Let $f(x)$ be a polynomial with rational coefficients, and let $\alpha$ be a real number. If \[\alpha^3-2019\alpha=(f(\alpha))^3-2019f(\alpha)=2021,\]prove that $(f^n(\alpha))^3-2019f^n(\alpha)=2021$ for any positive integer $n$. (Here, we define $f^n(x)=\underbrace{f(f(f\cdots f}_{n\text{ times}}(x)\cdots ))$.)
Problem
Source: 2021HKTST2 Q2
Tags: algebra, polynomial, roots
mathaddiction
31.10.2020 14:12
Let $Q(x)=x^3-2019x-2021$, it is easy to check that this polynomial has no rational root. Now if $f(\alpha)=\alpha$ then it is obvious, suppose $f(\alpha)\neq \alpha$. Now notice that $$Q(f(x))-Q(x)=(f(x)-x)(f(x)^2+xf(x)+x^2-2019)\qquad(1)$$Let $f(x)=P(x)Q(x)+R(x)$ for some $deg R\leq 2$. Then from $Q(\alpha)=0$ we have $f(\alpha)=R(\alpha)$ Define $$g(x)=R(x)^2+xR(x)+x^2-2019$$Then $g(\alpha)=0$. Let $$g(x)=P_1(x)Q(x)+R_1(x)$$CLAIM. $R_1\equiv 0$ Proof. Notice that $g(\alpha)=Q(\alpha)=0$ and hence $R_1(\alpha)=0$. Suppose $R_1$ is not zero. Let $$Q(x)=P_2(x)R_1(x)+R_2(x)$$Then $R_2(\alpha)=0$ and $\deg(R_2)=1$, hence $R_2\equiv 0$ since $\alpha$ is irrational. Therefore $$Q(x)=P_2(x)R_1(x)$$Now notice that $1\leq degR_1\leq 2$, hence one of the polynomials $P_2,R_1$ is linear, so $\alpha$ is rational, contradiction. $\blacksquare$ Now $$g(x)=P_1(x)Q(x)\qquad(2)$$From $(1),(2)$ we can see that if $Q(f^n(x))=0$ then $f(f^n(x))=R(f^n(x))$ and $g(f^n(x))=0$, hence $$Q(f(f^n(x))-Q(f^n(x))=0$$, this completes the proof.
grupyorum
31.10.2020 18:59
Fairly similar to this 1992 Shortlist problem.
Mr.C
31.10.2020 19:20
$a^3-2019a=f(a)^3-2019f(a)=2021$. clearly if $f(a)=a$ we are done. let the roots of $p(x)=x^3-2019x-2021$ be $x_1,x_2,x_3$. WLOG assume $a=x_1,f(x_1)=x_2$. now $p(x)-p(y)=(x-y)(x^2+y^2+xy-2019)$ so clearly again $f(a)^2+af(a)+a^2=2019$. as $x^3-2019x-2021$ also has the root $a$ we have $x^3-2019x-2021|f(x)^2+xf(x)+x^2-2019$ setting $x=x_2$ we get $f(x_2)^2+x_2f(x_2)+x_2^2=2019$. again the equation $y^2+y x_2+x_2^2=2019$ has roots $x_1,x_3$ . if $f(x_2)=x_1$ we reach a cycle and we are done. if $f(x_2)=x_3$ then again by above we need $f(x_3) \in \{x_1,x_2\}$ which gives us another cycle. . . nice problem. i am not sure how hard honk kong Tst actually is. but judging by china Tst problems this is way too easy .
mathaddiction
01.11.2020 02:07
Mr.C wrote: nice problem. Agree Mr.C wrote: i am not sure how hard honk kong Tst actually is. but judging by china Tst problems this is way too easy . This is just the second test, afterwards there are still six more tests namely the CHKMO, APMO and four selection tests consisting of IMO shortlists. Meanwhile, this test also aims at being an introduction to MO for new trainees, so it is easier. Indeed, the contestants in China are much much stronger than those in Hong Kong, so their tests are much harder indeed.