6.3 .9 Find all integer pairs $(a, b)$ such that there exists an integer-coefficient polynomial $p(x)$, satisfying: the product $\left(x^{2}+a x+b\right) p(x)$ has the form: $x^{n}+c_{n-1} x^{n-1}+\cdots+c_{0}$, where each $c_{i}$ is either 1 or -1.
Let $f(x)=x^{n}+c_{n-1} x^{n-1}+\cdots+c_{0}$. Suppose $z$ is a complex root of $f(x)$. If $|z|>1$, then
$$
\begin{aligned}
\left|z^{n}\right| & =\left|-c_{n-1} z^{n-1}-\cdots-c_{0}\right| \leqslant|z|^{n-1}+\cdots+1=\frac{|z|^{n}-1}{|z|-1} \\
& \Rightarrow|z|^{n}(|z|-2) \leqslant-1 \Rightarrow|z|<2,
\end{aligned}
$$

Therefore, the modulus of every complex root of $f(x)$ is less than 2.
Notice that $b \mid c_{0}$, so $b= \pm 1$. Let the two complex roots of $x^{2}+a x+b$ be $x_{1}, x_{2}$, then $\left|x_{1} x_{2}\right|=$ 1. Without loss of generality, assume $\left|x_{1}\right| \geqslant\left|x_{2}\right|$, then $\left|x_{2}\right| \leqslant 1$. Combining with $\left|x_{1}\right|<2$, we have
$$
|a|=\left|x_{1}+x_{2}\right| \leqslant\left|x_{1}\right|+\left|x_{2}\right|<2+1=3 .
$$

Thus, $a \in\{0, \pm 1, \pm 2\}$.
When $a= \pm 1$, take $p(x)=1$; when $a=0$, take $p(x)-x+1$; when $a= \pm 2$, the polynomials $x^{2} \pm 2 x-1$ both have a real root with absolute value greater than 2, which does not meet the requirement. For $x^{2} \pm 2 x+1$, take $p(x)=x \mp 1$ respectively.
The pairs $(a, b)$ that satisfy the conditions are $( \pm 1, \pm 1)$, $(0, \pm 1)$, and $( \pm 2,1)$.