# 2001 Pan African MO Problems/Problem 1

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## Problem

Find all positive integers $n$ such that: $$\dfrac{n^3+3}{n^2+7}$$ is a positive integer.

## Solution

Perform polynomial long division to get $\frac{n^3 + 3}{n^2 + 7} = n + \frac{3-7n}{n^2 + 7}$. Note that if $n^2 + 7 > |3-7n|$, then $\frac{3-7n}{n^2 + 7}$ can not be an integer. Thus, all of the solutions satisfy the inequality $|3-7n| \ge n^2 + 7$.

If $3-7n \ge 0$, then $n \le \frac73$. However, there are no positive integers in this case. If $3-7n < 0$, then $n > \frac37$ and $7n-3 \ge n^2 + 7$. Rearranging the second inequality results in $0 \ge n^2 - 7n + 10$. Factoring results in $0 \ge (n-5)(n-2)$, so $2 \le n \le 5$.

Now there are only four possible positive integers, so we can use trial and error to determine if $\frac{n^3 + 3}{n^2 + 7}$ is a positive integer. After doing trial and error, the only positive integers that make $\frac{n^3 + 3}{n^2 + 7}$ an integer are $\boxed{n = 2}$ or $\boxed{n = 5}$.

## See Also

 2001 Pan African MO (Problems) Preceded byFirst Problem 1 • 2 • 3 • 4 • 5 • 6 Followed byProblem 2 All Pan African MO Problems and Solutions
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