<s>[INST] Value of n for which f(n) = $\,n^2 + 9n + 30\,$ is a perfect square. I attempted this by setting $f(n) = \,m^2.\,$
So $\,n^2 + 9n + 30 = m ^2\,$.   
Then  $\,9(n + 10/3) = (m + n)(m - n)\,$.
So $m = 10/3$ and $n = -17/3$ which is incorrect.
 [/INST] If $n>0$ then
By comparison we have
$$ (n+4)^2 <n^2+9n+30 < (n+6)^2$$
Therefore
$$n^2+9n+30=(n+5)^2$$
Thus 
$$n=5$$
If $n\leq0$, similarly
$$(n+5)^2<n^2+9n+30< (n+4)^2$$ for all $n<-14$
Now we only need to consider the case $-14\leq n \leq 0$
Manually, it leads to $n=-14$, $n=-7$ and $n=-2$
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