When is this integer a perfect square: $n^2 + 20n + 12$ 
Let $S_n = n^2 + 20n + 12$, where  $n$ a positive integer. What is the sum of all possible values of $n$, for which $S_n$ is a perfect square?

Help me to solve it.

Use $$(n+4)^2<n^2+20n+12<(n+10)^2.$$
I got $n^2+20n+12=(n+8)^2$, which gives $n=13$ or
$n^2+20n+12=(n+6)^2$, which gives $n=3$.
Id est, the answer is $16$
because in cases $n^2+20n+12=(n+5)^2$, $n^2+20n+12=(n+7)^2$ and $n^2+20n+12=(n+9)^2$ we can not get solutions.
