<s>[INST] Determine all values of n such that this quadratic Determine all values of $n^2 + 19n + 99$ is a perfect square. I tried setting some square $b^2$ equal to the following, and then factoring as a Diophantine equation with $2$ variables... Didn't work.
 [/INST] $$
\begin{array}{c}
n^2 + 19n+99 = r^2 \\
(4n^2+76n) = 4r^2 - 396 \\
(2n+19)^2 -361 = 4r^2 - 396 \\
r^2 - (2n+19)^2 = 35 \\
(r+ (2n+19)) (r-(2n+19)) = 35
\end{array}
$$
Then if: $$
\begin{array}{ccc}
(r+ (2n+19)) =35, & (r-(2n+19)) = 1 & \longrightarrow n = -1 \\
(r+ (2n+19)) =7, & (r-(2n+19)) = 5 & \longrightarrow n = -9 \\
(r+ (2n+19)) =5, & (r-(2n+19)) = 7 & \longrightarrow n = -10 \\
(r+ (2n+19)) =1, & (r-(2n+19)) = 35 & \longrightarrow n = -18 \\
(r+ (2n+19)) =-35, & (r-(2n+19)) = -1 & \longrightarrow n = -18 \\
(r+ (2n+19)) =-7, & (r-(2n+19)) = -5 & \longrightarrow n = -10 \\
(r+ (2n+19)) =-5, & (r-(2n+19)) = -7 & \longrightarrow n = -9 \\
(r+ (2n+19)) =-1, & (r-(2n+19)) = -5 & \longrightarrow n = -1 \\
\end{array}
$$
So only those four integer solutions work.  $n$ must be negative, by the way.
</s>