Problem 1001

Proof: First, we prove that if $a$ can be computed, then $a \pm 1$ can also be computed. In fact, let $\boldsymbol{x}=a, \boldsymbol{y}=1$, then we can compute $a+1$. Let $u=a, y=-1$, then we can compute $-a-1$, thus we can compute $-a$. Then, let $u=-a$ and $y=-1$, we can compute $a-1$.

We use induction to prove the proposition. Clearly, let $u=v=1$, and we know that 1 can be computed, so by the proven result, for any integer $a_{0}, a$, they can be computed, i.e., the conclusion holds for $n=0$. Assume the conclusion holds for $x=k$. When $k=k+1$, for any $-k+1$-degree polynomial with integer coefficients
$$\begin{aligned} P_{k+1}(x) & =a_{0} x^{k+1}+a_{1} x^{2}+\cdots+a_{k} x \\ & +a_{k+1} \end{aligned}$$

Let $Q_{k}(x)=a_{0} x^{k}+a_{1} x^{2-1}+\cdots+a_{k-1} x + a_{k}-1$.

By the assumption, $Q_{k}(c)$ can be computed. Let $u=Q_{k}(c)$, $\boldsymbol{y}=c$, we get
$$\begin{aligned} Q_{k}(c) \cdot c + c & =a_{0} \cdot c^{k+1}+a_{1} \cdot c^{k}+\cdots \\ & +a_{k} \cdot c \end{aligned}$$

which can be computed. By the proven result, we have
$$P_{k+1}(c)=\left[Q_{k}(c) \cdot c + c\right] + a_{k+1}$$

which can also be computed. This proves that the conclusion holds for $n=k+1$. Therefore, for any $n$-degree polynomial with integer coefficients $P_{n}(x)$, $P_{n}(c)$ can be computed.
(Supplied by the Nankai University Mathematical Olympiad Research Group)