2006 IMO Problems/Problem 5

Problem

(Dan Schwarz, Romania) Let $P(x)$ be a polynomial of degree $n>1$ with integer coefficients, and let $k$ be a positive integer. Consider the polynomial $Q(x) = P( P ( \ldots P(P(x)) \ldots ))$ , where $P$ occurs $k$ times. Prove that there are at most $n$ integers $t$ such that $Q(t)=t$ .

Solution

We use the notation $P^k(x)$ for $Q(x)$ .

Lemma 1. The problem statement holds for $k=2$ .

Proof. Suppose that $a_1, \dotsc, a_k$ , $b_1, \dotsc, b_k$ are integers such that $P(a_j) = b_j$ and $P(b_j) = a_j$ for all indices $j$ . Let the set $\{ a_1, \dotsc, a_k, b_1, \dotsc, b_k \}$ have $m$ distinct elements. It suffices to show that $\deg (P) \ge m$ .

If $a_j = b_j$ for all indices $j$ , then the polynomial $P(x)-x$ has at least $m$ roots; since $P$ is not linear, it follows that $\deg P \ge m$ by the division algorithm.

Suppose on the other hand that $a_i \neq b_i$ , for some index $i$ . In this case, we claim that $a_j + b_j$ is constant for every index $j$ . Indeed, we note that $a_j - a_i \mid P(a_j) - P(a_i) = b_j - b_i \mid P(b_j) - P(b_i) = a_j - a_i ,$ so $\lvert a_j - a_i \rvert = \lvert b_j - b_i \rvert$ . Similarly, $a_j - b_i \mid P(a_j) - P(b_i) = b_j - a_i \mid P(b_j) - P(a_i) = a_j - b_i,$ so $\lvert a_j - b_i \rvert = \lvert b_j - a_i \rvert$ . It follows that $a_j + b_j = a_i + b_i$ .

This proves our claim. It follows that the polynomial $P(x) - \left( a_i + b_i - x \right)$ has at least $m$ roots. Since $P$ is not linear it follows again that $\deg P \ge m$ , as desired. Thus the lemma is proven. $\blacksquare$

Lemma 2. If $a$ is a positive integer such that $P^r(a)$ for some positive integer $r$ , then $P^2(a) = a$ .

Proof. Let us denote $a_0 = a$ , and $a_j = P^j(a)$ , for positive integers $j$ . Then $a_0 = a_r$ , and $\begin{align*} a_1 - a_0 &\mid P(a_1)- P(a_0) = a_2-a_1 \\ &\mid P(a_2) - P(a_1) = a_3 - a_2 \\ &\vdots \\ &\mid P(a_r)- P(a_{r-1}) = a_1 - a_0 . \end{align*}$ It follows that $\lvert a_{j+1} - a_j \rvert$ is constant for all indices $j$ ; let us abbreviate this quantity $d$ . Now, since $(a_1-a_0) + (a_2-a_1) + \dotsb + (a_r-a_{r-1}) = a_r-a_0 = 0,$ it follows that for some index $j$ , $a_j - a_{j+1} = -(a_{j+1} - a_{j+2}),$ or $a_j = a_{j+2} = P^2(a_j)$ . Since $a = a_r = P^{r-j}(a_j)$ , it then follows that $P^2(a) = a$ , as desired. $\blacksquare$

Now, if there are more than $n$ integers $t$ for which $Q(t) = t$ , then by Lemma 2, there are more than $n$ integers $t$ such that $P^2(t) = t$ , which is a contradiction by Lemma 1. Thus the problem is solved. $\blacksquare$

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

Hints

Try to use contradiction: $Q(x_1)=x_1,...,Q(x_{n+1})=x_{n+1}$ . WLOG suppose $x_1<...<x_{n+1}$ .

Think of using the theorem $a-b \mid f(a)-f(b)$ where $f(x)$ is a polynomial over $Z$ .

Try to prove that $P(x_1),...,P(x_{n+1})$ is either increasing or decreasing.

Construct a polynomial to give the final contradiction by keeping in mind the theorem: If the equation $f(x)=a$ ( $a$ is a constant) has more roots than $deg(f)$ , then $f(x)=a$ .

Solution 2

Note: We will denote $P(...(P(x))....)$ (where $P$ occurs $k$ times) as $P_k(x)$ throughout the proof, for any integer $k>1$ .

For the sake of contradiction, let the problem statement be false. For convinience's sake, let the stated equation thus have $n+1$ solutions: $P_k(x_1)$ = $x_1$ ,..., $P_k(x_{n+1})$ = $x_{n+1}$ .

Without loss of generality, suppose $x_1 < ... <x_{n+1}$ .

As $P(x)$ is defined over the integers, $x_i - x_j \mid P(x_i) - P(x_j)$ for all distinct $i,j$ . Also, $P(x_i) - P(x_j) \mid P_2(x_i) - P_2(x_j) \mid ... \mid P_k(x_i) - P_k(x_j) = x_i - x_j.$ Hence, $|P(x_i) - P(x_j)| = x_i - x_j$ for all $i>j$ .

Claim.

$P(x_i) \neq P(x_j)$ for any $i \neq j$ .

Proof.

For the sake of contradiction, suppose the claim statement is wrong.

Then, for some distinct $(i,j)$ , $P(x_i)=P(x_j) =>|P(x_i)-P(x_j)|=0=> x_i=x_j$ , a contradiction.

Hence, our claim is proved.

Claim.

$P(x_1),...,P(x_{n+1})$ is either increasing or decreasing.

Proof.

We will only prove the decreasing part, as the increasing part is similar.

For the sake of contradiction, suppose the claim statement is wrong. Then, $P(x_{i-1})>P(x_i)$ and $P(x_i)<P(x_{i+1})$ for some integer $i>1$ (that is, $P(x_1)>...>P(x_{i-1})>P(x_i)<P(x_{i+1})$ ).

For convinience's sake, suppose $i=2$ .

Then, $|P(x_1)-P(x_2)|=P(x_1)-P(x_2)=x_2-x_1$ and $|P(x_3)-P(x_2)|=P(x_3)-P(x_2)=x_3-x_2$ .

Adding, we get that $(P(x_1)+P(x_3))-2 \cdot P(x_2) = x_3 - x_1$ .

Now, $x_3 - x_1$ will either be substituted by $P(x_1)-P(x_3)$ or $P(x_3)-P(x_1)$ .

Either way, we get a contradiction to our previous claim.

Hence, $P(x_2)>P(x_3)$ . Similarly, $P(x_3)>P(x_4)$ ,..., $P(x_n)>P(x_{n+1})$ .

Then, $P(x_1)>...>P(x_{n+1})$ .

One may similarly prove the increasing part.

Hence, our claim is proved.

Case 1, The sequence $P(x_1),...,P(x_{n+1})$ is increasing:

Let $T(x)=(P(x)-x)-(P(x_1)-x_1)$ .

Then, $T(x)$ has $n+1$ roots: $x_1,...,x_{n+1}$ from our previous claim, while $deg(T)=deg(P)=n$ , a contradiction.

One may similarly show the contradiction if the sequence $P(x_1),...,P(x_{n+1})$ is decreasing, by just replacing the $x_1$ with $x_{n+1}$ in $T(x)$ .

Hence, we get our desired contradiction, and thus we are done.

Resources

1974 USAMO Problems/Problem 1, which implies a special case of this problem

2006 IMO (Problems) • Resources
Preceded by Problem 4	1 • 2 • 3 • 4 • 5 • 6	Followed by Problem 6
All IMO Problems and Solutions

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