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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 $a_{m+1}-1\mid a^2_m-1$

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What have you tried until now? Your question doesn't show that you put any effort into solving it. –  Beni Bogosel Sep 21 '12 at 14:58
    
With some small numbers examples or some effort the OP can probably answer this him/herself. –  mick Oct 11 '12 at 15:47
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1 Answer

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 $a_n$.

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Was this a bad answer? Why was it downvoted? –  Beni Bogosel Sep 22 '12 at 11:07
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