Should $f^0(\mathbb{Q}) = \mathbb{Q}$ instead of $f^0(\mathbb{Q}) = 0$?

Yes, allow the author of this problem to vouch that indeed $f^0(\mathbb{Q}) = \mathbb{Q}$. There is another obvious typo: one should read $f^{n+1}(\mathbb{Q}) = f(f^n(\mathbb{Q}))$. These sets are nothing else but the images of the iterates of $f$. EDIT. Now, that I have moderator powers, I applied these corrections to the original statement.

gouthamphilomath wrote: Given a polynomial $f(x)$ with rational coefficients, with degree $d \ge 2,$ we define a sequence of sets $f^0(\mathbb{Q}), f^1(\mathbb{Q}), \cdots$ as $f^0(\mathbb{Q})=0, f^{n+1}(\mathbb{Q})=f(f^{n}(\mathbb{Q}))$ for $n\ge 0$. (Given a set $S,$ we write $f(S)$ for the set $\{f(x), x\in S\})$ Let $f^{\omega}(\mathbb{Q})=\bigcap_{n=0}^{\infty} f^n(\mathbb{Q})$ be the set of numbers that are in all of the sets $f^n(\mathbb{Q}).$ Prove that $f^{\omega}(\mathbb{Q})$ is a finite set. In this proof, $\nu_p(q)$, where $p$ is prime, will denote the negative of the $p$-adic norm of $q$. Furthermore, the numerator and denominator of a rational number will refer to the numerator and denominator of that rational number when it is written in reduced form. Finally, we define $S = \{f^m(q) : m \in \mathbb{Z}^+, \exists k : f^k(q) = q \}$. Lemma 1: There exists an integer $n$ such that for all primes $p$ for which there exists a rational number $q$ satisfying $\nu_p(q) \leq \nu_p(f^m(q)), -1$ for some integer $m$, $p^{-\nu_p(q)} | n$. Proof: I claim that if there exists a prime $p$ for which there exists a rational $q$ such that $\nu_p(q) < \nu_p(f^m(q))$, then $p$ must divide the product of the numerators of the nonzero coefficients of $f$. If $p$ does not divide this product, then trivial bounds on $\nu_p$ combined with $\deg f > 1$ yield $0 > \nu_p(q) > \nu_p(f(q))$. It thus can easily be shown by induction that $\nu_p(q) > \nu_p(f^m(q))$ for any $q$, a contradiction. Furthermore, if $\nu_p(q) < C$ for some sufficiently small $C$, then trivial bounds on $\nu_p$ combined with $\deg f > 1$ again yield $C > \nu_p(Q) > \nu_p(f(q))$. Again, it can be shown by induction that $\nu_p(q) > \nu_p(f^m(q))$, a contradiction, so $\nu_p(q) \geq C$. Therefore,we may let $n$ be the product of the numerators of the nonzero coefficients of $f$ raised to the power of $-C$. $\blacksquare$ Lemma 2: $S$ is finite. Proof: Let $A$ be the set of rationals for which we can find some $m$ such that $f^m(q) = q$.. By lemma 1, there exists an integer $n$ such that all rationals in $A$ can be expressed in the form $\frac{b}{n}$, for some integer $b$. $f$ is a nonlinear polynomial, so $|f(x)| > |x|$ when $|x|$ is sufficiently large, whence $|f^m(x)| > |x|$ for sufficiently large $|x|$. It follows that $A$ is bounded from above and below, whence $nA = \{na : a \in A\}$ is bounded from above and below. But $nA$ is a set of integers, so it must be finite. Hence, $A$ is finite. It follows from the finiteness of $\{ f^i(a) : i \in \mathbb{Z}^+ \}$ for all $a \in A$ that $S$ is finite as well. $\blacksquare$ Take some $a \in f^{\omega}(\mathbb{Q})$. Let $n$ be the constant described in lemma 1, and let $d$ be the denominator of $a$. By definition, for any $m$ we can find some $r$ for which $f^m(r) = a$. If $\nu_p(r) > \nu_p(a)$, then $-\nu_p(r) \leq \nu_p(d) \leq \nu_p(nd)$. If $\nu_p(r) \leq \nu_p(a) = \nu_p(f^m(q))$, them $-\nu_p(r) \leq \nu_p(n) \leq \nu_p(nd)$, so the denominator of $r$ must always divide $nd$. Letting $w = nd$, we find that any rational $r$ satisfying $f^m(r) = a$ can be written in the form $\frac{b}{nd}$ for some integer $b$. Again noting that $|f^m(r)| \geq |r| > |a|$ for sufficiently large $r$, we see that the set $R$ of all $r$ for which $f^m(r) = a$ for some $m$ is bounded from above and below. Hence, $ndR$ is bounded from above and below. However, since it is a set of integers, we conclude that $ndR$ is finite, whence $R$ is finite. By the infinite pigeonhole principle, there exists some $r \in R$ for which we can find two distinct integers $m_1$ and $m_2$ with $m_1 < m_2$ such that $f^{m_1}(r) = f^{m_2}(r) = a$. But then, $f^{m_1}(r) = f^{m_2 - m_1}(f^{m_1}(r))$, so $a \in S$. Since $S$ is finite, our proof is complete. $\blacksquare$

