Similar to the case $x^4+y^4+z^4 = 1$ discussed in this MO post, define the system,

$$x^4+y^4+z^4+1 = (x+y+z+1)^4\tag1$$ $$\frac{x^2+x+1}{(x+y+1)(x+z+1)}=u\tag2$$ $$\frac{y^2+y+1}{(y+z+1)(y+x+1)}=v\tag3$$ $$\frac{z^2+z+1}{(z+x+1)(z+y+1)}=w\tag4$$

The variable $w$ is dependent via a rather complicated expression on $(u,v)$ so $(4)$ is superfluous. We use the first three equations to solve for the three unknowns $(x,y,z)$. After some algebra, it can be determined they are roots of quadratics, hence yields pairs of solutions. The discriminant of the quadratic is,

$$D^2 = -3(2 - u + u^2)^2 + 6(2 + u + 3u^2 - 3u^3 + u^4)v - 3(5 - 6u + 6u^2 - 2u^3 + u^4)v^2 + 6(1 - u)(1 - 2u - u^2)v^3 - 3(1 - u)^2v^4$$

So if there is rational $(u,v)$ such that $D$ is also rational, then the quartic in $v$ is birationally equivalent to an elliptic curve. For some desired $u$ of small height, we are looking for an initial solution $v_1$ from which an infinite more can be generated. Here are ten such $u$,

\begin{array}{|c|c|c|c|} \hline \text{#} & u & v_1 & \text{Discover}\\ \hline 1 & \dfrac{31}{6} & \dfrac{6619}{5550} & \text{Rouse} \\ \hline 2 & \dfrac{49}{24} & \dfrac{138551171933011575944603377}{41031556739549840108788225} & \text{MacLeod}\\ \hline 3 & \dfrac{67}{42} & ? & ?\\ \hline 4 & \dfrac{79}{54} & \dfrac{29549171683987}{25656103349287} & \text{MacLeod}\\ \hline 5 & \dfrac{97}{72} & ? & ?\\ \hline 6 & \dfrac{103}{78} & ? & ?\\ \hline 7 & \dfrac{121}{96} & \dfrac{6250987}{506400} & \text{Tomita}\\ \hline 8 & \dfrac{157}{150} & \dfrac{8467}{150} & \text{Rouse}\\ \hline 9 & \dfrac{181}{150} & \dfrac{277567}{31675} & \text{Tomita}\\ \hline 10 & \dfrac{193}{18} & \dfrac{619}{450} & \text{Wroblewski}\\ \hline \end{array}

In this 2014 MO post, Jeremy Rouse left open a lot of cases of $u$ including two found by Seiji Tomita in 2015 and two found by Allan MacLeod in 2017. In this 2017 paper, MacLeod gave the (huge!) solution for $u=\frac{49}{24}$ and $u=\frac{79}{54}$ in the last section of the paper. And while MacLeod didn't explicitly give the $v$ parameter, it was easy enough to reverse-engineer it from the $a^4+b^4+c^4+d^4 = (a+b+c+d)^4$ he gave.

Question:

1. Is it now possible to find $v$ for the three remaining $u$?
2. The height of MacLeod's two $v$ seems high. Can it be reduced?

Similar to the case $x^4+y^4+z^4 = 1$ discussed in this MO post, define the system,

$$x^4+y^4+z^4+1 = (x+y+z+1)^4\tag1$$ $$\frac{x^2+x+1}{(x+y+1)(x+z+1)}=u\tag2$$ $$\frac{y^2+y+1}{(y+z+1)(y+x+1)}=v\tag3$$ $$\frac{z^2+z+1}{(z+x+1)(z+y+1)}=w\tag4$$

The variable $w$ is dependent via a rather complicated expression on $(u,v)$ so $(4)$ is superfluous. We use the first three equations to solve for the three unknowns $(x,y,z)$. After some algebra, it can be determined they are roots of quadratics, hence yields pairs of solutions. The discriminant of the quadratic is,

$$D^2 = -3(2 - u + u^2)^2 + 6(2 + u + 3u^2 - 3u^3 + u^4)v - 3(5 - 6u + 6u^2 - 2u^3 + u^4)v^2 + 6(1 - u)(1 - 2u - u^2)v^3 - 3(1 - u)^2v^4$$

