Difficult elliptic curves for $a^4+b^4+c^4+d^4 = (a+b+c+d)^4$?

Question

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?

Answer 1

Regarding your first question, there are no solutions $v$ to your equation for the three values of $u$ given (at least assuming GRH). For $u = 97/72$ and $u = 103/78$, the $D^{2} = \text{ quartic in } v$ equation represents a $2$-torsion element of the Shafarevich-Tate group of its Jacobian, and so it has no rational points. For $u = 67/42$, this is not the case. However, if this curve had a point on it, that would force the Jacobian to have rank $> 1$. Running a $3$-descent on the Jacobian (which takes about an hour and a half on a 5 GHz Intel processor) shows the rank is $\leq 1$, which rules out the possibility of a solution for $u = 67/42$. All class group computations for descent were done assuming GRH.

Regarding your second question, the height of the points that MacLeod finds doesn't actually seem to me to be that high, and there don't appear to be smaller generators of the Mordell-Weil group in either instance. The canonical height on the Jacobian for $u = 79/54$ is about $123.6$ (so the numerator and denominator of the $x$-coordinate have size roughly that of $e^{123.6}$), and this is for an elliptic curve of conductor $\approx e^{40}$. It is not that hard to have elliptic curves with smaller conductor and points of infinite order on them with much larger canonical height, at least if the curve has rank $1$. At one point, I found a rational right triangle with area $n = 958957$. The elliptic curve $y^{2} = x^{3} - n^{2} x$ has conductor $\approx e^{31.7}$ and the canonical height of a generator is $\approx 40593$.

Answer 2

Let us consider first the case $$ \bbox[yellow]{\qquad u=\frac{97}{72}\ ,\qquad} $$ and put the hands on the objects related to it. The quartic associated to this value is: $$ (C)\ :\qquad D^2 = -\frac{625}{1728} V^{4} + \frac{454825}{62208} V^{3} - \frac{166900849}{8957952} V^{2} + \frac{127673449}{4478976} V - \frac{163660849}{8957952}\ . $$ Its Jacobian is an elliptic curve, for it we take the minimal model equation: $$ (E)\ :\qquad y^2 + x y = x^{3} - x^{2} + 2991122782378200 x - 38499783305693620227264\ . $$ The rank of $(E)$ is two, the/some generators $P,Q$ (with heights $\approx 4.68$ and $76.77$), together with the one $2$-torsion point $T$ are: $$ \small \begin{aligned} P&= \left(446716560 : 9509903521464 : 1\right)\ ,\\ Q&= \left(\frac{6218250553782059969701663001149942198563}{166433954039897933481343360701601} \ ,\right. \\ &\qquad\qquad \left.\frac{760326315424214638089012944601501635954931677722768267444581}{2147152527351741938513597837348439795652862731151}\right) \ . \\[3mm] T&=\left(12255888 \ ,\ -6127944\right)\ . \end{aligned} $$ The program to find them was quickly typed in an approximation of a train from Frankfurt to Wiesbaden, but i had to wait for the next day to get $P,Q$. It may be useful to see also the conclusion of the Simon-$2$-descent started in sage for the curve $E$:

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Summary of results:
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    rank(E) = 2
    #E(Q)/2E(Q) = 8

Information on III(E/Q):
    #III(E/Q)[phi']    <= 4
    #III(E/Q)[2]       = 4

Information on III(E'/Q):
    #phi'(III(E/Q)[2]) = 1
    #III(E'/Q)[phi]    = 1
    #III(E'/Q)[2]      = 1

-------------------------------------------------------

There is also an explicit map ($2$-covering) $\mu$ from $C$ to $E$, and we may want to solve for each linear combination $kP+lQ+tT$ (with small $k,l,t$) the algebraic system $\mu(D,V)=(kP+lQ+tT)$. (Alternatively, we can reshape the quartic to have a biquadratic form and identify it / search for it in the Selmer group computations...) No solutions!

