Question Number 208676 by RoseAli last updated on 20/Jun/24
Answered by Berbere last updated on 20/Jun/24
$$\sqrt[{\mathrm{4}}]{{a}}+\sqrt[{\mathrm{4}}]{{b}}=\sqrt{\mathrm{4}+\mathrm{3}\sqrt{\mathrm{2}}}=\sqrt[{\mathrm{4}}]{\mathrm{2}}.\sqrt{\mathrm{2}\sqrt{\mathrm{2}}+\mathrm{3}} \\ $$$$\Leftrightarrow\sqrt[{\mathrm{4}}]{\mathrm{2}{a}}+\sqrt[{\mathrm{4}}]{\mathrm{2}{b}}=\sqrt{\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)^{\mathrm{2}} }=\sqrt{\mathrm{2}}+\mathrm{1} \\ $$$$\Leftrightarrow\sqrt[{\mathrm{4}}]{\frac{{a}}{\mathrm{2}}}+\sqrt[{\mathrm{4}}]{\frac{{b}}{\mathrm{2}}}=\sqrt{\mathrm{2}}+\mathrm{1} \\ $$$${b}=\mathrm{0}\:\Rightarrow\sqrt[{\mathrm{4}}]{\frac{{a}}{\mathrm{2}}}=\sqrt{\mathrm{2}}+\mathrm{1}\:{no}\:{solution}\:\mathbb{N} \\ $$$${b}=\mathrm{1}\:\:{no}\:{solution} \\ $$$${b}=\mathrm{2}\Rightarrow{a}=\mathrm{8} \\ $$$${b}=\mathrm{3}\:{no}\:{solution} \\ $$$${b}=\mathrm{4}{no}\:{solution} \\ $$$${b}=\mathrm{8};{a}=\mathrm{2} \\ $$$$\left({a},{b}\right)\in\left\{\left(\mathrm{2},\mathrm{8}\right);\left(\mathrm{8},\mathrm{2}\right)\right\} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$