Finding more integer solutions to $x^4+y^4+z^4=3t^2$ below a bound?

Question

If $x^4+y^4+z^4=mt^2$ has one solution $xyzt\neq0$ then, using an elliptic curve, one can generally find infinitely many primitive solutions. We focus on $m=1,2,3$.

I. m = 1

Fauquembergue found that given the Pythagorean triple $a^2+b^2 = c^2$, then,

$$(ab)^4+(ac)^4+(bc)^4 = (a^4+a^2b^2+b^4)^2$$

II. m = 2

Proth (and others) found that if $a+b+c = 0$, then,

$$a^4+b^4+c^4 = 2(ab+ac+bc)^2$$

Ramanujan extended this to,

$$a^4(b-c)^4 + b^4(a-c)^4 + c^4(a-b)^4 = 2(ab+ac+bc)^\color{red}4$$

$$(a^2b+b^2c+c^2a)^4 + (ab^2+bc^2+ca^2)^4 + (3abc)^4 = 2(ab+ac+bc)^\color{red}6$$

and all $(ab+ac+bc)^\color{red}{2k}$.

Note: Infinitely many polynomial identities where $a+b+c\neq0$ are here and here.

III. m = 3

S. Realis found,

$$(5n^4 + 4n^3 + 9n^2 + 10n + 5)^4 + (5n^4 + 10n^3 + 9n^2 + 4n + 5)^4 + (5n^4 + 16n^3 + 27n^2 + 16n + 5)^4=3t^2$$

where $t$ is a palindromic octic polynomial. This has the general form,

$$(n^4 + an^3 + bn^2 + cn + 1)^4 + (n^4 + cn^3 + bn^2 + an + 1)^4 + (n^4 + dn^3 + en^2 + dn + 1)^4=3t^2$$

for five rational numbers $(a,b,c,d,e) = (4/5,\,9/5,\,10/5,\,16/5,\,27/5)$ though how Realis derived this I have no idea.

IV. Computer search

The fact that Realis found a quartic polynomial piqued my interest. So I did a search for $x^4+y^4+z^4=3t^2$ with $x\leq y\leq z \leq B\,$ below a bound $B$ and $\text{GCD}(x,y,z)=1$,

$$x,\,y,\,z\\ 1,\, 1,\, 1\\ 11,\, \color{red}{11},\, \color{green}{23}\\ \color{red}{11},\, 13,\, \color{red}{47}\\ \;\color{green}{23},\, \color{green}{85},\, 157\\ 37,\, \color{red}{47},\, \color{green}{85}\\ 41,\, \color{blue}{89},\, \color{blue}{115}\\ 55,\, 103,\, 149 \\ \color{blue}{89},\, 113,\, 259 \\ 91,\, \color{blue}{115},\, 293$$

Those repeating numbers seem a bit suspicious. We have the "cyclic",

$$\color{blue}{11}^4+\color{blue}{23}^4+11^4=3t_1^2\\ \color{blue}{23}^4+\color{blue}{85}^4+157^4=3t_2^2\\ \color{blue}{85}^4+\color{blue}{47}^4+37^4=3t_3^2\\ \color{blue}{47}^4+\color{blue}{11}^4+13^4=3t_4^2$$

which is reminiscent of the cubic,

$$ (\color{blue}{-9})^3 + \color{blue}{16}^3 + 33^3 = {34}^3\\ \color{blue}{16}^3 + \color{blue}{51}^3 + 213^3 = 214^3\\ \color{blue}{51}^3 + \color{blue}{82}^3 + 477^3 = {478}^3\\ \color{blue}{82}^3 + \color{blue}{57}^3 + 495^3 = 496^3\\ \color{blue}{57}^3 + \color{blue}{22}^3 + 255^3 = {256}^3\\ \color{blue}{22}^3 + (-9)^3 + 57^3 = 58^3 $$

though there is an infinite family responsible for the cubic version discussed in this post.

V. Question

If we brute-search $x^4+y^4+z^4=3t^2$ with the same conditions above but for a higher bound, will there be more repeating numbers?

