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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.

Tito Piezas III
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MathRevenge
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1 Answers1

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[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

emacs drives me nuts
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  • 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. – MathRevenge Jul 23 '22 at 08:57
  • @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. – emacs drives me nuts Jul 23 '22 at 09:17
  • @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. – Tito Piezas III Nov 16 '22 at 17:36
  • @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. – emacs drives me nuts Nov 17 '22 at 10:50
  • @emacsdrivesmenuts I've now asked it as a question. Kindly see https://math.stackexchange.com/questions/4578655/on-solutions-to-special-fourth-powers-a4b8-c4d8 – Tito Piezas III Nov 17 '22 at 11:13