Still more elliptic curves for $a^4+b^4+c^4=d^4$

404 Views Asked by At

There are 31 known primitive solutions to $a^4+b^4+c^4 = d^4$ with $d<10^{28}.$ (Update: As of Feb. 21, there are now 93. See this MSE table.) Old statistics are,

\begin{array}{|c|c|} \hline \text{Range} & \text{# of sol} \\ \hline 10^5-10^7 & \;3 \\ \hline 10^7-10^9 & 12 \\ \hline 10^9-10^{28} & 16 \\ \hline \text{Total} & 31 \\ \hline\end{array}

A brute-force search was done for $d<10^9$, and explains why it is more dense than the range $10^9-10^{28}$ (where solutions were found mostly using several elliptic curves). The objective of this post is to find more $d<10^{28}$ using other elliptic curves.


I. Curve 1

Given $a^4+b^4+c^4 = d^4$ in the form,

$$(15968 - 2334 v - 59v^2)^4 + (7068 + 3082 v + 10v^2)^4 + t^4 = (22628 + 54 v + 159v^2)^4$$

where,

$$4(110301312 + 10244932v - 1285119v^2 + 5299v^3 - 6260v^4) = t^2$$

For any $v$, the terms $(a,b,c,d)$ satisfy the simple relationship,

$$m_1=\frac{(a+b)^2-c^2-d^2}{a^2+ab+b^2+(a+b)d}=-\frac{9}{20}$$

For infinitely many $v$, then $t$ is also rational. "Smallish" solutions are,

$$v = \frac{77 }{9}, \frac{171808 }{16161}, \frac{2465138 }{293763}, \ \frac{5207881}{ 1383765}, \frac{13617482}{ 1280007}, \frac{1251197642}{314528967}$$

$$\color{red}{-v} = \frac{1022}{ 243}, \frac{50191 }{8685}, \frac{128416}{ 29685}, \frac{1116448 }{2565009}, \frac{10267558}{ 1775127}, \frac{237282598}{377952087}\quad$$

For example, let $v = -\frac{1022}{ 243}$ and $v=\frac{77 }{9}$, substituting $(v,t)$ then, after removing common factors, yields the $1$st and $2$nd smallest solutions,

$$95800^4+ 217519^4+ 414560^4= 422481^4\\ 673865^4 + 1390400^4 + 2767624^4 = 2813001^4$$

This curve has been well-explored and the 12 points yield 6 primitive $(a,b,c,d)$ found by Tomita with $d<10^{20}$. It may serve to show how the growth of the $v_k$ seems reasonably slow and it is hoped that the next curves will be similar.


II. Curve 2

$$(11980 - 1673 v - 54v^2)^4 + (-36 + 2321 v - 3v^2)^4 + t^4 = (24677 + 203 v + 71v^2)^4$$

where,

$$591800025 + 20030510v + 1671327v^2 + 92762v^3 - 4112v^4 = t^2$$

The terms satisfy,

$$m_2=\frac{(a+b)^2-c^2-d^2}{a^2+ab+b^2+(a+b)d}=-\frac{29}{12}$$

Only two small solutions are known. Can you find more?

$$v=-\frac{2020}{127}, \frac{76164}{2063}$$

After removing common factors, either $v$ will yield the $3$rd smallest,

$$1705575 ^4 + 5507880^4 + 8332208^4 = 8707481^4$$


III. Curve 3

$$(- 1058960 + 11203324v + 178500100v^2)^4 + (518320 + 16483396v - 294176372v^2)^4 + t^4 = (1304433 - 27003006v + 345712797v^2)^4$$

where,

$$-1251219988511 + 78204922436804v - 1649103906705762v^2 + 19988050672538996v^3 - 76026722992074935v^4=t^2$$

The terms satisfy,

$$m_3=\frac{(a+b)^2-c^2-d^2}{a^2+ab+b^2+(a+b)d}=-\frac{93}{80}$$

Only four solutions are known,

$$v=\frac{12040}{ 110133}, \frac{10691}{353335}, \frac{737109}{ 5187253}, \frac{7880680}{ 207097317}$$

