Problem 1186

4. Since the problem involves the values of the function $f(x)=a^{x}-p(x)$ at $n+2$ consecutive integers $x=0,1, \cdots, n+1$, we hope to estimate $f(x)$ using the difference formula to solve this problem. In fact, by definition, we have
$$\begin{array}{l} \Delta a^{x}=a^{x+1}-a^{x}=(a-1) a^{x}, \\ \Delta^{2} a^{x}=(a-1) \Delta a^{x}=(a-1)^{2} a^{x}, \\ \cdots \\ \Delta^{n-1} a^{x}=(a-1)^{n+1} a^{x} . \end{array}$$

Thus, $\left.\Delta^{n-1} a^{x}\right|_{x=0}=(a-1)^{n+1}$. Therefore, by the theorem, we have
$$\Delta^{n+1} f(0)=\sum_{k=0}^{n-1}(-1)^{n+1-k} C_{n+1}^{k}\left[a^{k}-p(k)\right] .$$

Also, $\Delta^{n+1} f(0)=\left.\Delta^{n+1} a^{x}\right|_{x=0}-\Delta^{n-1} p(0)=(a-1)^{n+1}$, so
$$(a-1)^{n+1}=\sum_{k=0}^{n+1}(-1)^{n-1-k} C_{n+1}^{k}\left[a^{k}-p(k)\right] .$$

Assume for contradiction that $\max _{0 \leqslant i \leqslant m-1}\left\{\left|a^{i}-p(i)\right|\right\}<1$, i.e., for all $i=0,1, \cdots, n+1,\left|a^{i}-p(i)\right|<1$. Thus, by $a \geqslant 3$, we get $2^{n+1} \leqslant(a-1)^{n+1}<\sum_{k=0}^{n+1} C_{n+1}^{*}=2^{n-1}$. This is a contradiction. Therefore, $\max _{0 \leqslant i \leqslant n+1}\left|a^{i}-p(i)\right| \geqslant 1$.