Find all integers $k$, such that there exists an integer sequence ${\{a_n\}}$ satisfies two conditions below (1) For all positive integers $n$,$a_{n+1}={a_n}^3+ka_n+1$ (2) $|a_n| \leq M$ holds for some real $M$
Problem
Source: 2024 CWMO P2
Tags: Sequence, algebra
CooperHqh
07.08.2024 12:03
Easy to find the answer is k=1,-1,-2,-4. Solution is also easy to find but this problem has no beauty.
sami1618
11.08.2024 01:20
Answer: $k=\boxed{-4,-2,-1,1}$ It is easy to show that $\{a_n\}$ is eventually periodic. It is also well known that the minimum period of repeatedly applying an integer polynomial is either $1$ or $2$. Thus it suffices to find all integers $k$ for which the following equation has an integer solution $(a,b)$. $$b=a^3+ka+1$$$$a=b^3+kb+1$$Adding the two equations together yields $$(a+b)=(a+b)(a^2+b^2-ab+k)+2$$Thus $(a+b)|2$ and is non zero so we split into cases.
Case 1: $a+b=2$
Substituting into the first equation gives $1=a(a^2+k+1)$ so $a,b=\pm 1$. The only pair that sums to $2$ is $(1,1)$ which works at $k=-1$.
Case 2: $a+b=1$
Substituting into the first equation gives $0=a(1+k+a^2)$. If one of $a,b$ is $0$ and the other $1$ we have $k=-2$ working. Otherwise $a^2=b^2=-(1+k)$ but then $a+b=1$ would make $a$ and $b$ not integral.
Case 3: $a+b=-1$
Substituting into the first equation gives $-2=a(a^2+k+1)$ so we must have $a,b=2,1,-1,-2$. The only pair that sums to $-1$ is $(1,-2)$ which works for $k=-4$.
Case 4: $a+b=-2$
Substituting into the first equation gives $-3=a(a^2+k+1)$ so we must have $a,b=3,1,-1,-3$ the pairs that sum to $-2$ are $(-1,-1)$ and $(1,-3)$ which first works at $k=1$ and second fails.