Could someone tell me which theory you require to learn for understanding this problem. In which branch of mathematics I am supposed to look? Thank you.

Let $f(x)=(a_dx^d+\cdots+a_0)/g$ for integers $a_d,\ldots,a_0,g$. If some rational number $s_0\in f^\omega(\mathbb{Q})$, then clearly we have an infinite sequence of integers $\{s_i\}_{i\ge0}$ satisfying $f(s_{i+1})=s_i$ for all nonnegative indices $i$ (since $f(x)=s_i$ has finitely many roots, this sequence can be defined inductively); also let $s_i=p_i/q_i$ for relatively prime $p_i,q_i$. First note that the sequence is clearly bounded in terms of $s_0$. Now we'll show that the $q_i$ are also bounded. Indeed, the rational root theorem on $f(s_1)=s_0$, or \[a_dq_0p_1^{d}+a_{d-1}q_0p_1^{d-1}q_1+\cdots+a_0q_0q_1^{d}=gp_0q_1^d,\]tells us that $q_1|a_dq_0$, so letting $l=a_dq_0/q_1$ and substituting into $f(s_2)=s_1$, we find that \[a_dq_0(a_dp_2^d+a_{d-1}p_2^{d-1}q_2+\cdots+a_0q_2^d)=gp_1lq_2^d,\]whence $d\ge2$ gives us (note that $\gcd(q_2,a_dq_0)|q_2$ on the RHS and $\gcd(q_2,a_dq_0)|a_dq_0$ on the LHS) \[q_2\gcd(q_2,a_dq_0) | a_d^2q_0 | a_d^2q_0^2.\](Taking modulo $q_2\gcd(q_2,a_dq_0)$, only the leading term on the LHS remains, and $q_2$ is relatively prime to $p_2$.) Thus \[\frac{q_2}{\gcd(q_2,a_dq_0)} \bigg{|} \left(\frac{a_dq_0}{\gcd(q_2,a_dq_0)}\right)^2\implies q_2=\gcd(q_2,a_dq_0)\implies q_2|a_dq_0.\](If $u|v^2$ and $\gcd(u,v)=1$, then $u=1$.) Continuing in this fashion (i.e. inducting), $q_i|a_dq_0$ for all $i$. Hence $\{a_dq_0s_i\}$ is a bounded sequence of integers and so the set of values that $\{s_i\}$ takes on is finite. By the pigeonhole principle, there exist indices $j<k$ with $s_j=s_k$, wherefore \[s_{k-j}=f^{j}(s_k)=f^{j}(s_j)=s_0\implies f^{k-j-t}(s_0)=f^{k-j-t}(s_{k-j})=s_t.\](With some trickier pigeonhole work, we can actually prove that $s_i$ must be periodic as a whole, but this is unnecessary here.) In particular, this means that for $0\le i\le k-j-1$, noting that $q_{k-j}=q_0$ and repeating the proof that $q_i|a_dq_0$ to show that $q_i|a_dq_1$, etc., we have \[q_i|a_d\gcd(q_0,q_1,\ldots,q_{k-j-1}).\]However, $f(s_{i+1})=s_i$, or \[a_dq_ip_{i+1}^{d}+a_{d-1}q_ip_{i+1}^{d-1}q_{i+1}+\cdots+a_0q_iq_{i+1}^{d}=gp_iq_{i+1}^d,\]implies from $d\ge2$ that for $0\le i\le k-j-1$, \[q_{i+1}\gcd(q_i,q_{i+1}) | a_dq_i | a_d^2\gcd(q_0,q_1,\ldots,q_{k-j-1}) | a_d^2\gcd(q_i,q_{i+1}).\]Then $i=k-j-1\implies q_0=q_{k-j}|a_d^2$. Yet for sufficiently large $|x|$, we have $|f(x)|>|x|$ since $d\ge2$, so because $f^{k-j}(s_0)=s_0$, $a_d^2s_0$ is a bounded integer. Therefore finitely many $s_0$ exist.

