Pranabesh Das, Pallab Kanti Dey, B. Maji, S. S. Rout
Published 2017-05-07Version 1
Recently, Bennett et al. found all perfect powers in the sum of $d$ consecutive cubes with $2 \leq d \leq 50$. In this paper, we consider the problem about finding out perfect powers in alternating sum of consecutive cubes. More precisely, we completely solve the Diophantine equation $(x+1)^3 - (x+2)^3 + \cdots - (x + 2d)^3 + (x + 2d + 1)^3 = z^p$, where $p$ is prime and $x,d,z$ are integers with $1 \leq d \leq 50$.
Perfect powers that are sums of consecutive cubes
Notes on the Diophantine Equation A^4+aB^4=C^4+aD^4
On the Diophantine equation $(x+1)^{k}+(x+2)^{k}+...+(2x)^{k}=y^n$
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