It's conjectured that there are no non-trivial solutions to the Diophantine equation: $$ A^8+B^8=C^8+D^8 $$ I was trying to play around with it using some substitutions.. In particular I first write it as: $A^8-D^8=C^8-B^8$. Then I replace $(A,D,C,B)$ with $(p+q,p-q,r+s,r-s)$ to obtein the following: $$ p q (p^2 + q^2) (p^4 + 6 p^2 q^2 + q^4) = r s (r^2 + s^2) (r^4 + 6 r^2 s^2 + s^4) $$ Then I replaced $(p,q,r,s)$ with $(ax+by,ax-by,cx+dy,cx-dy)$ to obtein: $$ a^8 x^8 - b^8 y^8 = c^8 x^8 - d^8 y^8 $$ or: $$ \frac{x^8}{y^8} = \frac{b^8-d^8}{a^8-c^8} $$ I was trying to substitute some values but I only manage to get trivial solutions.. I would like to find some values such that $(A,B)\not=(C,D)$. Any kind of suggestions would be much appreciated. Thanks in advance for your help!

Some references I have already looked at: https://arxiv.org/pdf/math/0505629v2.pdf, https://www.jstor.org/stable/pdf/2007700.pdf?refreqid=excelsior%3A139ca229e2c0550603c138b5ada591c3&ab_segments=&origin=&acceptTC=1, https://sites.google.com/site/tpiezas/013 .

EDIT 1: My last substituion is useless. I have just written $(A,D,C,B)$ as $(ax,by,cx,dy)$ in the end.. And there are no non-trivial solution in this case.

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    $\begingroup$ I don't think this is known: Generalized taxicab number. It doesn't say there isn't one, but if there were a solution in eighth powers, I imagine it would be plenty amazing enough to mention. $\endgroup$
    – Brian Tung
    Jul 22, 2022 at 17:37
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    $\begingroup$ what is known about $T^4 + U^4 = V^4 + W^4 \; ? \; \; \; $ $\endgroup$
    – Will Jagy
    Jul 22, 2022 at 17:38
  • $\begingroup$ @WillJagy: Euler found the smallest solution; see the link I provided. $\endgroup$
    – Brian Tung
    Jul 22, 2022 at 17:39
  • $\begingroup$ Yes, I know.. We don't know if there are some solutions to this diophantine. When k = 4 there are infinite solutions, and the smallest one has been found by Euler. Although I see some simmetry between k = 4 case and k = 8... I really think there must be a solution to $A^8+B^8=C^8+D^8$ .. but it will be very large. $\endgroup$ Jul 22, 2022 at 17:40
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    $\begingroup$ For the case $k=4$ you might find Finding formula that solves $w^4+x^4=y^4+z^4$ over the integers interesting, including some links provided there. $\endgroup$ Jul 22, 2022 at 17:49

1 Answer 1


[too long for a comment] The closest you can get is likely to be a sum with 8 terms like$^1$

$$966^8+539^8+81^8 = 954^8+725^8+481^8+310^8+158^8$$

Anything with less terms would violate the Lander, Parkin, and Selfridge Conjecture from 1966 which states that when Diophantibe equation

$$\sum_{i=1}^n a_i^k = \sum_{i=1}^m b_i^k $$

holds where $a_i$ and $b_i$ positive with $a_i\neq b_j$, then $m+n\geqslant k$.

It's still a conjecture, though, but given the amount of computational effort that's usually put in finding counter-examples to such conjectures, it's very unlikely that one finds such an example at home (if one exists).

$^1$Seen on EulerNet

  • $\begingroup$ Thank you for your reply. Anyway.. I would like to try some kind of brute-force softwares which solve diophantine equations. Do you know any of them? Thank you in advance. $\endgroup$ Jul 23, 2022 at 8:57
  • $\begingroup$ @Luca Onnis: No, I don't know any software for this. When I am doing such searches (for fun), I am using Python, Sage or C/C++ and code it. As there are parametric solutions for $k=4$, maybe a good starting point is to generate solutions for $k=4$ and check whether they also match $k=8$, i.e. if all bases are squares. IIUC, EulerNet is project that computes such relations as a distributed software, where you can download the software and contribute computation time. $\endgroup$ Jul 23, 2022 at 9:17
  • $\begingroup$ @emacsdrivesmenuts I'd be happy to find $$a^4+b^8 = c^4+d^8$$ The closest I found is $$257^4 + 16^8 = 292^4 + 193^4$$ $$17332^4 + 23^8 = 17236^4 + 6673^4$$ Do have a means to check Wroblewski's database here? I would do it myself but my computer's Excel is malfunctioning. If you can find one, I'll be happy to convert this comment to a question. $\endgroup$ Nov 16, 2022 at 17:36
  • $\begingroup$ @Tito Piezas III: You can do it with some lines of Python. Four of the quadruples have exactly one square, all other quadruples have no square. $\endgroup$ Nov 17, 2022 at 10:50
  • $\begingroup$ @emacsdrivesmenuts I've now asked it as a question. Kindly see math.stackexchange.com/questions/4578655/… $\endgroup$ Nov 17, 2022 at 11:13

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