@DmitryEzhov. Thanks! I will go over this. However, the constraint is $x\leq y\leq z$ so you missed $(11,,11,,23)$. There might be more like that. — Tito Piezas III, Apr 04 '25 at 09:06
For $x\leq y\leq z$ too many repeating entries, couldn't make good code — Dmitry Ezhov, Apr 04 '25 at 09:20
@DmitryEzhov That's ok. I just realized the equation $x^4+2y^4=3t^2$ can be transformed to an elliptic curve and should have infinitely many primitive solutions. — Tito Piezas III, Apr 04 '25 at 09:23
Is there any parametrization for $x^4+y^4+z^4=3t^2$ with polynomial of degree$ <4$ for $x,y,z$ ? — Guruprasad, Apr 04 '25 at 12:52
@DmitryEzhov In the link you gave, I see you've found 304 triples $(x,y,z)$. May I know how many integers have appeared at least twice? — Tito Piezas III, Apr 04 '25 at 12:53
@Guruprasad For $x^2+y^4+z^4=mt^2$, I know $m=2$ has quadratic parameterizations, but I don't know any yet for $m=3$. — Tito Piezas III, Apr 04 '25 at 12:56
@TitoPiezasIII Okay, I only know about finding solutions using Elliptic curve. — Guruprasad, Apr 04 '25 at 13:01
@TitoPiezasIII, for $x^4+2y^4=3t^2$ and $GCD(x,y)=1$ only $(x,y)$=(1,1),(23,11),(2375,6227) — Dmitry Ezhov, Apr 04 '25 at 14:58
@DmitryEzhov Others with $\text{GCD}(x,y)=1$ are, $$(x,y) = (40216177, 2628361)\ (x,y) = (611175635951, 3115654798201)$$ and infinitely more since an elliptic curve is involved, but they get large fast. — Tito Piezas III, Apr 04 '25 at 15:47

score 3 · Answer 1 · edited Apr 05 '25 at 03:45

3

Though incidental to your question, a similar parametric solution can be found like Realis'. $$x^4+y^4+z^4=nt^2 \tag{1} $$ Assume $(n,a_1,a_2,a_3,a_4)$ is a known solution of $(1).$

Substituting $x=a_1u+a, y=a_2u+b, z=a_3u,t=a_4u^2+gu+h$ into $(1)$ with $(n,a_1,a_2,a_3,a_4)=(3,1,1,1,1).$

We have $$(4b-6g+4a)u^3+(6a^2+6b^2-6h-3g^2)u^2+(4b^3-6hg+4a^3)u+a^4+b^4-3h^2=0 $$ Taking $(g,h)=\Bigl(\dfrac{2(a+b)}{3}, \dfrac{7a^2-4ab+7b^2}{9}\Bigr)$ to vanish the coefficient of $u^3$ and $u^2.$ Then, we have

$$ u = \dfrac{11a^4-28a^3b+57a^2b^2-28ab^3+11b^4}{6(2a^3-3a^2b-3ab^2+2b^3)}$$

Finally, we get a parametric solution.

$$x=23a^4-46a^3b+39a^2b^2-16ab^3+11b^4$$ $$y=11a^4-16a^3b+39a^2b^2-46ab^3+23b^4$$ $$z=11a^4-28a^3b+57a^2b^2-28ab^3+11b^4$$

edited Apr 05 '25 at 03:45

Tito Piezas III

60,745

answered Apr 05 '25 at 02:14

Tomita

3,209

+1 Very nice! I made some edits for aesthetics. This has the same form as by Realis, where $(x,y)$ are palindromic with each other, and $z$ is palindromic with itself. Great! – Tito Piezas III Apr 05 '25 at 03:48
By the way, what initial solution $(a_1,a_2,a_3,a_4)$ did Realis use? Or did he use a slightly different method? – Tito Piezas III Apr 05 '25 at 05:31
I have no idea, but from the results, it could be the same method I used. – Tomita Apr 05 '25 at 07:20
The new quartic post is ready. – Tito Piezas III Apr 05 '25 at 08:53

Jaap Scherphuis · Accepted Answer · 2025-04-04T13:56:25.880

I ran a slightly longer search, resulting in a few more triplets. Note that your list was not quite complete, maybe due to rounding errors. I don't see enough further repeats to suggest a pattern though.

