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

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 ?

• Unless you have a definite algorithm to do this by hand, maybe using some nice computer will be handy. – DonAntonio Apr 6 '17 at 23:22
• If someone has wolfram alpha premium, maybe they could give it a go – mrnovice Apr 6 '17 at 23:23
• Knowing the source of the number would be helpful. In particular, if this is a textbook problem, then perhaps you could mention some of the techniques covered for tackling these kinds of problems. – Blue Apr 6 '17 at 23:35
• Do you have a reason to believe the problem has a solution? The only fifth powers $\bmod {11}$ are $0,1,10$ so this can be $0-1-1$ or $10-1\pm0$ The fifth root of three times your number is about $9.15E7$ so you only have to check numbers that large. The $\bmod 11$ computation eliminates all but $1/20$ of the combinations. Maybe you can try some other moduli and make a greater reduction to make a computer search feasible. – Ross Millikan Apr 6 '17 at 23:35
• The "or difference" part is redundant; after all, if $n$ is a fifth power of an integer, then so is $-n$. Now, the questions of this kind are tricky, and we don't know the answers to some of those that seem to be way more elementary. For example, is 33 a sum of three cubes (possibly negative)? – Ivan Neretin Apr 7 '17 at 13:52

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 $282441633$?)
• I mean the computer search for the solution $282508861^5-282441633^5 + 0$ – didgogns Apr 7 '17 at 16:11
• Oh, once you somehow decide to consider that possibility, Mathematica can tell you the answer in about a second with Solve[x^5 + y^5 == 2140138088471960538384538519958130596908, {x, y}, Integers]. As mentioned in this answer, it's a finite problem. – Misha Lavrov Apr 7 '17 at 16:15