3. Determine all integers $n$ for which $\sqrt{n^{2}+4 n-5}$ is also an integer.
Solution. Let $\sqrt{n^{2}+4 n-5}=m$. According to this,

$$
\begin{aligned}
& \sqrt{n^{2}+4 n-5}=m \Rightarrow n^{2}+4 n-5=m^{2} \Rightarrow n^{2}+4 n+4-9=m^{2} \\
& \Rightarrow(n+2)^{2}-m^{2}=9 \quad \Rightarrow(n+2+m)(n+2-m)=9
\end{aligned}
$$

The two factors on the left side of the last equation are odd integers.

Since $m \geq 0$, the following cases are possible:

$$
\left\{\begin{array}{l}
n+2+m=9 \\
n+2-m=1
\end{array},\left\{\begin{array}{l}
n+2+m=3 \\
n+2-m=3
\end{array},\left\{\begin{array} { l } 
{ n + 2 + m = - 3 } \\
{ n + 2 - m = - 3 }
\end{array} \text { and } \left\{\begin{array}{l}
n+2+m=-1 \\
n+2-m=-9
\end{array}\right.\right.\right.\right.
$$

From this, we easily find that $n=3, n=1, n=-5, n=-7$.