Question Number 168672 by mathlove last updated on 15/Apr/22
$$\mathrm{2}^{{x}} ={y}^{\mathrm{2}} +\mathrm{7}\:\:\:\:\:\:\:\:\:{x},{y}\in{N} \\ $$$${x}=?\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{y}=? \\ $$
Commented by safojontoshtemirov last updated on 15/Apr/22
$${x}=\mathrm{3}\:\:\:{y}=\mathrm{1}\:\:\:\:{x}=\mathrm{7}\:\:\:\:{y}=\mathrm{11} \\ $$$${x}=\mathrm{4}\:\:\:{y}=\mathrm{3} \\ $$$${x}=\mathrm{5}\:\:\:{y}=\mathrm{5} \\ $$
Commented by Rasheed.Sindhi last updated on 16/Apr/22
$${Also} \\ $$$${x}=\mathrm{15},{y}=\mathrm{181} \\ $$$$... \\ $$$$.. \\ $$
Commented by Rasheed.Sindhi last updated on 16/Apr/22
$$\mathrm{2}^{{x}} ={y}^{\mathrm{2}} +\mathrm{7}\:\:\:\:\:\:\:\:\:{x},{y}\in{N} \\ $$$$\mathrm{2}^{{x}} ={y}^{\mathrm{2}} +\mathrm{7}\Rightarrow{y}^{\mathrm{2}} +\mathrm{7}\in\mathbb{E}\Rightarrow{y}^{\mathrm{2}} \in\mathbb{O}\Rightarrow{y}\in\mathbb{O} \\ $$