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Let us see now how the same game looks for the value $$ \bbox[yellow]{\qquad u=\frac {49}{24}\ ,\qquad} $$ known to have a suitable matching $v$-value. The quartic associated is: $$ (C)\ :\qquad D^2 = -\frac{625}{192}W^4 + \frac{104425}{2304}W^3 - \frac{6010129}{110592}W^2 + \frac{2783929}{55296}W - \frac{5650129}{110592}\ . $$ The Jacobian has a minimal model in the form of the elliptic curve $E$: $$ (E)\ :\qquad y^2 + xy = x^3 -x^2 + 4349022457680\,x -34039272045648040704 \ . $$ It has rank $2$, one easy point is $P=\left(6218643, 15277273491\right) $, and an other point linearly independent from $P$ is: $$ \tiny \left(\frac{517808174388951171727966023440810118966848667171857964954182533128977662731665916304557929039990304288234699}{148948384738283510952029721995772962646629114140628050665921509854328600324429613122797814978024172025}, \ \right. \\ \tiny \left.\frac{276152435357882425811329677183286170950464294252600560933746271599338131601642111352444210725127974718229549123022471103196921234056906449388418603609780236589342}{57484888966186929947708272827462213919536514493397945421025486983780591240028666893361511828465360130938302577535022079419652246609265185879054503909875}\right) \ . $$ This point $Q$ comes from the known $v$ value. If the system $(P,Q)$ of points can be reduced with a "smaller" $Q'$ instead of $Q$, then we would have $Q=kP+lQ'+rT$, but the first few $l$-division polynomials do not lead to a rational solution $Q'$. So we have to use $P$ and the "huge" $Q$. It turns out that $kP+lQ+rT$ can be traced back from $E$ to $C$ to a rational point, iff $k$ is even, $l$ is odd, and $r=0$ (no $T$) and the smaller values are experimentally: $$ \small \begin{array}{|l|c|} \hline (k, l) & v\\\hline (0, \pm 1) & \frac{138551171933011575944603377}{41031556739549840108788225} \\\hline (0, \pm 1) & \frac{89791364336280708026695801}{48759807596730867917907576} \\\hline (\pm 2, \pm 1) & \frac{2243265897338329761370142569}{181525207102693528111870200} \\\hline (\pm 2, \pm 1) & \frac{2424791104441023289482012769}{2061740690235636233258272369} \\\hline (\pm 2, \mp 1) & \frac{15833168155882680101878536241}{2002782091162783126124000641} \\\hline (\pm 2, \mp 1) & \frac{8917975123522731614001268441}{6915193032359948487877267800} \\\hline (\pm 4, \pm 1) & \frac{101684857089721023406104542510761}{46031479711585589842824272824536} \\\hline (\pm 4, \pm 1) & \frac{147716336801306613248928815335297}{55653377378135433563280269686225} \\\hline (\pm 4, \mp 1) & \frac{2831146523833963967640508335981481}{489913206966015071514017822723256} \\\hline (\pm 4, \mp 1) & \frac{3321059730799979039154526158704737}{2341233316867948896126490513258225} \\\hline (\pm 6, \pm 1) & \frac{19127606273739733372523475208279395964561}{1815738384043756176035314130644557564961} \\\hline (\pm 6, \pm 1) & \frac{10471672328891744774279394669461976764761}{8655933944847988598244080538817419199800} \\\hline (\pm 6, \mp 1) & \frac{1678161029905736496850080288614087393248009}{378742829297237862431620091186989944886200} \\\hline (\pm 6, \mp 1) & \frac{2056903859202974359281700379801077338134209}{1299418200608498634418460197427097448361809} \\\hline (\pm 8, \pm 1) & \frac{3653673223944866383234208261144829245645321105713561}{853700910998912188262453253886958932386625041537336} \\\hline (\pm 8, \pm 1) & \frac{4507374134943778571496661515031788178031946147250897}{2799972312945954194971755007257870313258696064176225} \\\hline (\pm 8, \mp 1) & \frac{3470566647074162639652358830556575873771413831660682577}{339662978552278967406756460701149247190131153777880225} \\\hline (\pm 8, \mp 1) & \frac{1905114812813220803529557645628862560480772492719281401}{1565451834260941836122801184927713313290641338941401176} \\\hline \end{array} $$ and we cannot do better. The $\pm$ and/or $\mp$ signs in the same line correspond to each other. In all cases we take $\pm Q$ and some small multiple of the small $P$.

