A note on the Diophantine equation \(2^{n-1}(2^{n}-1)=x^3+y^3+z^3\) (Q2631753)

It is no known if there is an odd perfect numbers but since Euclid we know that each even perfect number is of the form \(2^{p-1}(2^p-1)\) where \(p\) and \(2^p-1\) are primes. \par The author proves that if \(n\equiv 1\pmod 3\) or \(n\equiv 2\pmod 6\) then the equation \[ 2^{n-1}(2^n-1)=x^3+y^3+z^3 \] has at least one integer solution. If \(n\equiv \pm 1\pmod 6\) then this equation has at least two integer solutions. The result implies that any even perfect number can be written as a sum of three cubes. The author gives elegant elementary proofs. The paper is completed with numerical results, questions and conjectures.