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For each positive integer k > 1, define the sequence ${a_n}$ by $a_0 = 1$ and $a_n = kn + (−1)^na_{n−1}$ for each $n\ge 1$. Determine all values of k for which 2000 is a term of the sequence.

This is from BMO 1 2000, the values of k I got were (87,2001,667,3)

Are these all the solutions? how can I go about finding the solutions by induction in the simplest way possible?

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    Sorry I posted this before I finished (by accident) I have edited in the question. – 2017-02-05
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    I edited this to use MathJax. Please check to see if I did it correctly. – 2017-02-05

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The solution is $k \in \{2001, 667, 87, 23, 3\}$

We have

$a_{4n} = 4nk+1$, $a_{4n+1} = k - 1$, $a_{4n+2} = (4n+3)k - 1$, and $a_{4n+3} = 1$ which we can prove by induction.

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    Thanks! I think that it basically boils down to finding factors of 2001 which are congruent to 3 (mod 4) right? – 2017-02-05