1. Determine all integers $n$ for which the equation $x^{2}+n x+n+5=0$ has only integer solutions.
II/1. A quadratic equation can have integer solutions only if the discriminant is a perfect square. We can calculate $D=n^{2}-4(n+5)=(n-2)^{2}-24$. Therefore, for some non-negative integer $m$, we have $m^{2}=(n-2)^{2}-24$, or

$$
24=(n-2)^{2}-m^{2}=(n+m-2)(n-m-2)
$$

Since the numbers $n+m-2$ and $n-m-2$ have the same parity and $m \geq 0$, we have 4 possibilities.

If $n-m-2=-12$ and $n+m-2=-2$, then $n=-5$ (and $m=5$), and the quadratic equation has solutions $x_{1}=0$ and $x_{2}=5$. In the case of $n-m-2=-6$ and $n+m-2=-4$, $n=-3$ (and $m=1$), and the solutions of the quadratic equation are $x_{1}=1$ and $x_{2}=2$.

If $n-m-2=2$ and $n+m-2=12$, then $n=9$ and $m=5$, and the quadratic equation has solutions $x_{1}=-7$ and $x_{2}=-2$. In the case of $n-m-2=4$ and $n+m-2=6$, $n=7$ and $m=1$, and the solutions of the quadratic equation are $x_{1}=-4$ and $x_{2}=-3$.

Therefore, all possible values of the number $n$ are $-5, -3, 7$, and 9.

2. Method Let $x_{1}$ and $x_{2}$ be the integer solutions of this quadratic equation. Assume $x_{1} \leq x_{2}$. By Vieta's formulas, we have $x_{1}+x_{2}=-n$ and $x_{1} x_{2}=n+5$. If we add these equations, we get $x_{1} x_{2}+x_{1}+x_{2}=5$ and therefore

$$
\left(x_{1}+1\right)\left(x_{2}+1\right)=6
$$

Since $6=1 \cdot 6=2 \cdot 3=(-3) \cdot(-2)=(-6) \cdot(-1)$, we have four possibilities. In the first case, $x_{1}=0, x_{2}=5$, in the second $x_{1}=1, x_{2}=2$, in the third $x_{1}=-4, x_{2}=-3$, and in the last $x_{1}=-7, x_{2}=-2$. Using $x_{1}+x_{2}=-n$, we calculate that the number $n$ is equal to $-5, -3, 7$, or 9.