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The value
$$ \bbox[yellow]{\qquad u=\frac{79}{54}\qquad} $$ comes next. Same case as in the last case. We know a suitable $v$-value. The hyperelliptic equation is $$ (C)\ :\qquad D^2 = -\frac{625}{972} V^{4} + \frac{296425}{26244} V^{3} - \frac{62771749}{2834352} V^{2} + \frac{43119799}{1417176} V - \frac{60949249}{2834352} \ . $$ The associated elliptic curve is $$(E)\ :\qquad y^2 + x y = x^{3} - x^{2} + 564252197078925 x - 7585371394838754972939 \ . $$ Its rank is two, we have the easy generator $P=\left(17285613, 85623615006\right)$, and a complicate $Q$ that comes from $v$: $$ \tiny \begin{aligned} Q&= \left(\frac{13542339090743095765383787613370818125751777089959685743831}{17894311600476851474392191858272790472281668541025}, \right. \\ &\qquad \left. -\frac{1576734037239346407130938775293709409473216395039908874533272834524772999590119323029819}{75695923781673646265917100744359232105214325935926602241240920737773851375}\right)\ . \end{aligned} $$ Trying a division of linear combinations of $P,Q,T$, where $T$ is the torsion point $\left(11051238, -5525619\right)$, does not lead to a simpler $Q'$ instead of $Q$. Using $P,Q,T$ now to come back to $C$, we obtain the $v$-values: $$ \small \begin{array}{|l|c|} \hline (k, l) & v\\\hline (0, \pm 1) & \frac{27602637516637}{1946534167350} \\\hline (0, \pm 1) & \frac{29549171683987}{25656103349287} \\\hline (\pm 2, \mp1) & \frac{64561011034081}{8416055481925} \\\hline (\pm 2, \mp1) & \frac{36488533258003}{28072477776078} \\\hline (\pm 2, \pm 1) & \frac{170224582999332361}{38112012818115925} \\\hline (\pm 2, \pm 1) & \frac{104168297908724143}{66056285090608218} \\\hline (\pm 4, \mp1) & \frac{530423941611526573}{195285495631296150} \\\hline (\pm 4, \mp1) & \frac{725709437242822723}{335138445980230423} \\\hline (\pm 4, \pm 1) & \frac{5416112818419270589653403}{1754185172228497156428103} \\\hline (\pm 4, \pm 1) & \frac{3585148995323883873040753}{1830963823095386716612650} \\\hline (\pm 6, \mp1) & \frac{49567129992853983505848463}{9368590243773496713104538} \\\hline (\pm 6, \mp1) & \frac{58935720236627480218953001}{40198539749080486792743925} \\\hline (\pm 6, \pm 1) & \frac{950835079120185274825134105535046383}{101917415042396092102440766096630458} \\\hline (\pm 6, \pm 1) & \frac{1052752494162581366927574871631676841}{848917664077789182722693339438415925} \\\hline (\pm 8, \mp1) & \frac{46646981368168757173742500079395271467}{3034198862350222053285156708672959767} \\\hline (\pm 8, \mp1) & \frac{24840590115259489613513828394034115617}{21806391252909267560228671685361155850} \\\hline (\pm 8, \pm 1) & \frac{2367625305390720977515524471289060654055137295801107}{192662924257361228861961389694784554318025023410407} \\\hline (\pm 8, \pm 1) & \frac{1280144114824041103188742930491922604186581159605757}{1087481190566679874326781540797138049868556136195350} \\\hline \end{array} $$ the next points in the list have bigger heights, and we cannot do better regarding $v$. (The first other $v$-value for $(0,\pm1)$ is cosmetically smaller, but this is just a caprice.)

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One more point, $$ \bbox[yellow]{\qquad u= \frac{67}{42} \ .\qquad} $$ Same notations as above. $$ \begin{aligned} (C)&\ :\ & D^2&= -\frac{625}{588} V^{4} + \frac{208825}{12348} V^{3} - \frac{28173709}{1037232} V^{2} + \frac{17198059}{518616} V - \frac{27071209}{1037232} \ ,\\ (E)&\ :\ & y^2 + x y &= x^{3} - x^{2} + 123402731275035 x - 1426809590840149357419\ . \end{aligned} $$ The easy point is $P=\left(12875673, 47916944976\right)$. The root number is $-1$, so we expect rank $1$ or $3$. As mentioned in the Table 4.1, page 11, of MacLeod's arXiv experimental paper. On his page 8, we have a list of first $t$-values that lead to quartics that are locally soluble. But we do not have any rank-three-claim on it. I tried an ad-hoc $4$-descent, but my scripts are not working yet...

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... have to stop here. If any progress can be done, than this should happen on the structural side, either a better understanding of the (hyper)elliptic curves involved, or a connection between them, or... Just trying point for point if a descent works is not a systematic approach, and even small values of $u$ lead to elliptic curves with points of huge height. Of course, (rich) experimental data is always a profit, but so far i do not see the pattern and the price (in seconds) for each new case is high. I will still keep in touch with the problem some more days...