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Given,

$$a^4+b^4+c^4 = d^4\tag{0}$$

we have the identity,

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

where,

$$591800025 + 20030510 u + 1671327 u^2 + 92762 u^3 - 4112 u^4 = t^2\tag{1}$$

as well as a second,

$$(62697 + 5045 v - 242v^2)^4 + (-19200 - 9089 v + 46v^2)^4 + t^4 = (86825 - 27 v + 303v^2)^4$$

where,

$$-6422010512 + 412760610 v - 6214161 v^2 + 2027190 v^3 + 70673 v^4 = t^2\tag{2}$$

I know of only one rational solution of small height each to $(1)$ and $(2)$, namely,

$$ u =-2020/127$$

$$ v = -8251/94$$

Thus $(1)$ and $(2)$ can be transformed into elliptic curves, but they are distinct from the one used by Elkies to find the first solution to $(0)$, or the one that yields the smallest solution (found by R. Frye). From the initial rational point, I know how to generate others, but the numerators and denominators are huge.

Question: Anybody has software to find other rational solutions of small height to $(1)$ and $(2)$ ?

P.S. This question is related to the one I asked in MSE.

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    $\begingroup$ Using M.Stoll's ratpoints for about a minute on one head of the Sage cluster, I find that the first curve also has a pair of rational points at $u = 76164/2063$, and this is the only further point on either curve up to height $5 \cdot 10^5$. $\endgroup$ Oct 2, 2013 at 2:42
  • $\begingroup$ Thanks, Prof. Elkies! I am glad to know of a second point of small height, but both $u = -2020/127$ and $u=76164/2063$ yield the same $a,b,c,d$ (after sign changes, transposition, and removing common factors), namely $5507880^4 + 8332208 ^4 + 1705575 ^4 = 8707481 ^4$. Rats. $\endgroup$ Oct 2, 2013 at 2:53

1 Answer 1

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I have almost forgotten about this question, but I now know enough to answer it. Given the general form,

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


Identity 1. We have (after a minor change of signs),

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

where,

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

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

$$u_2=\frac{(a+b)^2-c^2-d^2}{a^2+ab+b^2+(a+b)d}=-\frac{29}{12}$$

Using an initial rational point $v=-\frac{2020}{127}$, Tomita in this 2024 post found that the quartic was birationally equivalent to the rank $3$ elliptic curve,

$$C_2: X^3+ 10660885666177X+ 13598881200848998978 = Y^2$$

yielding,

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

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

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

with $v=-\frac{2020}{127},\, \frac{1026427}{1526709},\, \frac{52784969}{6426498}$, respectively, and so on for infinitely many $v$.

Note 1: The first at $d = 8.7\times 10^6$ is in fact the third smallest in this list of 30 solutions, now 80 primitive solutions as of 2024.


Identity 2. We also have (after a minor change of signs),

$$(-62697 - 5045 v + 242v^2)^4 + (19200 + 9089 v - 46v^2)^4 + t^4 = (86825 - 27 v + 303v^2)^4$$

where,

$$-6422010512 + 412760610 v - 6214161 v^2 + 2027190 v^3 + 70673 v^4 = t^2$$

and the terms satisfy the similarly simple relation,

$$u_5=\frac{(a+b)^2-c^2-d^2}{a^2+ab+b^2+(a+b)d}=-\frac{41}{36}$$

Tomita found in this other 2024 post that one can use the rank $3$ elliptic curve

$$C_5:=X^3+2639323244332897X−20156152630838819347102=Y^2$$

to find rational $v$ yielding,

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

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

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

where $v =-\frac{8251}{94},\, \frac{12214947}{667183},\, -\frac{7693614747096}{171724585381},$ respectively, and so on for infinitely many $v$.

Note 2: For the parameter $u_k$, there are only about 17 known of small height. If you wish to help to find more, more details are in this recent MO post.

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