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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

international journal of mathematics and mathematical sciences

On the Diophantine equation<mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" alttext="$x^2+p^{2k+1}=4y^n$" id="E1"><mml:mrow><mml:msup><mml:mi>x</mml:mi><mml:mn>2</mml:mn></mml:msup><mml:mo>+</mml:mo><mml:msup><mml:mi>p</mml:mi><mml:mrow><mml:mn>2</mml:mn><mml:mi>k</mml:mi><mml:mo>+</mml:mo><mml:mn>1</mml:mn></mml:mrow></mml:msup><mml:mo>=</mml:mo><mml:mn>4</mml:mn><mml:msup><mml:mi>y</mml:mi><mml:mi>n</mml:mi></mml:msup></mml:mrow></mml:math>

On the Diophantine equationx2+p2k+1=4yn