The 4 $v_k$ will give 2 primitive $(a,b,c,d)$ with the first $v$ yielding the $4$th smallest,

$$5870000^4 + 8282543^4 + 11289040^4 = 12197457^4$$


V. Curve 4

$$(- 20150032 + 34614497v + 67829197v^2)^4 + (5444488 + 58502527v - 148896448v^2)^4 + t^4 = (30141789 - 124521519v + 197120781v^2)^4$$

where,

$$-812211871484873 + 7722674874928166v - 27864960882719827v^2 + 48860237516933014v^3 - 31577722089576368v^4 = t^2$$

The terms satisfy,

$$m_4=\frac{(a+b)^2-c^2-d^2}{a^2+ab+b^2+(a+b)d}=-\frac{136}{133}$$

Only four solutions are known,

$$v = \frac{17611}{74168}, \frac{438773}{1417384}, \frac{3258337}{6134331}, \frac{298041047}{414260301}$$

The 4 $v_k$ will also give 2 primitive $(a,b,c,d)$, with the second $v$ yielding the $5$th smallest,

$$4479031^4 + 12552200^4 + 14173720^4 = 16003017^4$$


V. Question

We skip Curve 1. Can more rational points to Curves 2,3,4 be found so it yields primitive solutions to $a^4+b^4+c^4 = d^4$ with the restriction $|d|<10^{28}$ so that it will fit in this table?

P.S. The list has no additions since 2015.

3

There are 3 best solutions below

7
On BEST ANSWER

Curve 2:

$$591800025 + 20030510v + 1671327v^2 + 92762v^3 - 4112v^4 = t^2$$ Quartic can be transformed to an elliptic curve as follows. $$E: Y^2= X^3+ 10660885666177X+ 13598881200848998978$$ $E$ has rank $3$ and generators are

$(X,Y)=((3102801, 8749246466),$

$(\cfrac{50932656665761}{437981184}, \cfrac{35310438807501467200625}{9166070218752}),$

$(\cfrac{1009080290902892989011849}{834120475619653369}, \cfrac{4050218635872140181838359935096759002}{761803867761923063287263347})).$

Small solutions were got using group law.

$$125777308440^4 + 894416022327^4 + 2032977944240^4 = 2051764828361^4$$

$$27546142170735^4+7908038161032^4+43940127884360^4=45556888578449^4$$

$$979639439973824041716201624^4+1202764577995511968722134745^4+699747942023521949377658320^4=1343040355606487633849060057^4$$

with $v =\dfrac{1026427}{1526709}$ , $v=\dfrac{52784969}{6426498}$, and $v = -\dfrac{5465694589546439}{428624205717762}$, respectively.


Curve 3:

$$-1251219988511 + 78204922436804v - 1649103906705762v^2 + 19988050672538996v^3 - 76026722992074935v^4=t^2$$ Quartic can be transformed to an elliptic curve as follows. $$E: Y^2= X^3+ 1687750917943790881X -294299265667029867546450078$$ $E$ has rank $2$ and generators are

$(X,Y)=((1057203921, 51687652673778),$

$(\cfrac{2216934161202292017321}{77272324441}, \cfrac{104489172057583918416101354132502}{21480083475784739})).$

One small solution was got using group law, $$ 15876595946759369395903^4+7188470920864810763360^4+12896301483090810351440^4=17503689286309573964097^4$$

and $v = \dfrac{68984390349}{671029381933}$.


Curve 4:

$$-812211871484873 + 7722674874928166v - 27864960882719827v^2 + 48860237516933014v^3 - 31577722089576368v^4 = t^2$$

Quartic can be transformed to an elliptic curve as follows. $$E: Y^2= X^3 -X^2 + 3971619223592226604X -1791602749600376016515086080$$

Though I could not decide the rank, two rational points were found below.

$(X,Y)=((\cfrac{19931851860497}{4624}, \cfrac{97128553676746820625}{314432}),$ $(\cfrac{16731825034017654073}{1552044816}, \cfrac{69552255025210190460581513125}{61144357571136}))$.

