Let $p$ be any (positive) prime number. Give all positive integers $b$ for which the roots of the quadratic equation $x^{2}-b x+b p = 0$ are integers.
Solution. Let the roots of the equation be $x_{1}$ and $x_{2}$, where $x_{1} \leq x_{2}$. According to Viète's formulas, $x_{1}+x_{2}=b$ and $x_{1} x_{2}=b p$. Substituting the first relationship into the second:

$$
\begin{gathered}
x_{1} x_{2}=\left(x_{1}+x_{2}\right) p \\
x_{1} x_{2}-p x_{1}-p x_{2}+p^{2}=p^{2} \\
\left(x_{1}-p\right)\left(x_{2}-p\right)=p^{2}
\end{gathered}
$$

Since $x_{1}-p \leq x_{2}-p$ are integers and $p$ is prime, only the following cases are possible.

1. Case: $x_{1}-p=1$ and $x_{2}-p=p^{2}$. Then $x_{1}=p+1, x_{2}=p^{2}+p$, so

$$
b=x_{1}+x_{2}=(p+1)^{2}
$$

2. Case: $x_{1}-p=-p^{2}$ and $x_{2}-p=-1$. Then

$$
b=x_{1}+x_{2}=\left(-p^{2}+p\right)+(p-1)=-(p-1)^{2}
$$

which does not satisfy the condition $b>0$.

3. Case: $x_{1}-p=p$ and $x_{2}-p=p$. Then $b=x_{1}+x_{2}=2 p+2 p=4 p$.
4. Case: $x_{1}-p=-p$ and $x_{2}-p=-p$. Then $b=x_{1}+x_{2}=0+0=0$, which is not positive.

Therefore, the value of $b$ can be $(p+1)^{2}$ or $4 p$.