$5 \cdot 28$ Give a set $M$, which consists of 7 consecutive natural numbers, and there exists a 5th-degree polynomial $P(x)$, such that: (1) each coefficient of the polynomial $P(x)$ is an integer;
(2) there are 5 numbers $k$ in $M$, including the largest and smallest numbers in $M$, that satisfy $P(k) = k$;
(3) there exists $k \in M$ such that $P(k) = 0$.
[Solution] Let the 7 consecutive numbers be $a, a+1, a+2, \cdots, a+6$. Construct
$$
P(x)=x+(x-a)(x-a-6)\left(x-b_{1}\right)\left(x_{2}-b_{2}\right)\left(x_{2}-b_{3}\right)
$$

where $\quad b_{1}, b_{2}, b_{3} \in\{a+1, a+2, \cdots, a+5\}$.
Satisfy $P(k)=0$ for $k \in\{a+1, a+2, \cdots, a+5\}$.
If $k=a+1$, then by (1)
$$
a+1+1 \cdot(-5) \cdot c_{1} \cdot c_{2} \cdot c_{3}=0 \text {, where } c_{1}, c_{2}, c_{3} \text { are from }-1,-2,-3 \text {, }
$$
-4.
Let $c_{1}=-1, c_{2}=-2, c_{3}=-3$.
Then $a+1+1 \cdot(-5) \cdot(-1) \cdot(-2) \cdot(-3)=0, a<0$, contradiction.
Similarly, $\left\{c_{1}, c_{2}, c_{3}\right\} \subset\{-1,-2,-3,-4\}$ all lead to $a<0$, contradiction.
If $k=a+2$, then by (1)
$a+2+2 \cdot(-4) \cdot c_{1} \cdot c_{2} \cdot c_{3}=0$, where
$$
\left\{c_{1}, c_{2}, c_{3}\right\} \subset\{1,-1,-2,-3\} \text {. }
$$

Take $c_{1}=1, c_{2}=-1, c_{3}=-2$,
then $a+2+2 \cdot(-4) \cdot 1 \cdot(-1) \cdot(-2)=0$.
$a=14, 7$ numbers are $14,15,16,17,18,19,20$.

Thus, $P(x)=x+(x-14)(x-20)(x-15)(x-17)(x-18)$
satisfies $\quad P(16)=16+2 \cdot(-4) \cdot 1 \cdot(-1) \cdot(-2)=0$,
$P(14)=14, P(20)=20, P(15)=15$,
$P(17)=17, P(18)=18$