International Journal of Mathematics and Mathematical Sciences
Volume 31 (2002), Issue 11, Pages 695-699
doi:10.1155/S0161171202106107
On the Diophantine equation x2+p2k+1=4yn
1Department of Mathematics, Girls College of Education, P.O. Box 22171, Riyadh 11495, Saudi Arabia
2Department of Mathematics, Girls College of Education, P.O. Box 56778, Riyadh 11564, Saudi Arabia
Received 12 June 2001
Copyright © 2002 S. Akhtar Arif and Amal S. Al-Ali. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
Abstract
It has been proved that if p is an odd prime, y>1, k≥0, n is an integer greater than or equal to 4, (n,3h)=1 where h is the class number of the field Q(−p), then the equation x2+p2k+1=4yn has exactly five families of solution in the positive integers x, y. It is further proved that when n=3 and p=3a2±4, then it has a unique solution k=0, y=a2±1.