<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] Hint:
Let $n^2+9n+30=(n+a)^2$ where $a$ is any integer
$\iff n=\dfrac{a^2-30}{9-2a}$ which has to be an integer
Now if integer $d(>0)$ divides both $a^2-30,9-2a;$
$d$ must divide $2(a^2-30)+a(9-2a)=9a-60$
$d$ must divide $-9(9-2a)-2(9a-60)=39$
So, a necessary condition for $(9-2a)|(a^2-30)$ is $9-2a$ must divide $39$
</s>