Question Number 162169 by Rasheed.Sindhi last updated on 27/Dec/21
$$ \\ $$$$\:\:\:\:\:\:\:\begin{array}{|c|}{\overset{\bullet} {\:\:\:\:\:\begin{array}{|c|}{\:\:\:\underset{{x}=?,{y}=?\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:} {\overset{{x},{y}\in\mathbb{Z}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:} {{x}+{y}+{x}^{\mathrm{2}} {y}^{\mathrm{2}} =\mathrm{586}}}\:\:}\\\hline\end{array}_{} ^{} }\:\:\:\:}\\\hline\end{array} \\ $$$$ \\ $$
Answered by naka3546 last updated on 27/Dec/21
$$\left({x},{y}\right)\:=\:\left(−\mathrm{2},\mathrm{12}\right)\:,\:\left(\mathrm{0},\:\mathrm{586}\right)\:,\:\left(\mathrm{12},\:−\mathrm{2}\right)\:,\:\left(\mathrm{586},\:\mathrm{0}\right) \\ $$
Commented by Rasheed.Sindhi last updated on 27/Dec/21
$$\mathrm{Also}\:\left(\mathrm{4},\mathrm{6}\right),\left(\mathrm{6},\mathrm{4}\right) \\ $$$$\mathbb{T}\mathrm{han}\Bbbk\:\mathrm{you}\:\mathrm{sir}!{But}\:{the}\:{process}? \\ $$
Answered by Rasheed.Sindhi last updated on 28/Dec/21
$${x}+{y}+{x}^{\mathrm{2}} {y}^{\mathrm{2}} =\mathrm{586} \\ $$$$\left({xy}\right)^{\mathrm{2}} =\mathrm{586}−\left({x}+{y}\right)\: \\ $$$$\Rightarrow\mathrm{586}−\left({x}+{y}\right)\:{is}\:{perfect}\:{square}. \\ $$$$\:\:\:\:\mathrm{586}−\left({x}+{y}\right)={a}^{\mathrm{2}} \\ $$$$\:\:\:\:{x}+{y}=\mathrm{586}−{a}^{\mathrm{2}} \\ $$$${For}\:{a}=\pm\mathrm{1},\pm\mathrm{2},\pm\mathrm{3},…\pm\mathrm{24} \\ $$$${x}+{y}=\mathrm{586}−\mathrm{1},\mathrm{586}−\mathrm{4},\mathrm{586}−\mathrm{9},…\mathrm{586}−\mathrm{576} \\ $$$$\begin{array}{|c|c|c|c|c|c|c|c|}{{a}}&\hline{{x}+{y}}&\hline{{x}^{\mathrm{2}} {y}^{\mathrm{2}} }&\hline{{xy}}&\hline{{z}^{\mathrm{2}} −\left({x}+{y}\right){z}+{xy}=\mathrm{0}}\\{\mathrm{0}}&\hline{\mathrm{586}}&\hline{\:\:\mathrm{0}}&\hline{\mathrm{0}}&\hline{{z}^{\mathrm{2}} −\mathrm{586}{z}=\mathrm{0}}\\{\pm\mathrm{1}}&\hline{\mathrm{585}}&\hline{\mathrm{1}}&\hline{\pm\mathrm{1}}&\hline{{z}^{\mathrm{2}} −\mathrm{585}{z}\pm\mathrm{1}=\mathrm{0}}\\{\pm\mathrm{2}}&\hline{\mathrm{582}}&\hline{\mathrm{4}}&\hline{\pm\mathrm{2}}&\hline{{z}^{\mathrm{2}} −\mathrm{582}{z}\pm\mathrm{2}=\mathrm{0}}\\{\pm\mathrm{3}}&\hline{\mathrm{577}}&\hline{\mathrm{9}}&\hline{\pm\mathrm{3}}&\hline{{z}^{\mathrm{2}} −\mathrm{577}{z}\pm\mathrm{3}=\mathrm{0}}\\{\pm\mathrm{4}}&\hline{\mathrm{570}}&\hline{\mathrm{16}}&\hline{\pm\mathrm{4}}&\hline{{z}^{\mathrm{2}} −\mathrm{570}\pm\mathrm{4}=\mathrm{0}}\\{…}&\hline{…}&\hline{}&\hline{}&\hline{}\\{\pm\mathrm{24}}&\hline{\mathrm{10}}&\hline{\mathrm{576}}&\hline{\pm\mathrm{24}}&\hline{{z}^{\mathrm{2}} −\mathrm{10}{z}\pm\mathrm{24}=\mathrm{0}}\\\hline\end{array} \\ $$$${Continue}… \\ $$