The great Euler conjected,
inter alia, that at least n powers of positive integers are needed to get a sum which is a n-th power as well.
It is up to you to find a set of integers {a,b,c,d,e}
such that
a^5+b^5+c^5+d^5=e^5.
REM: If unable to provide a proof, present the current state of known solutions for 4th and 5th powers.
AFAIK no solution for the 6th power exists so far.
WIKIPEDIA, in the article "Euler's Sum of Powers Conjecture" says:
"Euler's conjecture was disproven by L.J. Lander and T.R. Parkin in 1966 when, through a direct computer search on a CDC 6600, they found a counterexample for k=5 (fifth power case). This was published in a paper comprising just two sentences,
A total of three primitive (that is, in which the summands do NOT all have a common factor) counterexample are known:
--> 27^5+ 84^5+110^5+133^5=144^5 (Lander and Parkin, 1966)
--> (-220)^5+ 5027^5 + 6237^5+14068^5 = 14132^5 ( Scher and Seidl, 1996)
--> 55^5 + 3183^5+ 28969^5+85282^5 =85359^5 (Frye, 2004)
In 1988, Noam Elkies published a method to construct an infinite sequence of counter examples for the k=4 case. His smallest counterexample was:
2682440^4 + 15365639^4 + 18796760^4 =20615673^4
A particular case of Elkies' solution can be reduced to the identity:
(85v^2+484v-313)^4 +(68v^2 -586v+10)^4+ (2u)^4 = (357v^2 - 204v+363)^4
where, u^2= 22030 +28849v -56158v^2+36941v^3-31790v^4.
This is an elliptic curve with rational point at v_1 = -31/467. From this initial rational point, one can compute an infinite collection of others, Substituting v_1 into the identity and removing common factors gives the numerical example cited above.
In 1988, Roger Frye found the smallest possible counterexample
95800^4 + 217519^4 + 414560^4 = 422481^4
for k=4 by a direct computer search using techniques suggested by Elkies.This solution is the only one with the values of the variable below 1,000,000."
Edited on July 6, 2022, 11:22 pm
Current State of Known Solutions for 4th and 5th Powers
