Let the number $(22{)}^{2022}+(2022{)}^{22}$ leave the remainder $\mathrm{\alpha }$ when divided by 3 and $\mathrm{\beta }$ when divided by 7 . Then $\left({\mathrm{\alpha }}^2+{\mathrm{\beta }}^2\right)$ is equal to

$$\begin{gathered} (1+21{)}^{2022}+(2022{)}^{22} \\ \downarrow \downarrow \\ \begin{array}{c}3\mathrm{\lambda }+1\text{ divisible by '3' }\\ \therefore \mathrm{\alpha }=1\end{array} \\ \begin{array}{l}(1+21{)}^{2022}+(2023-1{)}^{22}\\ =(1+7\mathrm{\lambda })+(7\mathrm{\lambda }+1)\\ \therefore \mathrm{\beta }=2\\ {\mathrm{\alpha }}^2+{\mathrm{\beta }}^2=1+2^2=5\end{array} \end{gathered}$$