Zhero wrote: ... In this proof, $\nu_p(q)$, where $p$ is prime, will denote the negative of the $p$-adic norm of $q$. ... I think what you want to say is that $\nu_p(q)$ is the $p$-adic valuation. The $p$-adic norm has another definition (search on Google...). You shouldn't change the standard notations. This might confuse some readers.

Thanks for pointing that out. I've always thought that the norm referred to the negative of the exponent, not the prime raised to the negative of the exponent. This is what happens when I get too lazy to just define the p-adic valuation (which I wasn't certain was the standard name for $v_p$) myself. My apologies.

Zhero wrote: Thanks for pointing that out. I've always thought that the norm referred to the negative of the exponent, not the prime raised to the negative of the exponent. This is what happens when I get too lazy to just define the p-adic valuation (which I wasn't certain was the standard name for $v_p$) myself. My apologies. No problem It is important that you gave a correct solution to the problem.

Reduction to nearly imo 5 2006 a reformulation of the problem would be: If $X$ is a subset of $\mathbb{Q}$ with $f(X)=X$ then $X$ is finite. Let us assume that such an infinite set exists, Let $x$ be an element of $X$ first we will prove that there exists a positive integer $k$ such that $f^{(k)}(x)=x$ because $x$ is from $X$ for any integer $L$ there exist a finite number of rationals $a_{L_{1}}....a_L{_{K}}$ such that $f^{(L)}(a_{L_{K}})=x$, Now becaus of Konig's infinity lemma there exist $a_1.....a_K......$ such that $f(a_1)=x$,$f(a_2)=a_1$...$f(a_n)=a_{n-1}$... Lemma: The sequence $(a_n)_{n>0}$ becomes periodic. First the sequence is bounded, indeed if it wasn't there would exist $M$ such that $M>a_1$, $|f(X)|>|X|$ for every $X$ with $|X|>M$ which would clearely imply a contradiction.Now let us prove that there exist $N$ such that $Na_i$ is an integer for every i. let $kf(x) \in X[\mathbb{Z}]$ for some k, $kf(X)=f_nX^n+....f_0$ it is easy to see that we can chose $N=Kq$ where $a_1=\frac{p}{q}$ so the sequence $Na_i$ is bounded and integer valued and it's terms cand be found in a set with m elements, now in the sequence $a_i$ there exist a m+1-tuple $(a_i,.....,a_{i+m})$ which repets infinitely many times therefore there exists a pattern wich repeats infinitely many times...... so the sequence is periodic in particular there exists $f^{(r)}(a_1)=a_1$ by the lemma it follows that there exists a positive integer $k$ such that $f^{(k)}(x)=x$ Now we have arrived at IMO 5 2006 with $Q$ instead of $Z$, which can probably be proved in a similar way with imo 5.