One small solution was got using group law,

$$381461080909525552802665^4+168213921178037201816584^4+281048473715879152495040^4=409840652625395469143913^4$$

and $v = \dfrac{1535097779977}{2697183628776}$.

3
On

(Updated with solutions from this table.) This is an addendum to Tomita's two answers and places the whole post in context. With new points on Curves 1,2,3,4 such that $d<10^{28}$, we have a better idea of the growth of $d$ namely,

  1. Curve 0: $d\approx 10^{7},10^{8},10^{8},10^{9},10^{12},10^{16},10^{16},\color{blue}{10^{20}},10^{25},10^{26},10^{27}.$
  2. Curve 1: $d\approx 10^{5},10^{6},10^{9},10^{10},10^{13},10^{14},\color{blue}{10^{20}},10^{23},10^{24},10^{24}, 10^{25},10^{26}.$
  3. Curve 2: $d\approx 10^{7},10^{12},10^{13},\color{blue}{10^{27}}.$
  4. Curve 3: $d\approx 10^{7},10^{9},\color{blue}{10^{22}}.$
  5. Curve 4: $d\approx 10^{7},10^{8},\color{blue}{10^{23}}.$

Notice the "clustering" and "jumps" of exponents.


Note 1: Curve $0$ $\left(m = -\frac{5}{8}\right)$ was used by Noam Elkies to disprove Euler's conjecture on equal sums of like powers. Let,

$$(-313+484v+85v^2)^4+(10-586v+68v^2)^4+t^4=(-363+204v-357v^2)^4$$

where,

$$4(22030 + 28849 v - 56158 v^2 + 36941 v^3 - 31790 v^4) = t^2$$

Then the terms $(a,b,c,d)$ satisfy,

$$m_0=\frac{(a + b)^2 - c^2 - d^2}{a^2 + a b + b^2 + (a + b)d} =-\frac{5}{8}$$

Tomita found 7 solutions (using group law) for Curve 0 with $d<10^{20}$, the largest being $d\approx10^{16}.$ He also found an 8th one with $d\approx 10^{27}$, setting the bound for the table. Update: Bremner found 3 more solutions with $d< 10^{27}.$


Note 2: Curve 1 $\left(m = -\frac{9}{20}\right)$ also given in the post above,

$$4(110301312 + 10244932v - 1285119v^2 + 5299v^3 - 6260v^4) = t^2$$

yields the smallest counter-example to Euler's conjecture. Tomita found 6 points (using group law) with $d<10^{20}$, the largest being $d\approx10^{14}$. Update: Tomita found 4 more points in his second answer and Piezas found 2 more with $d<10^{27}$.

16
On

Curve 1:

$$4(110301312 + 10244932v - 1285119v^2 + 5299v^3 - 6260v^4) = t^2$$ Quartic can be transformed to an elliptic curve as follows. $$E: Y^2= X^3+ 2265722465761X -3154189403034549278$$ $E$ has rank $3$ and generators are

$(X,Y)=((1237921, 1244044242),$

$(\dfrac{101452934809}{9}, \dfrac{32314461714969994}{27}),$

$(\dfrac{98822808873914209}{20881117009}, \dfrac{32155995292015599252526194}{3017384051151527})).$

Small solutions were got using group law.

$$30248376090268690676600^4+118508989446504950664160^4+56915898438422390129561^4=120175486227071990769561^4$$ $$1758067984180618846616200^4+6632467268281371571709360^4+3057432874236989781768479^4=6714012701109174954871521^4$$ $$1165970778032514255823760^4+440517744543240750721000^4+59421842165791512201169^4=1171867103503245199920081^4$$ $$5328636655728999148343576^4+20991236668646283695879935^4+10137374115207940432133560^4=21291952935426564624339201^4$$

$$v = \dfrac{38262063182498}{4390437286923},\dfrac{10176189438546002}{1149740009812227},\dfrac{-3144800113847462}{545162208660903},\dfrac{428220031904}{49244280885}$$