19
$\begingroup$

(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.

$\endgroup$
  • $\begingroup$ Where can I find a list of the 20 known primitive solutions to (1)? $\endgroup$ – Kieren MacMillan Aug 20 '14 at 15:39
  • 3
    $\begingroup$ Leonid Durman's site at euler413.narod.ru It needs an update though. Andrew Bremner found a smaller #21 than Tomita's, namely $4707813440^4 + 7813353720^4 + 11988496761^4 = 12558554489^4$. $\endgroup$ – Tito Piezas III Aug 21 '14 at 14:16
8
$\begingroup$

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 $d<10^{28}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.