So I noticed this was basically a carbon copy of a WOOT Practice Olympiad problem, so here is my solution. It is fairly similar to Zhero's solution though. It is obvious the problem is showing there are finitely many rational sequences $q_1,q_2,...$ satisfying $q_i = f(q_{i+1})$ for all $i \ge 1$. I will show that only finitely many values can appear in sequences, which would obviously imply the problem. First we put some work into characterizing the sequences. Fix $q_1$ for now. Lemma 1: There exists a sufficiently large integer $N$ such that for all primes $q > N$, we have $v_q(q_n) \ge 0$ for all $n$. Proof: We have that if $v_q(q_n) < 0$ for big enough $q$ we have $v_q(q_1) < 0$ because for $q$ sufficiently large and $v_q(a) < 0$ we have $v_q(f(a)) = \deg f \cdot v_q(a)$. The result follows. $\square$ Lemma 2: For every prime $q$ there exists a non-negative integer $k$ such that $v_q(q_n) \ge -k$ for all $n$ unless all terms in the sequence are $0$. Proof : The proof of this is obvious when expanding $f$ out and looking at the $\frac{a^d}{q^{dz}} \cdot a_d$ term where $d = \deg f$ because then one can arrive at $v_q(q_i) < v_q(q_{i+1})$ because for $z$ large enough the $v_q$ of that term is clearly less than all the others so it determines $v_q$ of the whole expression, so then if you assume the contrary we have $v_q(q_i)$ is not bounded below from which it follows $v_q(q_i) = - \infty$ for all $i$. The only way this is possible is if $q_i = 0$ for all $i$ so we are done. $\square$ Lemma 3: There exists real numbers $A,B$ such that for any $n$ if $q_n = \frac{a}{b}$ we have $|a| < A, |b| < B$. Proof: Lemmas 1 and 2 combined show a $B$ exists. Now use the fact that if $|x|$ is big enough we have $|f(x)| > 2|x|$ when $\deg P > 1$ to show an $A$ exists because if an $A$ did not exist then $q_1$ would be unbounded, clearly false. $\square$ Now it follows only finitely many values of $q_i$ are possible. By pigeonhole we have there exists some rational number $a = q_{a_1} = q_{a_2} = ...$ for infinitely many $a_i$. Let $z$ be an integer such that $f^z(a) = a$ which obviously exists. Now I claim $q_{n+z} = q_n$. Take an $a_i > n+z$. Then $q_{n+z} = f^{a_i - n - z}(a)$ and $q_n = f^{a_i - n}(a)$. But as $a$'s orbit when applying $f$ has order dividing $z$ it follows $q_{n+z} = q_n$ so it follows the $q_i$ are periodic. But then this means $q_1$ falls into the finite interval of real numbers $x$ such that $|f(x)|$ does not become unbounded upon constant iterations of $f$. As the denominator is bounded as shown above (note that lemma 1, lemma 2 now give effective bounds on the denominator of $q_1$ rather than on $q_n$ for a fixed $q_1$), it immediately follows only finitely many values of $q_1$ are possible so we are done.

