<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] Let's assume that $n$ is positive. Observe that
$$(n+4)^2=n^2+8n+16<n^2+9n+30<n^2+12n+36=(n+6)^2$$ and the only possibility is $n^2+9n+30=(n+5)^2$.
Now, let's assume that $m=-n$ is positive and $m^2-9m+30$ is perfect square. Also observe that$$(m-5)^2=m^2-10m+25<m^2-9m+30$$
and when $m$ is large enough, $m^2-9m+30<m^2-8m+16=(m-4)^2$ and you can check the case $m^2-9m+30 \ge (m-4)^2$ manually.
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