2001 Pan African MO Problems/Problem 1

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 $\boxed{n = 5}$ .

2001 Pan African MO (Problems)
Preceded by First Problem	1 • 2 • 3 • 4 • 5 • 6	Followed by Problem 2
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2001 Pan African MO Problems/Problem 1

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