Suppose that $f(x)=\frac{c_nx^n+c_{n-1}x^{n-1}+\cdots+c_1x+c_0}N$ for integers $c_n\ne0,c_{n-1},\ldots,c_0,N\ne0$. First of all, note that we can consider the equivalent mapping $g(x)=kf\left(\frac xk\right)$ for any $k$, as $f\left(\frac xk\right)=\frac{g(x)}{k}$. Basically, we look at $kf^{\omega}(\mathbb Q)$ instead of $f^{\omega}(\mathbb Q)$. For some suitable choice of $k$, say $k=c_n$, we can make it so that $c_n=1$, i.e. the numerator is monic. We first claim that for every $x\in f^{\omega}(\mathbb Q)$, there exists a positive integer $m$ for which $f^m(x)=x$. If this is not true, then there is some infinite sequence $a_0,a_1,a_2,\ldots$ of rational numbers such that $f(a_k)=a_{k-1}$ for all $k\in\mathbb N$ and each rational number $a$ appears a finite number of times in the sequence. Otherwise, the sequence between every two occurrences of $a$ is uniquely determined to be $a,f(a),f^2(a),\ldots,f^m(a)=a$ in reverse, each of which satisfies $f^m(x)=x$. Furthermore, as $\deg f>1$, there exists $M$ such that $|x|>M\implies|f(x)|>|x|$. This means that if $a$ is the first term of the sequence with $|a|>M$, then the term before it satisfies $|f(a)|>|a|>M$, a contradiction. Hence, the sequence is bounded. Now, suppose that $qa_0\in\mathbb Z$. We claim that $qa_i\in\mathbb Z$ for all $i\in N$. Indeed, note that a rational root of $\frac{x^n+\cdots+c_0}N-\frac pq$, or equivalently $qx^n+\cdots+c_0-pN$ must have denominator dividing $q$ by the rational root theorem. But the sequence is bounded, so there are only finitely many values that can appear in it, contradicting the assumption that every number appears in it a finite amount of times. Thus, we must have that $f^m(x)=x$ for some $m$ for all $x\in f^{\omega}(\mathbb Q)$. Clearly, $|x|<M$, or otherwise $|x|<|f(x)|<|f^2(x)|<\cdots$. Now, suppose that $x\in f^{\omega}({\mathbb Q})$ is not an integer. If $v_p(x)<0$ for some prime $p$, then $v_p(f(x))=nv_p(x)-v_p(N)$, as $v_p(x^n)<v_p(c_{n-1}x^{n-1})<\cdots<v_p(c_0)$. But this means that $v_p(f(x))<v_p$ as $n>1$, so $v_p(x)>v_p(f(x))>v_p(f^2(x))>\cdots$, a contradiction. Hence, $f^{\omega}(\mathbb Q)$ is a subset of a set of bounded integers, which is finite, as desired.

Let $f(x)=\frac{a_0}{b_0}x^n+\dots +\frac{a_{n+1}}{b_{n+1}}$, where $a_i,b_i\in \mathbb{Z}$. Suppose all the primes that take part in the prime factorizations of integers $a_0, b_1,b_2,\dots,b_{n+1}$ are $p_1,p_2,\dots, p_k$ with the max powers correspondingly $\alpha_1,\alpha_2,\dots,\alpha_k$. A simple observation: Suppose we take some rational $x=a/b$ and a prime $p\,,\, p^{\alpha}\mid b$ , where $p$ either is distinct from $p_1,p_2,\dots,p_k$ or $p$ is equal to some $p_i$ , but $\alpha>\alpha_i$. Then the denominator of $f(x)$ is multiple of $p^{\alpha+1}$. Now, suppose $c_0/d_0 \in f^{\omega}(\mathbb{Q})$. It means, there exists a sequence $\frac{c_i}{d_i}\,,\,i=0,1,\dots$ , for which $f\left(\frac{c_{i+1}}{d_{i+1}}\right)=\frac{c_i}{d_i}$. Then according to the above observation, for all sufficiently large indices $i$ , the prime factorization of $d_i$ satisfies: $$(1)\,\,\,\,\,\,\,\,\,\, d_i=p_1^{\beta_1}p_2^{\beta_2}\cdots p_k^{\beta_k}\,,\,\beta_i\leq \alpha_i\,,\,i=1,2,\dots,k.$$On the other hand, since the degree of $f$ is at least $2$, there exists a positive constant $M$, such that $|f(x)|>|x|$ whenever $|x|\geq M$. It means all $c_i/d_i$ are inside $[-M,M]$. But inside that interval there are only finite number of rationals with denominators like in $(1)$. Denote that finite set of rationals by $R$. Since there are infinitely many $c_i/d_i$ , that hit rationals among $R$, that sequence should form a cycle inside $R$ , that's $c_0/d_0\in R$ and the result follows.

