Can 2140138088471960538384538519958130596908 be expressed as the sum, or difference, of three fifth powers?

Question

I have a large integer, 2140138088471960538384538519958130596908 , with 40 digits. I am looking for three fifth powers which when added together, or subtracted from one another, equal this number. This large integer is congruent to 9 ( mod 11 ) . Any ideas on how to make a start on this problem ?

Answer 1

Well, your number $N =2140138088471960538384538519958130596908$ is $$N=282508861^5-282441633^5 + 0$$ which I'm sure you were aware of but might have mentioned. In particular, if you're hoping for a second solution, this rules out any possible approach that takes the equation modulo $25$ or whatever and finds a contradiction.

Aside from that, well, there's an infinite search space, and the largest known solutions to $a^5+b^5+c^5+d^5=e^5$ or $a^5+b^5+c^5=d^5+e^5$ are orders of magnitude smaller, so I'm not too optimistic. I did search for other solutions using one small fifth power, and got nowhere.

(Can you explain how you chose $282508861$ and $282441633$?)