15*. Find all values of the natural number \(n\), for which numbers of the form: 1) \(n^{3}-n^{2}+n-1\); 2) \(n^{3}-6 n+4\), 3) \(n^{5}-2 n^{3}-n-\) -1 - would be prime.
15. 1) \(\mathrm{P}\) is a prime number. Factoring the polynomial, we get:

\[
n^{3}-n^{2}+n-1=(n-1)\left(n^{2}+1\right)
\]

If \(n=1\), the product is zero; if \(n=2\), the product is a prime number 5; if \(n>2\), each factor is greater than 1, so the product will be a composite number. The required number is 2; 2) \(n=3\); 3) \(n=2\).