In the recursive series:
- $a_{0}=60$
- $a_{n+1}=a_{n}^3-a_{n}$
I observe that:
- $n\equiv0\pmod2 \implies a_{n}\equiv+60\pmod{1000}$
- $n\equiv1\pmod2 \implies a_{n}\equiv-60\pmod{1000}$
How can I tackle and prove this observation?
I guess I might need to begin by converting this series from recursive to straightforward, but I'm not quite sure how to do that. Or is there a different way to approach this problem?
Thank you!
$$a_{n+1}=a_n^3-a_n=(a_n-1)(a_n)(a_n+1)$$ So if $a_k\equiv 60$ for some $k\in\mathbb{N}$ then $$a_{k+1}\equiv 59\cdot60\cdot61\equiv-60 \mod{1000}$$ $$a_{k+2}\equiv (-61)\cdot(-60)\cdot(-59)\equiv60 \mod{1000}$$ Hence as $a_0\equiv60\mod{1000}$ and $a_k\equiv a_{k+2} \text{ mod }1000$ we must have $$a_n\equiv\begin{cases}60\mod{1000}&n\equiv0\mod{2}\\-60\mod{1000}&n\equiv1\mod{2}\end{cases}$$