
Problem 9.3. Find all polynomials $f(x)=x^{n}+a_{n-1} x^{n-1}+\cdots+$ $a_{1} x+a_{0}, a_{0} \neq 0$ with integer coefficients such that $f\left(a_{i}\right)=0, i=$ $0, \ldots, n-1$.

Sava Grozdev


Solution: It is clear that $n>1$. Since $f(x)=\left(x-a_{0}\right)(x-$ $\left.a_{1}\right) \ldots\left(x-a_{n-1}\right)$, it follows from $f(0)=a_{0}=(-1)^{n} a_{0} \ldots a_{n-1}$ that $\left|a_{i}\right|=1$ for $i=1,2, \ldots, n-1$.

First case: $\left|a_{0}\right|=1$. Now $f(x)=(x-1)^{p}(x+1)^{q}, p+q=n>1$. We know that $(x-1)^{p}(x+1)^{q}=\left(x^{p}-p x^{p-1}+\ldots\right)\left(x^{q}+q x^{q-1}+\ldots\right)$ and comparing the coefficients in front of $x^{n-1}$ and $x^{n-2}$ we obtain

$$
\begin{aligned}
& q-p=a_{n-1}= \pm 1 \\
& \frac{q(q-1)}{2}+\frac{p(p-1)}{2}-p q=a_{n-2}= \pm 1
\end{aligned}
$$

It is easy to find now that $p+q=3$ and so $p=1, q=2$ or $p=2, q=1$. In the first case we get that $f(x)=x^{3}+x^{2}-x-1$, which is a solution and in the second one $f(x)=x^{3}-x^{2}-x+1$, which is not.

Second case: $\left|a_{0}\right| \geq 2$. Now $0=f\left(a_{0}\right)=\mid a_{0}^{n}+a_{n-1} a_{0}^{n-1}+\cdots+$ $a_{1} a_{0}+\left.a_{0}|\geq| a_{0}\right|^{n}-\left|a_{0}\right|^{n-1}-\cdots\left|a_{0}\right|^{2}-\left|a_{0}\right|-\left|a_{0}\right|=\frac{\left|a_{0}\right|\left(\left|a_{0}\right|-2\right)\left(\left|a_{0}\right|^{n-1}-1\right)}{\left|a_{0}\right|-1} \geq$ 0 . Therefore $\left|a_{0}\right|=2$. Moreover $a_{n-1} a_{0}^{n-1}, \ldots, a_{1} a_{0}, a_{0}$ have the same negativity-the opposite to those of $a_{0}^{n}$.

We conclude that $a_{0}=-2, n$ is an even number and $a_{i}=(-1)^{i+1}$ for $i=1,2, \ldots, n-1$. If $n>2$, then $a_{2}=-1$ and $0=f(-1)=$ $(-1)^{n}+(-1)^{n}(-1)^{n-1}+\cdots+(-1)^{2}(-1)-2=-n-2 \neq 0$, which is impossible. Therefore $n=2$ and $f(x)=x^{2}+x-2$, which is a solution.

Answer:

$$
f(x)=x^{2}+x-2 \quad \text { and } \quad f(x)=x^{3}+x^{2}-x-1
$$
