B1. Find all natural numbers $n$ for which

$$
\frac{100 n^{5}-n^{4}-50 n^{3}+2 n^{2}-290 n-2}{n^{2}-2}
$$

is an integer.

(20 points)
B1. If we divide the polynomial $100 n^{5}-n^{4}-50 n^{3}+2 n^{2}-290 n-2$ by the polynomial $n^{2}-2$, we get the result $100 n^{3}-n^{2}+150 n$ and the remainder $10 n-2$. Therefore,

$$
\frac{100 n^{5}-n^{4}-50 n^{3}+2 n^{2}-290 n-2}{n^{2}-2}=100 n^{3}-n^{2}+150 n+\frac{10 n-2}{n^{2}-2}
$$

This number will be an integer exactly when $\frac{10 n-2}{n^{2}-2}$ is an integer. We observe that for $n>10$ the numerator of this fraction is smaller than the denominator and both are positive, so $0<\frac{10 n-2}{n^{2}-2}<1$. Therefore, for $n>10$ the fraction $\frac{10 n-2}{n^{2}-2}$ is not an integer ..... 3 points
Calculated value of the expression $\frac{10 n-2}{n^{2}-2}$ for the first ten natural numbers ..... 3 points
Solution written as $n=1,2,3,10$ ..... 5 points