Let us define the sequence ${a_i}$ as $f^i(a_i) = a_0$ for fixed $a_0$. Then, we must have $a_n-a_{n+1} \mid P(a_n)-P(a_{n+1}) = a_{n-1}-a_n$. But now, note that is the denominator in simplest form of $a_0$ is $q$. Then the denominator of $a_i$ can be atmost $qa$ where a is the leading coefficient of the polynomial. Thus the denominator is lower bounded. But now $|a_i-a_{i+1}|$ is also bounded. So, it must eventually be constant. We take $2$ cases, i.e. it switches signs or it does not switch signs, and in either case we can see there are eventually a finite number of solutions for such $a_i$. Thus, the set $f^{\omega}( \mathbb{Q})$ is finite.

Let $S=f^\omega(\mathbb{Q})$ for brevity. Suppose that $f(x)=\frac{a_n}{b_n}x^n+\cdots+\frac{a_1}{b_1}x+\frac{a_0}{b_0}$ where $a_i \perp b_i$ for all $i$. If a prime $p$ doesn't divide $a_n(b_n\ldots b_0)$, then I claim that $t \in S \implies \nu_p(t) \geq 0$. If $\nu_p(x)\geq 0$ then $\nu_p(f(x)) \geq 0$. On the other hand, if $\nu_p(x)<0$ then $\nu_p(f(x))=d\nu_p(x)$, so it we have some $(x,y)$ such that $\nu_p(y)<0$ and $y=f^n(x)$ then $\nu_p(y)\leq d^n$, which implies the desired result. Consider an arbitrary $q_0 \in S$, and construct a sequence $\ldots,q_{-1},q_0,q_1,\ldots$ such that $q_i=f(q_{i+1})$ for all $i$, hence $\{\ldots,q_{-1},q_0,q_1,\ldots\} \subseteq S$ and $\nu_q(q_i) \geq 0$ for all $q \nmid a_n(b_n\ldots b_0)$. For each prime $p \mid a_n(b_n\ldots b_0)$, observe that $\nu_p(x)<-(\nu_p(a_n)+\max_i \nu_p(b_i)):=-T_p$ implies that $\nu_p(f(x))<\nu_p(x)$, since the leading $\frac{a_n}{b_n}x^n$ term will have the unique minimal $\nu_p$ amongst the terms of $f(x)$. Thus if $\nu_p(q_0)<-T_p$ then there exists some $N_p$ such that for all $i>N_p$ we have $\nu_p(q_i) \geq -T_p$. Since there exist finitely many such $p$, for all $i>\max_p N_p$ we can write $q_i=\frac{r_i}{\prod_p p^{T_p}}$ where $r_i \in \mathbb{Z}$. But now since $\deg f\geq 2$, there exists some $M>0$ such that every $x$ with $|f(x)|<|x|+1$ lies in the interval $[M,-M]$. If $x \not \in [M,-M]$ then $|f^n(x)| \geq |x|+n$. Suppose that $q_0 \not \in [M,-M]$ (in our above sequence), so $q_{-1},q_{-2},\ldots \not \in [M,-M]$ as well. Then there exists some $N$ such that for all $i>N$, $q_i \in [M,-M]$. But all such $q_i$ are also in $S$, hence there are finitely many of them by the above result. Hence there exist $i \neq j$ such that $q_i=q_j \in [M,-M]$, so the sequence is periodic, and thus some $k>0$ has $q_{-k}=q_i \in [M,-M]$: contradiction. Thus $q_0$ must lie in $[M,-M]$; since its denominator is bounded, it follows that there are finitely many choices for $q_0$, i.e. $S$ is finite. $\blacksquare$

