Bilge Peker, Selin, Cenberci
Published 2012-02-01Version 1
In this article, we consider the equation x^2+19^{m}=y^n, n>2, m>0. We find the solutions of the title equation for not only 2 \mid m but also 2\notdividesm.
On the Diophantine Equation $x^{2}+5^{a}\cdot 11^{b}=y^{n} $
On the Diophantine equation $(x+1)^{k}+(x+2)^{k}+...+(2x)^{k}=y^n$
Complete solution of the Diophantine Equation $x^{2}+5^{a}\cdot 11^{b}=y^{n}$
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