On why solutions to $x^4+y^4+z^4 = 1$ come in pairs

Question

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.)

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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 $v_k$.

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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.)

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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.

Answer 1

Solutions were found with $d<10^{26}.$

I. New u

1. $(u,v)=(\dfrac{1744}{495}, \dfrac{135}{1208})$ with rank $3$. Then $\pm \sqrt{D^2}$ gives the pair,

   $$372623278887^4+435210480720^4+369168502640^4=521084370137^4$$ $$4408757988560^4+5819035124295^4+5611660306848^4=7082388012473^4$$

2. $(u,v)=(-\dfrac{3168}{1553}, -\dfrac{857}{3696})$ with $2 \leq$ rank $ \leq4$. $$19031674138785^4+25762744660064^4+2054845288320^4=27497822498977^4$$ $$2927198165920^4+613935345969^4+6310500741600^4=6382441853233^4$$

3. $(u,v)=(-\dfrac{1376}{705}, \dfrac{14337}{340})$ with rank $3$.

$$36295982895^4+29676864960^4+11262039896^4=39871595729^4$$ $$148739531603136^4+32467583677535^4+220093974949320^4=230791363907489^4$$

4. $(u,v)=(-\dfrac{1152}{2345}, \dfrac{19005}{3688})$ with rank $2$.

$$443873167360^4 +142485966505^4+ 544848079888^4=597385645737^4$$

$$468405247415^4+1657554153472^4+801719896720^4=1682315502153^4$$

5. $(u,v)=(\dfrac{2265}{184}, -\dfrac{68256}{135125})$ with rank $3.$ $$78558599440^4+814295112544^4+337210257575^4=820234293081^4$$ $$7745659501403353894384^4+2120589250533219579335^4+11684173258429439467360^4=12214291847502204701241^4$$

6. $(u,v)=(-\dfrac{1245}{5012}, \dfrac{248521}{62784})$ with rank $2.$

$$39110088360^4+49796687200^4+71826977313^4=76973733409^4$$

$$12209879806944320496330055^4+26621272474250391413865480^4+30730370351168229154149048^4=34497456764264994703368889^4$$

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II. Old u with new v

1. From $(u,v_1)=(-\dfrac{1041}{320},-\dfrac{9}{20}) $ to get $v_2 =-\dfrac{2830405}{222976} $, with rank $4$.

$$34511786481280^4 + 56329979520665^4 + 26636493544576^4 = 58844817090201^4$$ $$1891988836723177605880960^4 + 985329200220584284726784^4 + 3003225858017812695181145^4 = 3122849928997768901912409^4$$

2. From $(u,v_1)=(-\dfrac{400}{37},-\dfrac{93}{80}) $ to get $v_2= \dfrac{1867333}{457280} $, with rank $3$.

$$1724845107301282006322000^4+574585584668340612894713^4+1018986340666195845813760^4=1779979592349189232414713^4$$

3. From $(u,v_1)=(-\dfrac{201}{4},-\dfrac{136}{133}) $ to get $v_2=\dfrac{4372152}{935219}$, with rank $3$.

$$8281143989708209432415360^4+1749772249172099623115896^4+6550300128305909879699935^4=8997319881974346759473697^4$$

4. From $(u,v_1)=(-\dfrac{125}{92},-\dfrac{936}{5281}) $ to get $v_2=\dfrac{8204718073}{2152051820}$, with rank $3$.

$$69320669852667799672^4+38320435200564613600^4\\ +56375727168307546985^4=77107030404994920297^4$$

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III. An elliptic curve example

For $(u,v)=(-\dfrac{125}{92}, -\dfrac{936}{5281}).$

$$V^2 = 2028135809U^4-10348419236U^3+15452504000U^2-19864419528U-2701875708$$ $$Y^2=X^3-X^2+9243195710310751148X -761969307339454319105751548$$

where,

$U = (4806380418543401982X-904918222778168849178608076600-160675940688120Y)/W$

$V = (-290502664258777889830361189336987740318300847654664Y+30351227342319833611479328476809250X^3-2575941095132966226374274732384734806159817640X^2-280542332655903002803396928758943793134265195041307240X-7890389152832211150695233323867214891900261612757200780457112240)/W^2$

$W = 906548763647395Y+161779100249526028403X-238333975114378921359218316974$

Answer 2

Addendum

Search results using group law with $d<10^{26}.$

u=233/60: (a,b,c,d)

(4707813440, 7813353720, 11988496761,12558554489)

(188195571677171463096, 335981923744570504065, 275897431444390465240,375075545025537358721)

(438980913824794665, 343651286746207896, 225712385669145920,481334894209428521)

u=-56/165:

(408600530760, 1047978087905, 1226022682752,1367141947873)

(84616109521023161865, 163180699054891578792, 210878774189729581880,228746036963039501833)

(772654695228940017240, 10320518856970101984393, 6653143628547990852040,10739931407728904606857)

u=1744/495:

(372623278887, 435210480720, 369168502640,521084370137)

(4408757988560, 5819035124295, 5611660306848,7082388012473)

u=-3168/1553:

(613935345969, 2927198165920, 6310500741600,6382441853233)

(25762744660064, 19031674138785, 2054845288320,27497822498977)

u=-1376/705:

(29676864960, 36295982895, 11262039896,39871595729)

(32467583677535, 148739531603136, 220093974949320,230791363907489)

(12036780855644297767488, 23415987016826083521705, 26432693245716083746520,29998124444432653523113)

(7599957410902753037705, 8407785501400674212160, 4280294741983707700872,9649219915259253551497)

Answer 3

Here comes a table of new solutions for $u$, searched by the scheme proposed above, compare also with https://oeis.org/A003828. The last columns shows the obtained solution $x,y,z,t$, possibly reordered so that $x<y<z<t$, that satisfies $x^4+y^4+z^4=t^4$.

Note the symmetry in the function joining $u,v$ with the constraint that $f(u,v)$ is a square, which is $$ f(u,v) = \begin{bmatrix} u^4 & u^3 & u^2 & u & 1 \end{bmatrix} \begin{bmatrix} 1 & -4 & 0 & -8 & 4 \\ -4 & 8 & -16 & 48 & -16 \\ 0 & -16 & 48 & -64 & 0 \\ -8 & 48 & -64 & 32 & 32 \\ 4 & -16 & 0 & 32 & -48 \end{bmatrix} \begin{bmatrix} v^4 \\ v^3 \\ v^2 \\ v \\ 1 \end{bmatrix} \ , $$ so $f(u,v)$ is a square iff $f(v,u)$ is.

Among the two pairs $(u,v)$ and $(v,u)$ one is "simpler", has a smaller denominator on the first component, and only this one is recorded in the table. Then the solutions are sorted in the table w.r.t. this $u$-denominator.

Some search is still running, so i may update. (This is a fair search, not one that uses the already known values, then traces back the $(u,v,w)$ triples.)

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$$ \begin{array}{|c|c|c|l|} \hline u & v & w & \text{Solution}\\\hline \frac{553}{80} & -\frac{33400}{19537} & -\frac{294473}{635180} & x = 24743080\\ & & & y = 3971389576\\ & & & z = 4657804375\\ & & & t = 5179020201\\\hline \frac{553}{80} & -\frac{33400}{19537} & -\frac{294473}{635180} & x = 1554532675059625\\ & & & y = 1841841620201576\\ & & & z = 5352683902805120\\ & & & t = 5380742305932201\\\hline \frac{1873}{200} & -\frac{51416}{9425} & -\frac{3599825}{50036084} & x = 4092004076331400\\ & & & y = 24975412054750025\\ & & & z = 103028409596553328\\ & & & t = 103117303193818953\\\hline \frac{1873}{200} & -\frac{51416}{9425} & -\frac{3599825}{50036084} & x = 487814048600\\ & & & y = 8528631804200\\ & & & z = 26901926181047\\ & & & t = 26969608212297\\\hline -\frac{97}{400} & \frac{78065}{484} & -\frac{61583704}{7959505} & x = 66822832760\\ & & & y = 1313903832425\\ & & & z = 6010589044544\\ & & & t = 6014017311081\\\hline -\frac{97}{400} & \frac{78065}{484} & -\frac{61583704}{7959505} & x = 16306696482461560\\ & & & y = 21794572772239369\\ & & & z = 87375622888246360\\ & & & t = 87486470529871881\\\hline -\frac{1425}{412} & -\frac{9}{20} & \frac{5728}{215} & x = 140976551\\ & & & y = 5821981400\\ & & & z = 15355831360\\ & & & t = 15434547801\\\hline -\frac{1425}{412} & -\frac{9}{20} & \frac{5728}{215} & x = 673865\\ & & & y = 1390400\\ & & & z = 2767624\\ & & & t = 2813001\\\hline \frac{1744}{495} & \frac{135}{1208} & -\frac{977657}{480240} & x = 369168502640\\ & & & y = 372623278887\\ & & & z = 435210480720\\ & & & t = 521084370137\\\hline \frac{1744}{495} & \frac{135}{1208} & -\frac{977657}{480240} & x = 4408757988560\\ & & & y = 5611660306848\\ & & & z = 5819035124295\\ & & & t = 7082388012473\\\hline -\frac{1376}{705} & \frac{14337}{340} & -\frac{81065}{89412} & x = 32467583677535\\ & & & y = 148739531603136\\ & & & z = 220093974949320\\ & & & t = 230791363907489\\\hline -\frac{1376}{705} & \frac{14337}{340} & -\frac{81065}{89412} & x = 11262039896\\ & & & y = 29676864960\\ & & & z = 36295982895\\ & & & t = 39871595729\\\hline \end{array} $$