So if there is rational $(u,v)$ such that $D$ is also rational, then the quartic in $v$ is birationally equivalent to an elliptic curve. For some desired $u$ of small height, we are looking for an initial solution $v_1$ from which an infinite more can be generated. Here are ten such $u$, part of a list from Jeremy Rouse,

\begin{array}{|c|c|c|c|} \hline \text{#} & u & v_1 & \text{Discover}\\ \hline 1 & \dfrac{31}{6} & \dfrac{6619}{5550} & \text{Rouse} \\ \hline 2 & \dfrac{49}{24} & \dfrac{138551171933011575944603377}{41031556739549840108788225} & \text{MacLeod}\\ \hline 3 & \dfrac{67}{42} & ? & ?\\ \hline 4 & \dfrac{79}{54} & \dfrac{29549171683987}{25656103349287} & \text{MacLeod}\\ \hline 5 & \dfrac{97}{72} & ? & ?\\ \hline 6 & \dfrac{103}{78} & ? & ?\\ \hline 7 & \dfrac{121}{96} & \dfrac{6250987}{506400} & \text{Tomita}\\ \hline 8 & \dfrac{157}{150} & \dfrac{8467}{150} & \text{Rouse}\\ \hline 9 & \dfrac{181}{150} & \dfrac{277567}{31675} & \text{Tomita}\\ \hline 10 & \dfrac{193}{18} & \dfrac{619}{450} & \text{Wroblewski}\\ \hline \end{array}

In this 2014 MO post, Rouse left open a lot of cases of $u$ including two solved by Seiji Tomita in 2015 and two solved by Allan MacLeod in 2017. In this 2017 paper, MacLeod gave the (huge!) solution for $u=\frac{49}{24}$ and $u=\frac{79}{54}$ in the last section of the paper. And while he didn't explicitly give the $v$ parameter, it was easy enough to reverse-engineer it from the $a^4+b^4+c^4+d^4 = (a+b+c+d)^4$ he gave.

Question:

1. Is it now possible to find $v$ for the three remaining $u$?
2. The height of MacLeod's two $v$ seems high. Can it be reduced by other points of smaller height?

Similar to the case $x^4+y^4+z^4 = 1$ discussed in this MO post, define the system,

$$x^4+y^4+z^4+1 = (x+y+z+1)^4\tag1$$ $$\frac{x^2+x+1}{(x+y+1)(x+z+1)}=u\tag2$$ $$\frac{y^2+y+1}{(y+z+1)(y+x+1)}=v\tag3$$ $$\frac{z^2+z+1}{(z+x+1)(z+y+1)}=w\tag4$$

The variable $w$ is dependent via a rather complicated expression on $(u,v)$ so $(4)$ is superfluous. We use the first three equations to solve for the three unknowns $(x,y,z)$. After some algebra, it can be determined they are roots of quadratics, hence yields pairs of solutions. The discriminant of the quadratic is,

$$D^2 = -3(2 - u + u^2)^2 + 6(2 + u + 3u^2 - 3u^3 + u^4)v - 3(5 - 6u + 6u^2 - 2u^3 + u^4)v^2 + 6(1 - u)(1 - 2u - u^2)v^3 - 3(1 - u)^2v^4$$

So if there is rational $(u,v)$ such that $D$ is also rational, then the quartic in $v$ is birationally equivalent to an elliptic curve. For some desired $u$ of small height, we are looking for an initial solution $v_1$ from which an infinite more can be generated. Here are seven such $u$,

\begin{array}{|c|c|c|} \hline \text{#} & u & v_1 \\ \hline 1 & \dfrac{31}{6} & \dfrac{6619}{5550} \\ \hline 2 & \dfrac{49}{24} & \dfrac{138551171933011575944603377}{41031556739549840108788225} \\ \hline 3 & \dfrac{67}{42} & ? \\ \hline 4 & \dfrac{79}{54} & \dfrac{29549171683987}{25656103349287} \\ \hline 5 & \dfrac{97}{72} & ? \\ \hline 6 & \dfrac{103}{78} & ? \\ \hline 7 & \dfrac{193}{18} & \dfrac{619}{450} \\ \hline \end{array}

