the least value for $m\ge 2005$ such that $a_{m+1}-1\mid a^2_m-1$

Question

If we assume that $a_1=3$ and for every $n\ge1$ we assume that $a_{n+1}=(n+1)\cdot a_n-n$.How to find the least value for $m\ge 2005$ such that $a_{m+1}-1\mid a^2_m-1$

Answer 1

Note that $$ a_{n+1}-1=(n+1)(a_n-1) $$ so the question becomes:

Find the least value $m \geq 2005 $ such that $m+1 | a_m+1$

With a small amount of effort, the first formula can give you an explicit formula for $a_n$.