $v_1$, one can find infinitely many other $v_k$.
IV. Question
So given,
$$D^2 = -3(2 - u + u^2)^2 + 6(2 + u + 3u^2 - 3u^3 + u^4)v - 3(5 - 6u + 6u^2 - 2u^3 + u^4)v^2 + 6(1 - u)(1 - 2u - u^2)v^3 - 3(1 - u)^2v^4$$
can you find a $u$ of small height not in the table above?
