3. Determine polynomials $P$ and $Q$ such that for all real numbers $x$,

$$
Q\left(x^{2}\right)=(x+1)^{4}-x(P(x))^{2}
$$

The school - written part of the 1st round of category B takes place

## on Tuesday, January 23, 2001

so that it starts in the morning and the contestants have 4 hours of pure time to solve the problems. For each problem, a contestant can earn 6 points, and a successful solver is a student who scores 10 points or more. These details will be communicated to the students before the start of the competition.
3. Suppose that $P$ is a polynomial of degree $n$. The terms of the polynomial $Q\left(x^{2}\right)$ contain only even powers of $x$, and the polynomial $x \cdot P^{2}(x)$ is of odd degree $2 n+1$. Since it must hold that $Q\left(x^{2}\right)=(x+1)^{4}-x \cdot P^{2}(x)$, we see that $2 n+1$ cannot be greater than 4. Therefore, $n \leq 1$ and

$$
P(x)=a x+b, \quad Q\left(x^{2}\right)=(x+1)^{4}-x(a x+b)^{2} .
$$

![](https://cdn.mathpix.com/cropped/2024_04_17_fc67ddbe7f68431451b0g-2.jpg?height=523&width=368&top_left_y=1061&top_left_x=1472)

Fig. 1

After simplification, we get $Q\left(x^{2}\right)=x^{4}+\left(4-a^{2}\right) x^{3}+(6-2 a b) x^{2}+\left(4-b^{2}\right) x+1$. The coefficients of the odd powers of $x$ are zero, so $a, b \in\{-2,2\}$ and $Q\left(x^{2}\right)=x^{4}+(6-2 a b) x^{2}+1$. By substituting each of the four possible pairs of numbers $a, b$ into (1), we find all four solutions to the problem:

$$
\begin{aligned}
& P(x)=2 x+2 \text { and } Q(x)=x^{2}-2 x+1, \\
& P(x)=2 x-2 \text { and } Q(x)=x^{2}+14 x+1, \\
& P(x)=-2 x+2 \text { and } Q(x)=x^{2}+14 x+1, \\
& P(x)=-2 x-2 \text { and } Q(x)=x^{2}-2 x+1
\end{aligned}
$$