3. Solve the equation $x!+y!+z!=u!$ in $\mathbb{N}$.

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Solution. Let the quadruple $(x, y, z, u)$ be a solution to the equation. Let $v=\max \{x, y, z\}$, then $1 \leq v2=2$ !. Similarly, the equation has no solution when $u=1$. Therefore, the only solution is $x=y=z=2, u=3$.