On the simultaneous Diophantine equations $$ m \cdot (x_1^k+x_2^k+ \cdots + x_{t_1}^k)=n \cdot (y_1^k+y_2^k+ \cdots y_{t_2}^k)$$ m · ( x 1 k + x 2 k + ⋯ + x t 1 k ) = n · ( y 1 k + y 2 k + ⋯ y t 2 k ) ; $$k=1,3$$ k = 1 , 3
Crossref DOI link: https://doi.org/10.1007/s10998-017-0183-2
Published Online: 2017-06-21
Published Print: 2017-12
Update policy: https://doi.org/10.1007/springer_crossmark_policy
Izadi, Farzali
Baghalaghdam, Mehdi
License valid from 2017-06-21