Find all $n\in\mathbb N$ such that $\sqrt{n^2+8n-5}$ is an integer. $n^2+8n-5$ has to be a perfect square.
How to find all $n$?

Note that $n^2 + 8x - 5 = (n + 4)^2 - 21$. Now this number will certainly be no square if it is bigger than $(n + 3)^2$ (since there are no squares strictly between $(n + 3)^2$ and $(n + 4)^2$), i.e. if
$$\begin{align*}
&n^2 + 8n - 5 > (n + 3)^2 \\
\iff& n^2 + 8n - 5 > n^2 + 6n + 9 \\
\iff& 2n > 14 \\
\iff& n > 7
\end{align*}$$
So we know that $n^2 + 8n - 5$ is no square if $n$ is greater than $7$. But this leaves only $n = 1, 2, \ldots, 7$ to check manually. This shows that $n = 1$ and $n = 7$ are the only numbers for which $n^2 + 8n - 5$ is a square.
