4. 170 Find all positive integers $w, x, y, z$ that satisfy $w!=x!+y!+z!$.

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［Solution］Assume without loss of generality that $w>x \geqslant y \geqslant z$.
If $y>z$, then dividing both sides of the equation by $z!$ yields
$$
\begin{aligned}
& w(w-1) \cdots(z+1) \\
= & x(x-1) \cdots(z+1)+y(y-1) \cdots(z+1)+1 .
\end{aligned}
$$

Clearly, $z+1$ divides the left side of the equation, but does not divide the right side. This contradiction implies that we must have $y=z$.
If $x>y=z$, then dividing both sides of the equation by $z!$ yields
$$
w(w-1) \cdots(z+1)=x(x-1) \cdots(z+1)+2 \text {, }
$$

Thus, $z+1$ divides 2, so $z+1=2$. Therefore, the equation simplifies to
$$
w(w-1) \cdots 3=x(x-1) \cdots 3+1 \text {, }
$$

which is impossible. Therefore, we must have $x=y=z$. Thus, we obtain
$$
w!=3 z! \text {, }
$$

which implies $w=3, x=y=z=2$.