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Find (many, a family of) non-trivial solutions of the diophantine equation

$$\frac{x^4 + y^4}{z^4 + t^4}= u^2\ \ (*)$$

Notes:

  1. If the $\{x,y\}$, $\{z,t\}$ are proportional then we have a trivial solution. So we are interested in non-trivial solutions

  2. The equation $\frac{x^4 + y^4}{z^4 + t^4}= 1$ is known as the generalized taxicab problem, see this question. The smallest known solutions are fairly large.

  3. Perhaps one could try to solve it for some concrete $u>1$, like $u=2$.

  4. Origin of the problem: I was looking for squares $a^2$, $b^2$, $c^2$, such that $a^2 + b^2$, $b^2 + c^2$, $a^2 + b^2 + c^2$ are also square. Let's write

$$\begin{aligned}a &= (x t)^2 - (y z)^2 \\ b &= 2 x y z t\\ c&= (x z)^2 - (y t)^2\end{aligned}$$

Now $a^2 + b^2 = ((x t)^2 + (y z)^2)^2$, $b^2 + c^2 = ((x z)^2+ (y t)^2)^2$, and $a^2 + b^2 + c^2 = x^4 t^4 + y^4 z^4 + x^4 z^4 + y^4 t^4= (x^4 + y^4)(z^4 + t^4)$. Therefore, we need $(x^4 + y^4)(z^4 + t^4)$ to be a square. Note that we need non-proportional solutions, otherwise we get $a=0$, or $b=0$.

  1. The equation can be rewritten as $$(x^4 + y^4)(z^4 + t^4) = v^2\ \ (**)$$

  2. Per Fermat, the equation $x^4 + y^4 = w^2$ has no trivial solutions.

  3. It would be nice to have a family of solutions.

  4. Possible approach to finding solutions to the equation in form $(**)$. List pairs $x<y$ relatively prime and determine the square free part of $x^4 + y^4$. Then look for concidences $s(x^4 + y^4) = s(z^4 + t^4)$ ( perhaps a simplistic approach).

Any feedback would be appreciated! Thank you for your attention!

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  • $\begingroup$ See for example here. $\endgroup$
    – Dietrich Burde
    Jun 20, 2022 at 14:32

2 Answers 2

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Here is a table with small solutions ($x \le y \le 7300$, ordered by $y$) for the equation $$x^4+y^4 = u^2(z^4+t^4);$$

it is divided into $3$ parts:

  • $u=1$;
  • $u=41$;
  • other values for $u$.

\begin{array}{|r|r|r|r|r|} \hline x & y & z & t & u \\ \hline 59 & 158 & 133 & 134 & 1 \\ 7 & 239 & 157 & 227 & 1 \\ 193 & 292 & 256 & 257 & 1 \\ 271 & 502 & 298 & 497 & 1 \\ 103 & 542 & 359 & 514 & 1 \\ 222 & 631 & 503 & 558 & 1 \\ 76 & 1203 & 653 & 1176 & 1 \\ 878 & 1381 & 997 & 1342 & 1 \\ 1324 & 2189 & 1784 & 1997 & 1 \\ 1042 & 2461 & 2026 & 2141 & 1 \\ 248 & 2797 & 2131 & 2524 & 1 \\ 1034 & 2949 & 1797 & 2854 & 1 \\ 1577 & 3190 & 2345 & 2986 & 1 \\ 1623 & 3494 & 2338 & 3351 & 1 \\ 661 & 3537 & 2767 & 3147 & 1 \\ 3364 & 4849 & 4288 & 4303 & 1 \\ 2694 & 4883 & 3966 & 4397 & 1 \\ 604 & 5053 & 1283 & 5048 & 1 \\ 2027 & 6140 & 4840 & 5461 & 1 \\ 274 & 6619 & 5093 & 5942 & 1 \\ 2707 & 6730 & 3070 & 6701 & 1 \\ 498 & 6761 & 5222 & 6057 & 1 \\ \cdots & \cdots & \cdots & \cdots & \cdots \\ \hline 2 & 13 & 1 & 2 & 41 \\ 364 & 685 & 88 & 95 & 41 \\ 1223 & 2977 & 149 & 467 & 41 \\ 1417 & 5078 & 346 & 787 & 41 \\ 5146 & 5803 & 169 & 1022 & 41 \\ 8206 & 9633 & 894 & 1637 & 41 \\ ? & ? & ? & ? & ? \\ \hline 38 & 43 & 1 & 2 & 569 \\ 121 & 205 & 5 & 7 & 809 \\ 159 & 269 & 9 & 13 & 409 \\ 151 & 388 & 7 & 8 & 1889 \\ 9 & 437 & 1 & 3 & 21089 \\ 314 & 863 & 1 & 2 & 182209 \\ 859 & 1186 & 1 & 2 & 385241 \\ 10 & 1261 & 5 & 22 & 3281 \\ 1112 & 1323 & 7 & 12 & 14089 \\ 908 & 1559 & 7 & 8 & 31841 \\ 1041 & 2219 & 3 & 7 & 101201 \\ 2277 & 2521 & 1 & 3 & 905761 \\ 1166 & 2589 & 9 & 38 & 4729 \\ 2148 & 2723 & 12 & 23 & 15929 \\ 1383 & 2936 & 7 & 12 & 58049 \\ 1006 & 3177 & 2 & 3 & 1029961 \\ 911 & 3633 & 21 & 59 & 3769 \\ 475 & 3823 & 5 & 7 & 265721 \\ 1951 & 5777 & 79 & 103 & 2729 \\ 1897 & 5983 & 71 & 127 & 2129 \\ 2962 & 6741 & 22 & 57 & 14089 \\ 4969 & 6987 & 77 & 111 & 4001 \\ 4719 & 7187 & 1 & 3 & 6211649 \\ \cdots & \cdots & \cdots & \cdots & \cdots \end{array}

