\[\begin{array}{l}
{\text{Решите в натуральных числах уравнение:}} \hfill \\
{{\text{5}}^{k!}} + {3^{k!}} + 1 = {m^3}. \hfill \\
\end{array}\]
\[\begin{array}{l}
k = 1 \Rightarrow 9 = {m^3},{\text{ }}\emptyset \hfill \\
k = 2 \Rightarrow 35 = {m^3},{\text{ }}\emptyset \hfill \\
k = 3 \Rightarrow 16355 = {m^3},{\text{ }}\emptyset \hfill \\
{\text{Пусть }}k > 3. \hfill \\
{{\text{5}}^{k!}} = {5^{1 \cdot 2 \cdot 3 \cdot 4 \cdot ... \cdot k}} = {\left( {{5^{2 \cdot 3}}} \right)^{4 \cdot 5 \cdot ... \cdot k}} \equiv 1\left( {\bmod 7} \right) \hfill \\
{3^{k!}} = {\left( {{3^{2 \cdot 3}}} \right)^{4 \cdot 5 \cdot ... \cdot k}} \equiv 1\left( {\bmod 7} \right) \hfill \\
\left. \begin{array}{l}
{{\text{5}}^{k!}} + {3^{k!}} + 1 \equiv 3\left( {\bmod 7} \right) \hfill \\
{m^3}\cancel{ \equiv }3\left( {\bmod 7} \right) \hfill \\
\end{array} \right\} \Rightarrow {{\text{5}}^{k!}} + {3^{k!}} + 1 \ne {m^3}. \hfill \\
\end{array}\]
\[\emptyset \]