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 $a_{m+1}-1\mid a^2_m-1$
the least value for $m\ge 2005$ such that $a_{m+1}-1\mid a^2_m-1$
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number-theory
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0What have you tried until now? Your question doesn't show that you put any effort into solving it. – 2012-09-21
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0With some small numbers examples or some effort the OP can probably answer this him/herself. – 2012-10-11