Let's find the polynomial $p(x)$ of the lowest degree that satisfies the following conditions:
a) the coefficients of $p(x)$ are integers,
b) $p(x)$ can be factored into the product of linear polynomials,
c) the roots of $p(x)$ are integers,
d) $p(0)=-1$,
e) $p(3)=128$.
Solution. Let the degree of the sought polynomial be $n$. From condition b), it follows that the polynomial has $n$ roots. Then the polynomial can be written in the following form:

$$
p(x)=a\left(x-x_{1}\right) \cdot\left(x-x_{2}\right) \cdot \ldots \cdot\left(x-x_{n}\right)
$$

where $a$ is the leading coefficient of the polynomial $p(x)$ and $x_{1}, x_{2}, \ldots, x_{n}$ are the roots of the polynomial. According to condition d),

$$
\begin{gathered}
-1=p(0)=a\left(0-x_{1}\right) \cdot\left(0-x_{2}\right) \cdot \ldots \cdot\left(0-x_{n}\right)=(-1)^{n} \cdot a \cdot x_{1} \cdot x_{2} \cdot \ldots \cdot x_{n} \\
\text { therefore } \quad(-1)^{n+1}=a \cdot x_{1} \cdot x_{2} \cdot \ldots \cdot x_{n}
\end{gathered}
$$

Since the factors on the right side are integers according to a) and c), the absolute value of each factor is 1. Thus, the roots of the polynomial can only be $a+1$ and $a-1$, and $a$ can also be +1 or -1. Let the polynomial have a root of $(+1)$ with multiplicity $l$ and a root of $(-1)$ with multiplicity $k$, where $k+l=n$. Then

$$
p(x)=a(x-1)^{l} \cdot(x+1)^{k}
$$

thus, according to e),

$$
128=p(3)=a \cdot 2^{l} \cdot 4^{k}
$$

Since the product $2^{l} \cdot 4^{k}$ is positive, the value of $a$ must be +1. Therefore,

$$
128=2^{7}=2^{l} \cdot 4^{k}=2^{l+2 k}
$$

from which

$$
7=l+2 k=(l+k)+k=n+k
$$

Since $l$ and $k$ are natural numbers,

$$
7 \geqq 2 k
$$

that is, $3 \geqq k$:

Thus, $k$ can be at most 3, so $n$ must be at least 4. Therefore, the polynomial of the lowest degree $p(x)$ that satisfies all 5 conditions is a fourth-degree polynomial:

$$
p(x)=(x-1)(x+1)^{3}=x^{4}+2 x^{3}-2 x-1
$$

Ujváry-Menyhárt Zoltán (Bp., Fazekas M. Gyak. Gimn., II. o. t.)

Note: It can be seen that the conditions of the problem are satisfied by as many polynomials as there are non-negative integer solutions to the equation

$$
7=l+2 k
$$

This has three possibilities: $l=1, k=3 ; l=3, k=2$ and $l=5, k=1$.

The corresponding polynomials are:

$$
\begin{aligned}
& p_{1}(x)=(x-1)(x+1)^{3} \\
& p_{2}(x)=(x-1)^{3}(x+1)^{2} \\
& p_{3}(x)=(x-1)^{5}(x+1)
\end{aligned}
$$