Find all prime numbers $p$ for which there exist positive integers $x, y$ and $z$ such that the number

$$
x^{p}+y^{p}+z^{p}-x-y-z
$$

is a product of exactly three distinct prime numbers.
Let \( A = x^p + y^p + z^p - x - y - z \). For \( p = 2 \), we take \( x = y = 4 \) and \( z = 3 \). Then \( A = 30 = 2 \cdot 3 \cdot 5 \). For \( p = 3 \) we can take \( x = 3 \) and \( y = 2 \) and \( z = 1 \). Then again \( A = 30 = 2 \cdot 3 \cdot 5 \). For \( p = 5 \) we can take \( x = 2 \) and \( y = 1 \) and \( z = 1 \). Again \( A = 30 = 2 \cdot 3 \cdot 5 \).

Assume now that \( p \geq 7 \). Working modulo 2 and modulo 3 we see that \( A \) is divisible by both 2 and 3. Moreover, by Fermat's Little Theorem, we have

\[
x^p + y^p + z^p - x - y - z \equiv x + y + z - x - y - z = 0 \bmod p.
\]

Therefore, by the given condition, we have to solve the equation

\[
x^p + y^p + z^p - x - y - z = 6p.
\]

If one of the numbers \( x, y \) and \( z \) is bigger than or equal to 2, let's say \( x \geq 2 \), then

\[
6p \geq x^p - x = x(x^{p-1} - 1) \geq 2(2^{p-1} - 1) = 2^p - 2.
\]

It is easy to check by induction that \( 2^n - 2 > 6n \) for all natural numbers \( n \geq 6 \). This contradiction shows that there are no more values of \( p \) which satisfy the required property.

Remark. There are a couple of other ways to prove that \( 2^p - 2 > 6p \) for \( p \geq 7 \). For example, we can use the Binomial Theorem as follows:

\[
2^p - 2 \geq 1 + p + \frac{p(p-1)}{2} + \frac{p(p-1)(p-2)}{6} - 2 \geq 1 + p + 3p + 5p - 2 > 6p.
\]

We can also use Bernoulli's Inequality as follows:

\[
2^p - 2 = 8(1+1)^{p-3} - 2 \geq 8(1 + (p-3)) - 2 = 8p - 18 > 6p.
\]

The last inequality is true for \( p \geq 11 \). For \( p = 7 \) we can see directly that \( 2^p - 2 > 6p \).

One can also use calculus to show that \( f(x) = 2^x - 6x \) is increasing for \( x \geq 5 \).