# Proving a modular character of a given recursive series

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 $$a_n\equiv\begin{cases}60\mod{1000}&n\equiv0\mod{2}\\-60\mod{1000}&n\equiv1\mod{2}\end{cases}$$