4. Find the reduced residue system modulo $m=7^{2} \cdot 11^{2}$ in the form of equation (25) and equation (30) of §3.
$$\text { 4. } \begin{array}{l}
-17 \cdot 121 \cdot 3^{\gamma^{(1)}}+42 \cdot 49 \cdot 7^{(2)}, 0 \leqslant \gamma^{(1)} \leqslant 41,0 \leqslant \gamma^{(2)} \leqslant 109 . \\
\widetilde{g}_{1}=-17 \cdot 121 \cdot 3+42 \cdot 49=-4113 \\
\widetilde{g}_{2}=-17 \cdot 121+42 \cdot 49 \cdot 7=12349 .
\end{array}$$

Since $7^{2} \cdot 11^{2}=5929$, we can take $\widetilde{g}_{1}=1816, \widetilde{g}_{2}=491$.
$$1816^{r^{(1)}} \cdot 491^{r^{(2)}}, \quad 0 \leqslant \gamma^{(1)} \leqslant 41, \quad 0 \leqslant \gamma^{(2)} \leqslant 109$$