I'm curious why $u=41$ is relatively rich with solutions (after $u=1$). I have doubts if there are infinitely many solutions for $u=41$.
And, of course, I see no parametrization of solutions (to have a family of them).

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    $\begingroup$ Thank you, that is very interesting! So after $u=1$, it seems that the next $u$ with solutions is $41$? The only thing I know about $41$ is that it is prime, and also that polynomial $x^2 - x + 41$. It seems there are many solutions for $u=1$ ( infinitely many?) $\endgroup$
    – orangeskid
    Jul 5, 2022 at 17:27
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    $\begingroup$ @orangeskid, yes, it looks like there are infinitely many solutions for $u=1$: see please Jaroslaw Wroblewski results: math.uni.wroc.pl/~jwr/422 , math.uni.wroc.pl/~jwr/422/422-10m.txt , math.uni.wroc.pl/~jwr/422/422.zip $\endgroup$
    – Oleg567
    Jul 5, 2022 at 20:21
  • $\begingroup$ Oh, thank you! That is impressive indeed. Now, Dietrich Burde was kind to provide this link. I wonder if they are able to get any solutions from elliptic curves. Maybe it could show infinity of solutions ( I don't have access to the article). $\endgroup$
    – orangeskid
    Jul 5, 2022 at 20:44
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Consider the equation,

$$x^4+y^4 = u^2(z^4+t^4)$$

then there is a solution for infinitely many $u$ with special properties. Let,

\begin{align}u &=\,4(p^4+q^4)^4-3(p^4-q^4)^4\\[6pt] &=\frac{(p^2-q^2)^8+(2pq)^8+(p^2+q^2)^8-\tfrac32(2pq)^8}{2}\\[6pt] &=\frac{2(p^8 + 14p^4q^4 + q^8)^2 - \tfrac32(2pq)^8}{2} \end{align}

Note the Pythagorean triple $(p^2-q^2,\, 2pq,\, p^2+q^2)$ and the appearance of the polynomial $(p^8 + 14p^4q^4 + q^8)$ which is an invariant of the octahedron,

Octahedron

Then we have Carmichael's identity,

$$p^4(p^8 - 6p^4q^4 - 3q^8)^4 + q^4(3p^8 + 6p^4q^4 - q^8)^4 = (p^4 +q^4)u^2$$

more noted for his Carmichael numbers. For example, let $(p,q)=(1,2)$, then,

$$314^4 + 863^4 = (1^4 + 2^4)182209^2$$

which appears in the table in the first answer. Some remarks: First, while $u$ gets large very fast, it does prove there are infinitely many of them. Second, for the special case $u=1$,

$$x^4+y^4 = z^4+t^4$$

this has been discussed in several MSE posts and there are infinitely many polynomial parameterizations for $(x,y,z,t)$ by using an elliptic curve, as first done by Euler. Third, whether there are polynomial parameterizations for $u=41$ and others remains to be seen.

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    $\begingroup$ I am impressed with the fact that there are families of solutions. Will have to fire up some CAS and check. Amazing again, thank you! $\endgroup$
    – orangeskid
    Aug 19 at 15:40
  • $\begingroup$ Nice solution. Because of the showed non-homogenous of $36$-degree Carmichael's identity, probably it gives ALL the solutions. Anyway, difficult to find out a solution don't having this identity as a source. $\endgroup$
    – Piquito
    Aug 19 at 19:02
  • $\begingroup$ @Piquito Thanks. But no, it doesn't give all solutions. It does show that if you give me any two integers $(p,q)$, then I can quickly find two integers $(r,s)$ such that $$\frac{r^4+s^4}{p^4+q^4}=\text{square}$$ $\endgroup$
    – Tito Piezas III
    Aug 19 at 19:14
  • $\begingroup$ @TitoPiezasIII Yes I see, when $p,q$ are not coprimes. $\endgroup$
    – Piquito
    Aug 19 at 20:18
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    $\begingroup$ @orangeskid I assume you like unusual connections. Turns out Carmichael's parameterization was hiding an octahedron. Dunno why though. Probably has to do with all the $4$th and $8$th powers. $\endgroup$
    – Tito Piezas III
    Aug 20 at 7:12

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