Let's determine all integer values of $x$ for which $x^{2}+19 x+95$ is a perfect square.
Let's introduce the notation

$$
y^{2}=x^{2}+19 x+95
$$

From the resulting equation, we solve for $x$:

$$
x_{1,2}=\frac{-19 \pm \sqrt{19^{2}-4 \cdot\left(95-y^{2}\right)}}{2}=\frac{-19 \pm \sqrt{4 y^{2}-19}}{2}
$$

Given that $x, y$ are integers, this is only possible if $\sqrt{4 y^{2}-19}$ is also an integer, meaning $4 y^{2}-19$ must be a perfect square. Denote its value by $z^{2}$ (where $z$ is a non-negative integer),

$$
4 y^{2}-z^{2}=19(2 y+z)(2 y-z)=19
$$

Here, $y, z$ are integers, so $2 y+z$ and $2 y-z$ must be divisors of 19, i.e., their possible values are $\pm 1, \pm 19$. Since $z \geq 0$, we have $2 y+z \geq 2 y-z$. Therefore, $2 y+z$ can be 19 or -1. Considering these two cases:

$$
\begin{aligned}
& 2 y+z \\
& =19 \\
& 2 y-z \\
& =1
\end{aligned}
$$

$$
\Rightarrow \quad 4 y=20 \quad \Rightarrow \quad y=5, z=9
$$

9. $2 y+z=-12 y-z=-19\} \Rightarrow 4 y=-20 \Rightarrow y=-5, z=$

$\langle-5-14$. This shows that the solution is valid, as the thought process indicates that it is feasible. Substituting back into $(1)$, we get $x=\frac{-19 \pm \sqrt{4 \cdot 5^{2}-19}}{2}=\frac{-19 \pm \sqrt{81}}{2}=$ Based on the work of Penkalo Alex (Zalaegerszeg, Zrínyi M. Gymnasium, 2nd grade)

Remark. Another method of solving could be the following. The relationship $x^{2}+19 x+95=(x+9)^{2}+x+14=(x+10)^{2}-x-5$ shows that for $x>-5$, $(x+9)^{2}<x^{2}+19 x+95<(x+10)^{2}$, while for $x<-14$, $(x+10)^{2}<$ $x^{2}+19 x+95<(x+9)^{2}$. Based on this, only $x=-14,-13, \ldots,-5$ are possible, and only the two extreme values prove to be valid.

Based on the work of Juhász András (Budapest Fazekas M. Secondary School and Gymnasium, 8th grade)