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.
[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