TITLE: True or false : $\gcd(a_m, a_n) = a_{\gcd(m, n)}$?
QUESTION [12 upvotes]: Define a sequence $\left(a_0, a_1, a_2, \ldots\right)$ of integers by $a_0= 0, a_{n+1}=P(a_n)$, where $P$ is a polynomial with POSITIVE integer coefficients.
I guess that $\gcd(a_m, a_n) = a_{\gcd(m, n)} $.

REPLY [7 votes]: Hint $ $ Mimic the $\overbrace{\text{(subtractive) Euclidean algorithm}}^{\rm \Large\qquad\!  (n,m)\ =\ (n,\,m-n)}\,$ with inductive step being
$$\rm \begin{align}\rm (\overbrace{P^n(0),\,P^m(0)}^{\Large (a(n),\,\ a(m))\ })\, &=\,\rm (\color{#0a0}{P^n(0)},\,P^{m-n}(\color{#c00}{P^n(0)})\\
\rm  &=\, \rm (\underbrace{P^n(0),\,P^{m-n}(\color{#c00}0))}_{\Large (a(n),\,\ a(m-n))}\end{align}\qquad\qquad\qquad $$
via Polynomial Congruence Rule: $\ \ \ \rm\begin{align}\rm\bmod\, \color{#0a0}{P^n(0)}\!:\,\  &\rm\color{#c00}{P^n(0)\,\equiv\, 0}\\\,\rm\Rightarrow\ \ \ P^{m-n}(&\rm\color{#c00}{P^n(0)})\equiv P^{m-n}(\color{#c00}0)\end{align}$ 
As does the Congruence Rule, the proof works for any polynomial  with integer coefficients.