Problem 1080

[Proof] For $n \in N$, use induction to prove a more general conclusion: If an $n$-degree polynomial satisfies, when
$$k=n+2, n+3, \cdots, 2 n+2$$

then, $P(k)=a_{k}$, then $P(2 n+3)=a_{2 n+3}-1$. When $n=1$, we have $P(3)=2$, $P(4)=3$, thus $P(x) \equiv x-1$, and
$$P(5)=4=a_{5}-1 .$$

Now assume the conclusion holds for $n-1$. We will prove it also holds for $n$. Let the polynomial $P(x)$ have degree $n$, and when $k=n+2, k=n+3, \cdots, 2 n+2$, we have
$$P(k)=a_{k} .$$

Consider the polynomial
$$Q(x)=P(x+2)-P(x+1) .$$

Clearly, its degree is no more than $n-1$, because when $k=n+1, n+2, \cdots, 2 n$,
$$\begin{aligned} Q(k) & =P(k+2)-P(k+1) \\ & =a_{k+2}-a_{k+1} \\ & =a_{k} . \end{aligned}$$

So $Q(x)$ satisfies, when $k=n+1, n+2, \cdots, 2 n$, $Q(k)=a_{k}$. By the induction hypothesis, we have
$$Q(2 n+1)=a_{2 n+1}-1,$$

but
$$Q(2 n+1)=P(2 n+3)-P(2 n+2),$$

thus
$$\begin{aligned} P(2 n+3) & =P(2 n+2)+Q(2 n+1) \\ & =a_{2 n+2}+a_{2 n+1}-1 \\ & =a_{2 n+3}-1 \end{aligned}$$