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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!

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$$a_{n+1}=a_n^3-a_n=(a_n-1)(a_n)(a_n+1)$$ 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 $$a_n\equiv\begin{cases}60\mod{1000}&n\equiv0\mod{2}\\-60\mod{1000}&n\equiv1\mod{2}\end{cases}$$

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