Question Number 80053 by mr W last updated on 30/Jan/20
$${Find}\:{integer}\:{x},\:{y}\:{such}\:{that} \\ $$$$\mathrm{2}^{{x}} −{y}^{\mathrm{2}} =\mathrm{615} \\ $$
Commented by john santu last updated on 30/Jan/20
$$\left(\sqrt{\mathrm{2}^{\mathrm{x}} }\right)^{\mathrm{2}} −\left(\mathrm{y}\right)^{\mathrm{2}} =\mathrm{615} \\ $$$$\left(\sqrt{\mathrm{2}^{\mathrm{x}} \:}−\mathrm{y}\right)\left(\sqrt{\mathrm{2}^{\mathrm{x}} }\:+\mathrm{y}\right)=\mathrm{615}\:=\:\mathrm{123}×\mathrm{5} \\ $$$$\sqrt{\mathrm{2}^{\mathrm{x}} }\:+\mathrm{y}=\mathrm{123}\:\wedge\:\sqrt{\mathrm{2}^{\mathrm{x}} }\:−\mathrm{y}=\mathrm{5} \\ $$$$\sqrt{\mathrm{2}^{\mathrm{x}} }\:=\:\mathrm{64}=\mathrm{2}^{\mathrm{6}} \:\Rightarrow\mathrm{x}=\mathrm{12}\:,\:\mathrm{y}\:=\:\mathrm{59} \\ $$
Commented by mr W last updated on 30/Jan/20
$${very}\:{nice}\:{solution}!\:{thanks}! \\ $$
Commented by john santu last updated on 30/Jan/20
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{mister} \\ $$
Commented by MJS last updated on 30/Jan/20
$${x}=\mathrm{12}\wedge{y}=\pm\mathrm{59} \\ $$