Problem 1329

Solution:

The only solutions are the polynomials $P(x)=1$ and $P(x)=x-c$, where $c$ is a constant.

Suppose that $P(x)=(x-c)\left(x-x_{2}\right) \cdots\left(x-x_{n}\right)$ is a nonconstant polynomial, where $c$ is its complex root whose real part is the smallest.

By the condition of the problem, $x+1-c$ divides $P(x)^2-1$, from which it follows that $P(c-1)= \pm 1$. On the other hand, for $2 \leqslant i \leqslant n$ we have $\left|c-1-x_{i}\right| \geqslant 1$, so $|P(c-1)|=\prod_{i=2}^{n}\left|c-1-x_{i}\right| \geqslant 1$, and this is only possible if $\left|c-1-x_{i}\right|=1$, i.e. $x_{i}=c$ for all $i$.

It follows that $P(x)=(x-c)^{n}$. However, $P(x+c)^2-1=x^{2 n}-1$ is not divisible by $P(x+c+1)=(x+1)^{n}$ if $n \geqslant 2$, so the only possibility is $n=1$.