On counterexamples to Euler's conjecture using Bremner's and Durman's elliptic curves

Question

Noam Elkies found the first counter-example to Euler's sum of powers conjecture that,

$$a^4+b^4+c^4 = d^4$$

was not solvable by expressing the equation as an intersection of two quadric surfaces dependent on a parameter $m$. There are only eleven known rational $m$ of small height, with 4 found by Andrew Bremner discussed in this post. Smaller $m$ tend to yield $(a,b,c,d)$ as polynomials with small coefficients.

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I. Bremner's elliptic curve

Find rational $(v,z)$ such that,

$$z^2 = 42856039590241 + 4301879366236v - 65877950554v^2 - 710638564v^3 - 109887359v^4 $$

then we have,

$$10^4(690689 - 17642v - 407v^2)^4 + 10^4(260257 + 43910v + 473v^2)^4+ z^4 = 3^4(2676749 + 26902v + 3549v^2)^4$$

For any $v$, the terms satisfy the nice relation,

$$m=\frac{(a + b)^2 - c^2 - d^2}{a^2 + a b + b^2 + (a + b)d} =-\frac{5}{44}$$

with $m =-\frac{5}{44}$ being one of Bremner's. They also obviously obey,

$$-c^4+d^4 = 0 \text{ mod }5^4$$

a general property of such Diophantine equations. There are only two known solutions $v$ with "smallish" height,

$$v = -\frac{103605703}{47433977},\; \frac{2950708837949}{171081882189}$$

both yielding, after removing common factors, the same,

$$2024155336530384440^4+ 585715960903147640^4 + 2556827383749699103^4= 2778996090487120353^4$$

where $d \approx 2.77\times10^{18}$.

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II. Durman's elliptic curve

Find rational $(v,z)$ such that,

$$z^2 = -583937117447 + 1322131490860v - 1113337123194v^2 + 437040946060v^3 - 64494011447v^4 $$

then we have,

$$12^4(-50071 + 28490v + 2829v^2)^4 + 12^4(15601 + 21202v - 14883v^2)^4+ z^4 = (829109 - 804482v + 268253v^2)^4$$

and for any $v$,

$$m=\frac{(a + b)^2 - c^2 - d^2}{a^2 + a b + b^2 + (a + b)d} =-\frac{41}{36}$$

There are only two known solutions $v$,

$$v = \frac{416887}{178391},\; \frac{149943493}{118146461}$$

both yielding the same,

$$588903336^4 + 859396455^4 + 1166705840^4 = 1259768473^4$$

where $d \approx 1.25\times10^{9}$.

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Question:

Compared to the five related elliptic curves of this post, it seems Bremner's first pair of $v$ is unusually "large", while Durman's pair is not quite so. For both, are there $v$ of smaller height, or at least others such that it yields $d < 10^{28}$?

Answer 1

I. Bremner's elliptic curve

$$t^2 = 42856039590241 + 4301879366236v - 65877950554v^2 - 710638564v^3 - 109887359v^4$$ Quartic can be transformed to an elliptic curve as follows. $$E: Y^2= X^3 -X^2 + 349942184229228X -11167797929528591502588$$ $E$ has rank $3$ and generators are

$(X,Y)=((17487516, 17312851170),$

$(\dfrac{133466957657960755098}{1953596039521}, \dfrac{1572415798225425554647417475088}{2730562673994936431}),$

$(\dfrac{62931098782149520555151851692}{3627110466854477131489}, \dfrac{2458757717248798994666837091789787683070446}{218444529876451696484690059970513})).$

Small solutions were got using group law.

$19399184483902029008^4+2329747842666412840^4+12696186158476139705^4=20234461127553384633^4$

$1519814187310380835480^4+23896480714429616100215^4+13623248018235893097232^4=24504057146788194291849^4$

$2877363855098380947880^4+444897078221606141840^4+2016612085130087009647^4=3037467718844497770129^4$

$$v = \dfrac{-354638588051}{87849185509}, \dfrac{299059640905}{13394405401}, \dfrac{-37187846905}{10856642831}$$

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II. Durman's elliptic curve

$$z^2 = -583937117447 + 1322131490860v - 1113337123194v^2 + 437040946060v^3 - 64494011447v^4$$ Quartic can be transformed to an elliptic curve as follows. $$E: Y^2= X^3 + 2639323244332897X -20156152630838819347102$$ $E$ has rank $3$ and generators are

$(X,Y)=((\dfrac{150485239969}{1681}, \dfrac{66591661732011250}{68921},$

$(\dfrac{406882921075439}{32706961}, \dfrac{22603757491909733280000}{187051109959}),$

$(\dfrac{692289872015705012161}{91114336616689}, \dfrac{15945857110396198598969382871250}{869721239797220696887})).$

Small solutions were got using group law.

$18125123544^4+41714673255^4+34169217200^4=46055390617^4$

$10539980352556633840239^4+7799922278924748599160^4+4141571237269338150920^4=11305555143522867817873^4$

$13226266181198583365544^4+16841033682021117865520^4+4780632380106855105975^4=18276027741543869996617^4$

$47172089378698523207965335^4+26409847035187091768472744^4+94680315476024517009462320^4=96242977191578497031965033^4$

$$v = \dfrac{192013}{146001}, \dfrac{121355464559}{45273476043}, \dfrac{66997185029}{63378288733}, \dfrac{152066946716871}{84713675309443}$$