In this 2014 MO post, Jeremy Rouse left open some cases of $u$ including the two found by Allan MacLeod a few years later. In this 2017 paper, MacLeod gave the (huge!) solution for $u=\frac{49}{24}$ and $u=\frac{79}{54}$ in the last section of the paper. And while MacLeod didn't explicitly give the $v$ parameter, it was easy enough to reverse-engineer it from the $a^4+b^4+c^4+d^4 = (a+b+c+d)^4$ he gave.

Question:

1. Is it now possible to find $v$ for the three remaining $u$?
2. The height of MacLeod's two $v$ seems high. Can it be reduced?

Similar to the case $x^4+y^4+z^4 = 1$ discussed in this MO post, define the system,

$$x^4+y^4+z^4+1 = (x+y+z+1)^4\tag1$$ $$\frac{x^2+x+1}{(x+y+1)(x+z+1)}=u\tag2$$ $$\frac{y^2+y+1}{(y+z+1)(y+x+1)}=v\tag3$$ $$\frac{z^2+z+1}{(z+x+1)(z+y+1)}=w\tag4$$

The variable $w$ is dependent via a rather complicated expression on $(u,v)$ so $(4)$ is superfluous. We use the first three equations to solve for the three unknowns $(x,y,z)$. After some algebra, it can be determined they are roots of quadratics, hence yields pairs of solutions. The discriminant of the quadratic is,

$$D^2 = -3(2 - u + u^2)^2 + 6(2 + u + 3u^2 - 3u^3 + u^4)v - 3(5 - 6u + 6u^2 - 2u^3 + u^4)v^2 + 6(1 - u)(1 - 2u - u^2)v^3 - 3(1 - u)^2v^4$$

So if there is rational $(u,v)$ such that $D$ is also rational, then the quartic in $v$ is birationally equivalent to an elliptic curve. For some desired $u$ of small height, we are looking for an initial solution $v_1$ from which an infinite more can be generated. Here are ten such $u$,

\begin{array}{|c|c|c|c|} \hline \text{#} & u & v_1 & \text{Discover}\\ \hline 1 & \dfrac{31}{6} & \dfrac{6619}{5550} & \text{Rouse} \\ \hline 2 & \dfrac{49}{24} & \dfrac{138551171933011575944603377}{41031556739549840108788225} & \text{MacLeod}\\ \hline 3 & \dfrac{67}{42} & ? & ?\\ \hline 4 & \dfrac{79}{54} & \dfrac{29549171683987}{25656103349287} & \text{MacLeod}\\ \hline 5 & \dfrac{97}{72} & ? & ?\\ \hline 6 & \dfrac{103}{78} & ? & ?\\ \hline 7 & \dfrac{121}{96} & \dfrac{6250987}{506400} & \text{Tomita}\\ \hline 8 & \dfrac{157}{150} & \dfrac{8467}{150} & \text{Rouse}\\ \hline 9 & \dfrac{181}{150} & \dfrac{277567}{31675} & \text{Tomita}\\ \hline 10 & \dfrac{193}{18} & \dfrac{619}{450} & \text{Wroblewski}\\ \hline \end{array}

In this 2014 MO post, Jeremy Rouse left open a lot of cases of $u$ including two found by Seiji Tomita in 2015 and two found by Allan MacLeod in 2017. In this 2017 paper, MacLeod gave the (huge!) solution for $u=\frac{49}{24}$ and $u=\frac{79}{54}$ in the last section of the paper. And while MacLeod didn't explicitly give the $v$ parameter, it was easy enough to reverse-engineer it from the $a^4+b^4+c^4+d^4 = (a+b+c+d)^4$ he gave.

Question:

1. Is it now possible to find $v$ for the three remaining $u$?
2. The height of MacLeod's two $v$ seems high. Can it be reduced?