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(Note: This has been updated to be similar with this MO post.)

There are exactly 31 known primitive solutions to,

$$a^4+b^4+c^4 = d^4\tag{1}$$

with $d<10^{28}$. Noam Elkies showed that $(1)$ as,

$$(p + r)^4 + (p - r)^4 + s^4 = q^4\tag{2}$$

can be completely solved as an intersection of two quadric surfaces,

$$-(3 m^2 - 8m + 6) p^2 + 2 (m^2 - 2) p q - 2 m q^2 = (m^2 + 2) r^2\tag{3}$$

$$-4 (m^2 - 2) p^2 + 8 m p q + (m^2 - 2) q^2 = (m^2 + 2) s^2\tag{4}$$

for some constant $m$. Given a known solution to $(1)$, $m$ can be recovered as,

$$m = \frac{4p^2-q^2-s^2}{3p^2-2pq+r^2} = \frac{(a+b)^2-c^2-d^2}{a^2+ab+b^2-(a+b)d}\tag5$$

One can reverse-engineer the known solutions and find that there are only eight known rational $m$ of small height such that $(3),(4)$ can be rationally solved, namely,

$$m_k = -\frac{5}{8},\;-\frac{9}{20},\;-\frac{29}{12},\;-\frac{41}{36},\;-\frac{93}{80},\;-\frac{136}{133}$$

and positive,

$$m_k = \frac{201}{4},\;\frac{233}{60}$$

with the last one recently found by Andrew Bremner and yielding #21 mentioned in the comments below. The first three $m_k$ give rise to the conditional equations,

  1. $(-313+484v+85v^2)^4+(10-586v+68v^2)^4+(2t)^4=(363-204v+357v^2)^4$
  2. $(-15968 + 2334 v+59v^2)^4+(7068 + 3082 v + 10v^2)^4+(2t)^4 = (22628 + 54 v + 159v^2)^4$
  3. $(-11980 + 1673 v + 54v^2)^4+(36 - 2321 v + 3v^2)^4+(t)^4 = (24677 + 203 v + 71v^2 )^4$

with small solutions,

  1. $v = -31/467,\;\;-3015/9707,\;\;18247/19530,\;\;30671/229738$
  2. $v = 77/9,\;\;-1022/243,\;\; -50191/8685 $
  3. $v = -2020/127$

For example, factoring the first equation yields the elliptic curve condition,

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

An initial point is $v=-31/467$ from which one can find an infinite more. These small points $v_i$ explain some of the 20 solutions with $d<10^{10}$, while the rest have unwieldy $m$. (The 3rd family might still have rational points that yield $d$ within that range.)

Question: What other $m$ is there of small height not in the list of eight above?

P.S: My thanks to Noam Elkies for help with a further family in the old version of this post.

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  • $\begingroup$ Where can I find a list of the 20 known primitive solutions to (1)? $\endgroup$ Commented Aug 20, 2014 at 15:39
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    $\begingroup$ Leonid Durman's site at euler413.narod.ru It needs an update though. Andrew Bremner found a smaller #21 than Tomita's, namely $4707813440^4 + 7813353720^4 + 11988496761^4 = 12558554489^4$. $\endgroup$ Commented Aug 21, 2014 at 14:16

1 Answer 1

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Andrew Bremner found the $k$th $m_k$ of small height for $k=8,9,10,11$. Given,

$$a^4+b^4+c^4 = d^4\tag1$$

$$(p + r)^4 + (p - r)^4 + s^4 = q^4$$

where $R_k = p,q,r,s$,

$$R_8 = 6260583580,\; 12558554489,\; -1552770140,\; 11988496761$$

$$R_9 = -1456578618665,\; 2734283895746,\; -639377557145,\; 2452045365504$$

$$R_{10} = -3142543344652846743,\;\, 5557992180974240706,\; -1971111422846551463, 4048310673060768880$$

$$R_{11} = -2361164981843721467350575,\; 62586521087452988953161234,\; 5241104489910083087 0860865,\; 16178554328069755572637088$$

Define,

$$m_k = \frac{4p^2-q^2-s^2}{3p^2-2pq+r^2}$$

Then,

$$m_8 = \frac{233}{60}$$

$$m_9 = -\frac{56}{165}$$

$$m_{10} = -\frac{5}{44}$$

$$m_{11} = -\frac{125}{92}$$

These solutions would be #21, #23, #29, and #30 in Leonid Durman's list, moving Tomita's from #27 to #31. There are now $31$ primitive solutions to $(1)$ with $d<10^{28}$.

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