12-42 Find all positive integers $w, x, y, z$ that satisfy the equation $w!=x!+y!+z!$.
(15th Canadian Mathematics Competition, 1983)
[Solution 1] Without loss of generality, let $w > x \geqslant y \geqslant z$.
It is clear that $w \geqslant x+1$, thus we have

that is
$$
\begin{array}{c}
(x+1)! \leqslant w! = x! + y! + z! \leqslant 3 x!, \\
(x+1)! \leqslant 3 x!, \\
x + 1 \leqslant 3, \\
x \leqslant 2 .
\end{array}
$$
(1) If $x=2, y=2, z=2$, then
$$
w! = 2! + 2! + 2! = 6 = 3!,
$$

so
$$
w = 3.
$$
(2) If $x=2, y=2, z=1$, then
$$
w! = 2! + 2! + 1! = 5.
$$

Since 5 is not equal to the factorial of any positive integer, there is no solution in this case.
Similarly, it can be shown that there are no solutions when $x=2, y=1, z=1$ and when $x=1, y=1, z=1$.
Therefore, the only solution is
$$
x=2, \quad y=2, \quad z=2, \quad w=3.
$$
[Solution 2] Without loss of generality, let $w > x \geqslant y \geqslant z$.
If $y > z$, then dividing both sides of the equation by $z!$, we get
$$
\begin{aligned}
& w(w-1) \cdots(z+1) \\
= & x(x-1) \cdots(z+1) + y(y-1) \cdots(z+1) + 1.
\end{aligned}
$$

Since $z+1 > 1$ divides the left side of the equation but does not divide the right side, the equation clearly does not hold.
Therefore, it must be that
$$
y = z.
$$

If $x > y = z$, then dividing both sides of the equation by $y! = z!$, we get
$$
w(w-1) \cdots(z+1) = x(x-1) \cdots(z+1) + 2.
$$

Then $z+1$ must be a divisor of 2, so $z+1 = 2$.
Further, dividing by $z+1 = 2$, we get
$$
w(w-1) \cdots 3 = x(x-1) \cdots 3 + 1.
$$

This is clearly not possible, so it must be that

that is
$$
x = y = z.
$$
$$
w! = 3 x!
$$

So
$$
w = 3, \quad x = y = z = 2.
$$