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Tito Piezas III
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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,

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

Then forFor any $v$, the terms satisfy the simple relation,

$$m_2=\frac{(a+b)^2-c^2-d^2}{a^2+ab+b^2+(a+b)d}=-\frac{29}{12}$$$$u_2=\frac{(a+b)^2-c^2-d^2}{a^2+ab+b^2+(a+b)d}=-\frac{29}{12}$$

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,

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

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

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

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,

$$m_5=\frac{(a+b)^2-c^2-d^2}{a^2+ab+b^2+(a+b)d}=-\frac{41}{36}$$$$u_5=\frac{(a+b)^2-c^2-d^2}{a^2+ab+b^2+(a+b)d}=-\frac{41}{36}$$

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 infinitely many 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.

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,

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

Then for any $v$, the terms satisfy the simple relation,

$$m_2=\frac{(a+b)^2-c^2-d^2}{a^2+ab+b^2+(a+b)d}=-\frac{29}{12}$$

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: 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 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,

$$m_5=\frac{(a+b)^2-c^2-d^2}{a^2+ab+b^2+(a+b)d}=-\frac{41}{36}$$

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 infinitely many 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 $v$.

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.

Source Link
Tito Piezas III
  • 12.6k
  • 1
  • 39
  • 89

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,

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

Then for any $v$, the terms satisfy the simple relation,

$$m_2=\frac{(a+b)^2-c^2-d^2}{a^2+ab+b^2+(a+b)d}=-\frac{29}{12}$$

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: 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 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,

$$m_5=\frac{(a+b)^2-c^2-d^2}{a^2+ab+b^2+(a+b)d}=-\frac{41}{36}$$

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 infinitely many 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 $v$.