$$11, 11, 23\\ 11, 13, 47\\ 37, 47, 85\\ 41, 89, 115\\ 55, 103, 149\\ 107, 121, 155\\ 23, 85, 157\\ 35, 157, 169\\ 115, 139, 187\\ 29, 185, 223\\ 89, 113, 259\\ 145, 167, 263\\ 91, 115, 293\\ 11, 25, 299\\ 29, 221, 307\\ 173, 209, 353\\ 131, 155, 359\\ 13, 143, 421\\ 29, 247, 425\\ 37, 433, 467\\ 103, 305, 533\\ 253, 289, 541\\ 17, 89, 547\\ 17, 91, 593\\ 193, 397, 637\\ 527, 601, 649\\ 29, 161, 689\\ 367, 451, 703\\ 109, 287, 755\\ 169, 323, 755\\ 373, 611, 817\\ 683, 899, 935\\ 73, 923, 947\\ 721, 877, 997\\ 53, 91, 1027\\ 91, 533, 1061\\ 643, 667, 1085\\ 713, 1087, 1109\\ 629, 773, 1133\\ 91, 907, 1171\\ 1103, 1153, 1175\\ 781, 925, 1177\\ 67, 1183, 1241\\ 493, 769, 1261\\ 163, 271, 1279\\ 469, 731, 1345\\ 533, 1013, 1421\\ 343, 715, 1433\\ 1205, 1409, 1433\\ 935, 1093, 1451\\ 197, 535, 1519\\ 901, 1129, 1573\\ 865, 1079, 1583\\ 595, 713, 1601\\ 689, 1229, 1615\\ 377, 1097, 1675\\ 13, 1189, 1679\\ 533, 713, 1721\\ 1561, 1669, 1763\\ 407, 1667, 1859\\ 1081, 1367, 2017\\ 535, 1949, 2047\\ 329, 809, 2059\\ 1417, 1979, 2149\\ 1283, 1885, 2161\\ 349, 2111, 2269\\ 671, 793, 2329\\ 553, 1585, 2329\\ 85, 2267, 2351\\ 821, 1721, 2383\\ 433, 1057, 2399\\ 53, 1363, 2405\\ 997, 1921, 2413\\ 163, 221, 2491\\ 221, 1783, 2513\\ 583, 1363, 2585\\ 1213, 1897, 2603\\ 877, 1427, 2665\\ 683, 913, 2725\\ 2347, 2561, 2753\\ 2417, 2669, 2767\\ 833, 1721, 2861\\ 61, 2593, 2867\\ 1489, 2833, 2977\\ 323, 505, 3013\\ 1555, 2357, 3019\\ 1921, 2639, 3023\\ 1987, 3029, 3043\\ 533, 2969, 3079\\ 341, 463, 3127\\ 2605, 2759, 3133\\ 2021, 2191, 3137\\ 277, 1573, 3143\\ 127, 2539, 3185\\ 1051, 1085, 3197\\ 35, 1451, 3263\\ 1883, 2483, 3263\\ 467, 2833, 3275\\ 2353, 2869, 3349\\ 499, 1385, 3377\\ 601, 1547, 3397\\ 2357, 3427, 3461\\ 845, 2107, 3499\\ 1169, 3295, 3499\\ 739, 2623, 3521\\ 689, 2335, 3547\\ 1681, 1969, 3553\\ 1351, 3389, 3559\\ 1909, 2053, 3575\\ 1961, 2539, 3593\\ 2035, 3031, 3641\\ 1879, 1975, 3653\\ 715, 799, 3679\\ 41, 101, 3689\\ 2935, 2947, 3713\\ 2399, 2579, 3731\\ 769, 3397, 3757\\ 221, 2827, 3773\\ 263, 2507, 3851\\ 2833, 2903, 3863\\ 493, 2219, 3887\\ 173, 1471, 3895\\ 1711, 1949, 3895\\ 1849, 3061, 3899\\ 493, 1187, 3911\\ 565, 853, 3913\\ 1153, 1549, 3925\\ 755, 1979, 3959\\ 59, 3805, 3959\\ 869, 2441, 3967\\ 143, 1417, 3985\\ 475, 2143, 3991\\ 169, 3083, 3995\\ 1271, 1693, 4045\\ 569, 1001, 4061\\ 2077, 2819, 4079\\ 533, 823, 4087\\ 113, 1685, 4121\\ 77, 2059, 4133\\ 1435, 2599, 4159\\ 1543, 1781, 4183\\ 1087, 4147, 4183\\ ...$$

This is where I stopped the search. It produced 142 triples (not including 1,1,1) in the range $1\le x\le y\le z\le 4183$. I see now I did not include $x=0$ as a possibility in the search.

In this range there are 65 numbers that occur more than once, and they are:

2x: $17$, $23$, $35$, $37$, $41$, $47$, $53$, $103$, $113$, $143$, $155$, $157$, $163$, $173$, $263$, $323$, $433$, $467$, $535$, $601$, $683$, $715$, $769$, $877$, $935$, $997$, $1085$, $1087$, $1153$, $1363$, $1417$, $1433$, $1451$, $1573$, $1921$, $1949$, $1979$, $2059$, $2329$, $2357$, $2399$, $2539$, $3263$, $3397$, $3499$, $3895$, $3959$, $4183$
3x: $13$, $85$, $89$, $115$, $169$, $493$, $689$, $713$, $755$, $1721$, $2833$
4x: $11$, $29$, $221$
5x: $91$
6x: $533$

Edit: I see in the comments that Dmitry Ezhov has done a much longer search.

Thanks. I've edited the post to given an example of cyclic "repeated" entries, though the cubic version has an infinite family behind it. — Tito Piezas III, Apr 04 '25 at 09:42
I see you've updated your answer. May I know how many triples $(x,y,z)$ have you found so far, and how integers have appeared at least twice? — Tito Piezas III, Apr 04 '25 at 12:50
So $142$ triples $(x,y,z)$ and $62$ integers that appear more than once. That’s interesting. For cubes, the phenomenon was more visible and could be explained by polynomial parameterizations. But $4$th powers is more challenging. — Tito Piezas III, Apr 04 '25 at 14:35
Here's another post about common terms in a quartic system but this one has an explanation, — Tito Piezas III, Apr 05 '25 at 08:52

Finding more integer solutions to $x^4+y^4+z^4=3t^2$ below a bound?

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