Problem 2. Find all pairs $(P, Q)$ of polynomials with real coefficients such that

$$
\frac{P(x)}{Q(x)}-\frac{P(x+1)}{Q(x+1)}=\frac{1}{x(x+2)}
$$

for infinitely many $x \in \mathbb{R}$.

Nikolai Nikolov, Oleg Mushkarov
2. First solution. It suffices to consider the case when $P$ and $Q(\not \equiv 0)$ are relatively prime polynomials and the leading coefficient of $Q$ equals 1 . We have

$$
x(x+2)(P(x) Q(x+1)-Q(x) P(x+1))=Q(x) Q(x+1)
$$

for infinitely many $x$, i.e. for every $x$. Thus the polynomials $Q(x)$ and $Q(x+1)$ divide $x(x+2) Q(x+1)$ and $x(x+2) Q(x)$ respectively.

Therefore $S(x) Q(x)=x(x+2) Q(x+1)$ and $T(X) Q(x+1)=x(x+2) Q(x)$, where $S$ and $T$ are quadratic polynomials with leading coefficients 1 . Hence, $S(x) T(x)=x^{2}(x+2)^{2}$. There are three cases to be considered.

Case 1. $S(x)=T(x)=x(x+2)$. Then $Q(x+1)=Q(x)$, i.e. $Q \equiv 1$ and the condition of the problem shows that this is impossible.

Case 2. $S(x)=x^{2}$ and $T(x)=(x+2)^{2}$. Then $x Q(x)=(x+2) Q(x+1)$. Therefore $Q(1)=0$ and it follows by induction that $Q(n)=0$ for all $n \in \mathbb{N}$. Hence $Q \equiv 0$, a contradiction.

Case 3. $S(x)=(x+2)^{2}$ and $T(x)=x^{2}$. Then $(x+2) Q(x)=x Q(x+1)$ and therefore $x$ divides $Q(x)$ and $x+2$ divides $Q(x+1)$, i.e. $x+1$ divides $Q(x)$. It follows that $Q(x)=x(x+1) Q_{1}(x)$, where $Q_{1}$ has leading coefficient 1 and $Q_{1}(x+1)=Q(x)$. We conclude that $Q_{1}(x)=1$ and $Q(x)=x(x+1)$. Now plugging $Q(x)$ in (1) gives

$$
(x+2) P(x)-x P(x+1)=x+1
$$

Setting $x=0$ and $x=-1$ we obtain $P(0)=\frac{1}{2}$ and $P(-1)=-\frac{1}{2}$. Therefore $P(x)=\frac{1}{2}+x+x(x+1) P_{1}(x)$, where $P_{1}$ is a polynomial. Now (2) implies that $P_{1}(x+1)=P(x)$ and therefore $P_{1}$ is a constant.

We conclude that if the polynomials $P$ and $Q$ are relatively prime and $a_{0}=1$, then $Q(x)=x(x+1)$ and $P(x)=\frac{1}{2}+x+c x(x+1)$.

Therefore the answer is

$$
Q(x)=x(x+1) R(x) \text { and } P(x)=\left(\frac{1}{2}+x+c x(x+1)\right) R(x)
$$

where $R$ is an arbitrary nonzero polynomial and $c$ is a constant.

Second solution. The given identity can be written as

$$
\frac{P(x)}{Q(x)}-\frac{1}{2}\left(\frac{1}{x}+\frac{1}{x+1}\right)=\frac{P(x+1)}{Q(x+1)}-\frac{1}{2}\left(\frac{1}{x+1}+\frac{1}{x+2}\right)
$$

Hence it follows by induction that

$$
\frac{P(x)}{Q(x)}-\frac{1}{2}\left(\frac{1}{x}+\frac{1}{x+1}\right)=\frac{P(x+n)}{Q(x+n)}-\frac{1}{2}\left(\frac{1}{x+n}+\frac{1}{x+n+1}\right)
$$

Fixing $x$ and letting $n \rightarrow \infty$ we see that $\frac{P(x)}{Q(x)}-\frac{1}{2}\left(\frac{1}{x}+\frac{1}{x+1}\right)=c$, where $c$ is a constant. Now it is easy to conclude that $Q(x)=x(x+1) R(x)$ and $P(x)=\left(\frac{1}{2}+x+c x(x+1)\right) R(x)$.

Remark. One can prove in a similar manner the following assertion:

Let $P$ and $Q$ be polynomials with real coefficients such that

$$
\frac{P(x)}{Q(x)}-\frac{P(x+1)}{Q(x+1)}=\frac{1}{x(x+a)}
$$

for infinitely many $x \in \mathbb{R}$, where $a$ is a real number.

Then $a \in \mathbb{Z}$ and $a \neq 0$. Moreover let $n \in \mathbb{N}$,

$$
S_{n}(x)=\frac{1}{0!n}+\frac{x}{1!(n-1)}+\frac{x(x+1)}{2!(n-2)}+\cdots+\frac{x(x+1) \ldots(x+n-2)}{(n-1)!1}
$$

and $T_{n}(x)=x(x+1) \ldots(x+n-1)$. If $a=n$ then

$$
Q(x)=T_{n}(x) R(x) \text { и } P(x)=\left(S_{n}(x)+c T_{n}(x)\right) R(x)
$$

where $R$ is an arbitrary nonzero polynomial and $c$ is a constant. The case $a=-n$ is treated in the same way after substitutions $P_{1}(x)=-P(1-x)$ and $Q_{1}(x)=-Q(1-x)$.