4. Find all integers n for which the fraction

$$
\frac{n^{3}+2010}{n^{2}+2010}
$$

is equal to an integer.
SOLUTION. Fraction

$$
\frac{n^{3}+2010}{n^{2}+2010}=n-\frac{2010(n-1)}{n^{2}+2010}
$$

is an integer precisely when $n^{2}+2010$ is a divisor of the number $2010(n-1)=2 \cdot 3 \cdot 5 \cdot 67(n-1)$.

If $n$ is not a multiple of the prime number 67, then the numbers $n^{2}+2010$ and 67 are coprime, so $n^{2}+2010$ must be a divisor of the number $30(n-1)$. Since $|30(n-1)|<n^{2}+2010$, only $n=1$ satisfies this.

Let $n=67 m$, where $m$ is an integer. Then

$$
\frac{2010(n-1)}{n^{2}+2010}=\frac{30(67 m-1)}{67 m^{2}+30} .
$$

If $m$ is not a multiple of five, the number $67 m^{2}+30$ must be a divisor of the number $6(67 m-1)$. However, for $|m| \leqq 4$ this is not the case, and for $|m| \geqq 6$, $|6(67 m-1)|<67 m^{2}+30$. Therefore, $m=5 k$, where $k$ is an integer. Then

$$
\frac{30(67 m-1)}{67 m^{2}+30}=\frac{6(335 k-1)}{335 k^{2}+6} .
$$

For $|k| \geqq 7$, the absolute value of this fraction is non-zero and less than 1. Of the remaining numbers, $k=0$ and $k=-6$ satisfy the condition.

The number $\frac{n^{3}+2010}{n^{2}+2010}$ is therefore an integer precisely when the integer $n$ is one of the numbers 0, 1, or -2010.

## GUIDING AND SUPPLEMENTARY PROBLEMS:

N1. Find the smallest natural number $n$ for which the quotient $\frac{n^{2}+15 n}{33000}$ is a natural number. [56-B-S-3]

N2. Find all pairs $(p, q)$ of real numbers such that the polynomial $x^{2}+p x q$ is a divisor of the polynomial $x^{4}+p x^{2}+q$. $[56-\mathrm{B}-\mathrm{I}-5]$