IMO Shortlist 1990 problem 26
Dodao/la:
arhiva2. travnja 2012. Let
![p(x)](/media/m/3/9/3/393afccb4b82415d2114a3ff957b444f.png)
be a cubic polynomial with rational coefficients.
![q_1](/media/m/6/7/c/67c86efc63a51050f6ef0e84a91cec5f.png)
,
![q_2](/media/m/6/e/6/6e6f9ae8f824a18aa7d1970e0029987b.png)
,
![q_3](/media/m/9/2/0/920b9c864b86045a4b2a9458e49015d2.png)
, ... is a sequence of rationals such that
![q_n = p(q_{n + 1})](/media/m/a/7/1/a715bb6d835e988eccec7f7d980fd248.png)
for all positive
![n](/media/m/a/e/5/ae594d7d1e46f4b979494cf8a815232b.png)
. Show that for some
![k](/media/m/f/1/3/f135be660b73381aa6bec048f0f79afc.png)
, we have
![q_{n + k} = q_n](/media/m/2/5/d/25d41a107fc9d849e546f85e8c9a0cb5.png)
for all positive
![n](/media/m/a/e/5/ae594d7d1e46f4b979494cf8a815232b.png)
.
%V0
Let $p(x)$ be a cubic polynomial with rational coefficients. $q_1$, $q_2$, $q_3$, ... is a sequence of rationals such that $q_n = p(q_{n + 1})$ for all positive $n$. Show that for some $k$, we have $q_{n + k} = q_n$ for all positive $n$.
Izvor: Međunarodna matematička olimpijada, shortlist 1990