Mock AIME 2 2006-2007 Problems/Problem 2

Problem

The set $S$ consists of all integers from $1$ to $2007$ , inclusive. For how many elements $n$ in $S$ is $f(n) = \frac{2n^3+n^2-n-2}{n^2-1}$ an integer?

Solution

$f(n) = \frac{2n^3+n^2-n-2}{n^2-1} = \frac{(n - 1)(2n^2 + 3n + 2)}{(n - 1)(n + 1)} = \frac{2n^2 + 3n + 2}{n + 1} = 2n + 1 + \frac1{n+1}$ . So in fact, there are 0 such elements of $S$ .

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Mock AIME 2 2006-2007

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Mock AIME 2 2006-2007 Problems/Problem 2

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