Prove that if $\exists m\in \mathbb{Z}^+$ for which $a_m=0\implies a_1$ or $a_2=0.$

Question

Let $f(x)$ be a polynomial with integer coefficients. Define a sequence of integers $\{a_n\}_{n\in\mathbb{N}_0}$ such that $a_0=0$ and $a_{n+1}=f(a_n).$ Prove that if $\exists m\ne 0$ such that $a_m=0\implies a_1$ or $a_2=0.$

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$$ f(x)=b_l x^2+b_{l-1} x^{l-1}+\cdots+b_1 x+b_0 $$ If $a_1=b_0=0,$ then $a_n=0\;\forall n \geqslant 2.$ Now, say $\exists m$ such that $a_m=0.$

We see \begin{align} a_{n+1}-a_n=\; b_l&\left[a_n^l-a_{n-1}^l\right]+\\ b_{l-1}&\left[a_n^{l-1}-a_{n-1}^{l-1}\right]+\cdots+\\ b_1&\left[a_n-a_{n-1}\right]. \end{align} So, \begin{align} a_2-a_1 &\mid a_3-a_2 \\ a_3-a_2 &\mid a_4-a_3 \\ &\,\vdots \\ a_m-a_{m-1} &\mid a_{m+1}-a_m \\ \implies a_2-a_1 &\mid a_{m+1}-a_m \\ \implies a_2-b_0 &\mid a_{m+1}\left[\because a_m=0, a_1=b_0\right] \\ a_{m+1}=f\left(a_m\right)\;&\!=f(0)=b_0 \\ a_2-b_0 &\mid b_0 \\ b_0 &\mid a_2. \end{align} I feel like I'm close to solving it, but I have to show $a_2=0$ somehow. Can someone please provide a possible continuation to my solution?

Answer 1

Let us put $c_n=a_{n+1}-a_n$. Then, as you've observed, $c_n \mid c_{n+1}$. At the same time, as you've also observed, $c_m=c_0$, and actually since $a_{m+t}=a_t$ for any $t$, sequence $c_n$ is also periodic with period $m$.

Therefore $|c_0| \leqslant |c_1| \leqslant \ldots \leqslant |c_m|=|c_0|$, so actually we have equality everywhere, and $c_n = \pm c_0$ for any $n$.

If $c_1=-c_0$, then $a_2-a_1=a_0-a_1$, so $a_2=0$. Suppose $c_1=c_0$. If all $c_n$ are equal, then $a_n$ is an arithmetic sequence. But $a_0=a_m=0$, so it has to be identically zero in this case.

If they are not all equal, then at some point we have $c_n=-c_{n+1}$. But this means $a_{n+1}-a_n = a_{n+1}-a_{n+2}$, so $a_n = a_{n+2}$. This situation can be ruled out easily (it means $a_{n+t}$ is periodic with period $2$, so $m \mid 2$), hence we are done.