Find all non-constant polynomials $P(x)$ and $Q(x)$ with real coefficients satisfying the equality $P(Q(x))=P(x)Q(x)-P(x)$. (I. Voronovich)
Problem
Source: 2019 Belarusian National Olympiad 10.5
Tags: algebra, polynomial
RagvaloD
01.09.2019 12:02
Let $p,q$ are degrees of $P,Q$ then $pq=p+q \to (p-1)(q-1)=1,p=2,1=2$ $P(x)=ax^2+bx+c,Q(x)=kx^2+mx+n$ For coeficient near $x^4$ is $ak^2=ak \to k=1$ Let $x_1+x_2=-m$ and $Q(x_1) \neq 1$. Then $Q(x_1)=Q(x_2)$ and so $P(x_1)(Q(x_1)-1)=P(x_2)(Q(x_2)-1) \to P(x_1)=P(x_2) \to x_1+x_2=-\frac{b}{a} \to m=\frac{b}{a}$ So $P(x)=aQ(x)-an+c$ $aQ(x)^2+bQ(x)+c=(aQ(x)-an+c)(Q(x)-1)$ $(b+an-c+a)Q(x)+2c-an=0 \to n=\frac{2c}{a},b=c-an-a=-c-a$ $P(x)=ax^2-(a+c)x+c,Q(x)=x^2-\frac{a+c}{a}x+\frac{2c}{a}$ Let $c=da$ then Answer: $P(x)=a(x^2-(1+d)x+d),Q(x)=x^2-(1+d)x+2d$
NikoIsLife
01.09.2019 12:02
Let the degrees of $P$ and $Q$ be $p$ and $q$ respectively. We have
$$pq=p+q\implies(p-1)(q-1)=1\implies p=q=2$$This shows that $P$ and $Q$ are both quadratic functions. Let $P(x)=ax^2+bx+c$ and $Q(x)=dx^2+ex+f$.
Let $r$ be a (not necessarily real) root of $Q(x)-1=0$. Substituting $x=r$ gives
$$P(1)=0$$Therefore, $1$ is a root of $P$. We have $P(1)=a+b+c\implies c=-a-b$. This gives us:
$$P(x)=ax^2+bx-a-b=(x-1)(ax+a+b)$$We have
$$(Q(x)-1)(aQ(x)+a+b)=(x-1)(ax+a+b)(Q(x)-1)$$Dividing both sides by $(Q(x)-1)$,
$$aQ(x)+a+b=ax^2+bx-a-b$$$$Q(x)=x^2+\frac bax-\frac{2a+2b}a$$Therefore, all such solutions are:
$$\boxed{\begin{cases}P(x)=ax^2+bx-a-b\\
Q(x)=x^2+\frac bax-\frac{2a+2b}a\end{cases}}$$where $a$ and $b$ are any real numbers such that $a\ne0$.