3. For how many natural numbers $n$ is the value of the expression $n^{3}-14 n^{2}+64 n-93$ a prime number?
III/3. Let's try to factor the expression. Since $93=3 \cdot 31$, the integer roots of the polynomial $p(x) = x^{3}-14 x^{2}+64 x-93$ can only be $\pm 1, \pm 3, \pm 31$ and $\pm 93$. By mental calculation, we verify that neither 1 nor -1 are roots. Using Horner's algorithm, we determine that 3 is a root:

|  | 1 | -14 | 64 | -93 |
| ---: | ---: | ---: | ---: | ---: |
|  |  | 3 | -33 | 93 |
| 3 | 1 | -11 | 31 |  |

We write $n^{3}-14 n^{2}+64 n-93=(n-3)\left(n^{2}-11 n+31\right)$. The value of the expression is a prime number if one of the factors is 1 and the other is a prime number. First, assume that $n-3=1$. Then $n=4$, the first factor is 1, and the second is $4^{2}-11 \cdot 4+31=3$, which is a prime number.

Next, assume that $n^{2}-11 n+31=1$. Then $n^{2}-11 n+30=0=(n-5)(n-6)$. The factor $n^{2}-11 n+31$ is equal to 1 for $n=5$ and $n=6$. For both values of $n$, the factor $n-3$ is a prime number, specifically 2 and 3, respectively.

The value of the expression $n^{3}-14 n^{2}+64 n-93$ is a prime number for three values of the natural number $n$, namely 4, 5, and 6.

Writing the expression in the form of a product: 3 points

Solution $n=4$ (linear factor equal to 1): 2 points

Solutions $n=5$ and $n=6$ (quadratic factor equal to 1): 2 points