## Task B-3.3.

Determine all natural numbers $n$ for which $n^{3}-10 n^{2}+28 n-19$ is a prime number.
## Solution.

We will factor the expression:

$$
\begin{aligned}
& n^{3}-10 n^{2}+28 n-19= \\
= & n^{3}-n^{2}-9 n^{2}+19 n+9 n-19= \\
= & n^{2}(n-1)-9 n(n-1)+19(n-1)= \\
= & (n-1)\left(n^{2}-9 n+19\right)
\end{aligned}
$$

This product can only be a prime number if one of the factors is 1 and the other factor is a prime number.

If $n-1=1 \Rightarrow n=2$. Then $n^{2}-9 n+19=4-18+19=5$, which is a prime number. 1 point

If $n^{2}-9 n+19=1$, then $n^{2}-9 n+18=0 \Rightarrow(n-3)(n-6)=0$

$\Rightarrow n=3$ or $n=6$.

1 point

For these numbers, the first factor is $3-1=2$, or $6-1=5$, which are prime numbers.

1 point

Therefore, the solutions are the numbers 2, 3, and 6.