35 Let $n$ be a positive integer such that $n^{2}+19 n+48$ is a perfect square. Find the value of $n$.
35. Answer: 33
$n^{2}+19 n+48=(n+3)(n+16)$. If $n+3=13 k$ for some integer $k$, then $n+16=$ $13(k+1)$. However, $(n+3)(n+16)=169 k(k+1)$ cannot be a square. So 13 must not divide $n+3 \Rightarrow \operatorname{gcd}(n+3, n+16)=1 \Rightarrow$ they should be both perfect squares. Let $n+3=m^{2}$. Since $n+16 \geq(m+1)^{2}=m^{2}+2 m+1=n+3+2 m+1$, we have $m \leq 6$. Checking $m=2,3,4,5$ and $6, n+16=m^{2}+13=17,22,29,38$ and 49 respectively. Hence $m=6$ and $n=33$.