Problem 1002

Solution. Let $P(x)=a_{n} x^{n}+a_{n-1} x^{n-1}+\ldots+a_{1} x+a_{0}$. We observe that

$$\frac{P(a)-P(b)}{a-b}=\frac{a_{n}\left(a^{n}-b^{n}\right)+a_{n-1}\left(a^{n-1}-b^{n-1}\right)+\ldots+a_{1}(a-b)}{a-b}$$

and this is an integer because the coefficients $a_{k}$ are integers and $a^{k}-b^{k}$ is divisible by $a-b$ for $k=1,2,3, \ldots, n$. For the same reasons, $\frac{P(b)-P(c)}{b-c}$ and $\frac{P(c)-P(a)}{c-a}$ are also integers.

Assume for the polynomial $P$ that $P(a)=b, P(b)=c$, and $P(c)=a$, where $a, b$, and $c$ are distinct integers. From the previous discussion, we have that $\frac{b-c}{a-b}, \frac{c-a}{b-c}$, and $\frac{a-b}{c-a}$ are integers. Since their product is equal to 1, each of them must be equal to 1 or -1. If the first number is equal to -1, we get $a=c$, which contradicts the assumption. For the same reasons, the other two numbers cannot be equal to -1. Therefore, all three numbers are equal to 1. From this, we get

$$\begin{aligned} & b-c=a-b \\ & c-a=b-c \\ & a-b=c-a \end{aligned}$$

Subtracting the second equation from the first, we get $b=c$, which contradicts the assumption. Therefore, there is no polynomial with the desired properties.