1. Provide an example of a polynomial of the lowest possible positive degree that, when divided by both the polynomial $x^{2}-4$ and the polynomial $x^{3}-8 x+8$, gives a remainder of 1.
1. Let $P$ be the sought polynomial. The results of its division by both given polynomials mean that for suitable polynomials $A$ and $B$, the following equalities hold:

$$
P(x)=A(x)\left(x^{2}-4\right)+1 \quad \text { and } \quad P(x)=B(x)\left(x^{3}-8 x+8\right)+1 .
$$

Here, $A$ and $B$ are not zero polynomials, since $P$ must have a positive degree, which is at least 3.

By comparing the right-hand sides, we obtain the polynomial equality

$$
A(x)\left(x^{2}-4\right)=B(x)\left(x^{3}-8 x+8\right) .
$$

Among the roots of the polynomial on the left side, the numbers $x=2$ and $x=-2$ are certainly included, of which only the first is a root of the cubic factor $x^{3}-8 x+8$ on the right side (this can be easily verified by substitution). Therefore, $x=-2$ must be a root of the polynomial $B$, which thus has a degree of at least one, so the sought polynomial $P$ has a degree of at least 4. Moreover, $P$ will have a degree of four precisely when the corresponding $B$ is of the form $B(x)=a(x+2)$ for a suitable number $a \neq 0$. (The corresponding $A(x)=a\left(x^{3}-8 x+8\right) /(x-2)$ does not need to be calculated, but it is provided: $A(x)=a\left(x^{2}+2 x-4\right)$.)

All polynomials $P$ satisfying the conditions of the problem are of the form

$$
P(x)=\underbrace{a(x+2)}_{B(x)}\left(x^{3}-8 x+8\right)+1=a\left(x^{4}+2 x^{3}-8 x^{2}-8 x+16\right)+1 .
$$

(Verification is not necessary in the given process; for a complete solution, it suffices to provide one example, such as choosing $a=1$.)

For a complete solution, award 6 points. For creating the necessary equality, award 2 points. For proving that the sought polynomial has a degree of at least 4, award 3 points. For completing the solution, award 1 point.