The integers $x, y, z$, and $p$ satisfy the equation $x^{2}+x z-x y-y z=-p$. Given that $p$ is prime, what is the value of $|y+z|$ in terms of $p$ ? Your answer should refer to the variable $p$ so that it works for every prime number $p$.
Factoring $x^{2}+x z-x y-y z$, we have $x(x+z)-y(x+z)$ or $(x-y)(x+z)$.

Therefore, the integers $x, y, z$, and $p$ satisfy $(x-y)(x+z)=-p$ or $(y-x)(x+z)=p$.

Because $p$ is a prime number and $y-x$ and $x+z$ are integers, it must be that $(y-x, x+z)$ is a factor pair of $p$.

Observe that $y+z=(y-x)+(x+z)$, so $y+z$ must be equal to the sum of the integers in a factor pair of $p$.

Since $p$ is a prime number, there are only two ways to express $p$ as the product of two integers, and they are $p=(-1) \times(-p)$ and $p=1 \times p$.

The sum of the factors in these products are $-p-1=-(p+1)$ and $p+1$, respectively.

Therefore, $y+z= \pm(p+1)$, so $|y+z|=p+1$.

ANSWER: $p+1$