We will use (without proof) the following lemma: Let $f\in\mathbb{Z}[x]$ be a polynomial of degree $d$. Then there is a constant $C$ dependent on $f$ s.t. $$H(f(q))\geq C H^d(q)$$for all rational numbers $q$, where $H\left(\frac{a}{b}\right)=\max\{|a|,|b|\}$ and $\gcd(a,b)=1$. With this in mind, consider $q_0\in f^{\omega}(\mathbb{Q})$, and a sequence of rationals s.t. $q_0=f(q_1)=\dots =f^{(n)}(q_n)=\dots$. Then using the lemma, $$H(q_0)=H(f^{(n)}(q_n))\geq C^{1+d+\dots +d^{n-1}}H^{d^n}(q_n)\implies H^{\frac{1}{d^n}}(q_0)\geq C^{\frac{1}{d}+\dots +\frac{1}{d^n}}H(q_n).$$Therefore, the sequence $H(q_n)$ is bounded, and so must be the sequence $q_n$. Thus, there is a rational number appearing infinitely many times (say, $q$). We thus have $$q_0=f^{(i_1)}(q)=f^{(i_2)}(q)=\dots.$$Substituting $q_0$ and writing $j_l=i_{l+1}-i_l$ we get $$q_0=f^{(j_1)}(q_0)=f^{(j_2)}(q_0)=\dots.$$With the same argument as above, $H(q_0)\geq C^{\frac{d^n-1}{d-1}}H^{d^n}(q_0)$, which implies $H(q_0)\leq C^{-\frac{1}{d-1}}$, so $q_0$ may take finitely many values. As $q_0$ was randomly selected from $f^{\omega}(\mathbb{Q})$, we have proven that $|f^{\omega}(\mathbb{Q})|$ is finite.

Silly problem with silly solution. Let $F = f^{\omega}({\mathbb Q})$. Define the height of a rational $r \ne 0$ as \[ h(r) = \min_p (\nu_p(r)) \]over all primes $p$. We define $h(0) = +\infty$. Fix a polynomial $f(x) = \frac{a_n}{b_n} x^n + \frac{a_{n-1}}{b_{n-1}} x^{n-1} + \dots + \frac{a_0}{b_0}$ where $a_n \ne 0$ and $\gcd(a_i, b_i) = 1$ unless $a_i = 0, b_i = 1$. Claim: For each prime $p$, there exists an integer $k_p$ if $\nu_p(r) < k_p$ then $\nu_p(f(r)) \le \nu_p(r)$. Furthermore, $k_p = 0$ if $p$ is coprime to all nonzero numerators. Proof. Take $k_p$ such that $\nu_p(a_nx^n)$ is smaller than all the other terms. If $p$ is coprime to all nonzero numerators, then this is just true. $\blacksquare$ Let $N = \prod_p p^{k_p}$ Define $V \subset {\mathbb Q}$ as the set of rationals such that $Nr$ is not an integer. This means that if $r \in V$, then $h(f(r)) < r$. Claim: There exists $R$ such that if $|x| > R$ then $|f(r)| > |r|$. Proof. Follows asymptotically. $\blacksquare$ Claim: If $r \in V$, then $r$ is not in $F$. Proof. For a rational $q$, define $t(q)$ as the minimum $k$ such that $f^k(q) \in V$, and define it as $-1$ if no such $k$ exists. Now, we claim that \[ n > |h(r)| + 2 + \max_{|x| \le \max\{|R|, |r|\}} t(x). \]works. As such, it follows that [(i)]If $q \in V$, then $h(f^n(q)) < (h(r) - 1) < h(r)$ so $f^n(q) \ne r$. If $|q| \ge \max\{|R|, |r|\}$, then $|f^n(q)| > |q| > |r|$ similarly. If $|q| \le \max\{|R|, |r|\}$ and $q \not\in V$, then consider $t(q)$. If $t(q) = -1$, then $f^n(q) \not\in V$ and is thus not $r$. Else, $h(f^n(q)) < (h(r) - 1) + h(f^{t(q)}(q)) < h(r)$ and thus $f^n(q) \ne r$ again. $\blacksquare$ Claim: If $|r| > R$ then $r$ is not in $F$. Proof. Define $t(q)$ as the minimum $k$ such that $f^k(q) = r$, elsewise $-1$. Then repeat the same as above as $q \in V, |q| > |r|$ rule themselves out. Then the remaining finite set $[-|r|, |r|] \cap (\mathbb{Q} \setminus V)$ is finite and gets ruled out by taking $n$ larger than all $t(q)$. $\blacksquare$ As such, since there are finitely many values in $[-R, R] \cap (\mathbb{Q} \setminus V)$ it necessarily follows that $F$ is finite.