Question Number 102491 by pticantor last updated on 09/Jul/20
$$ \\ $$$$\boldsymbol{{L}}{e}\boldsymbol{{ts}}\:\boldsymbol{{p}}\:\in\mathbb{N}\:\boldsymbol{{and}}\:\boldsymbol{{n}}\in\mathbb{N}^{\ast} \\ $$$$\boldsymbol{{A}}_{{n}} =\mathrm{2}^{\boldsymbol{{n}}} +\boldsymbol{{p}}\:\boldsymbol{{and}}\:\boldsymbol{{d}}_{\boldsymbol{{n}}} =\boldsymbol{{P}}{GCD}\left(\boldsymbol{{A}}_{{n}} ,\boldsymbol{{A}}_{\boldsymbol{{n}}+\mathrm{1}} \right) \\ $$$$\left.\mathrm{1}\right)\:\boldsymbol{{show}}\:\boldsymbol{{that}}\:\:\boldsymbol{{d}}_{\boldsymbol{{n}}} /\mathrm{2}^{\boldsymbol{{n}}} \\ $$$$\left.\mathrm{2}\right)\boldsymbol{{determine}}\:\boldsymbol{{the}}\:\boldsymbol{{parity}}\:{of}\:\boldsymbol{{A}}_{{n}} \:{as}\:{a}\:{function}\:{of}\:{that}\:\boldsymbol{{of}}\:\boldsymbol{{p}} \\ $$$$\left.\mathrm{3}\right){determine}\:{the}\:{parity}\:{of}\:\boldsymbol{{d}}_{\boldsymbol{{n}}} \:{as}\:{a}\:{function}\:{of}\:{that}\:{of}\:\boldsymbol{{p}} \\ $$$$\left.\mathrm{4}\right){deduce}\:{pgcd}\left(\mathrm{2}^{\mathrm{2009}} +\mathrm{2009},\mathrm{2}^{\mathrm{2010}} +\mathrm{2009}\right) \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$