33. For $n=1,2,3, \ldots$, let $a_{n}=n^{2}+100$, and let $d_{n}$ denote the greatest common divisor of $a_{n}$ and $a_{n+1}$. Find the maximum value of $d_{n}$ as $n$ ranges over all positive integers.
33. Answer: 401
For each $n=1,2,3, \ldots$, since $d_{n}=\operatorname{gcd}\left(a_{n}, a_{k+1}\right)$, we have $d_{n} \mid a_{n}$ and $d_{n} \mid a_{n+1}$. Thus $d_{n} \mid a_{n+1}-a_{n}$, that is, $d_{n} \mid(n+1)^{2}+100-\left(n^{2}+100\right)$, which gives $d_{n} \mid 2 n+1$. Hence $d_{n} \mid 2\left(n^{2}+100\right)-n(2 n+1)$, and we obtain $d_{n} \mid 200-n$. It follows that $d_{n} \mid 2(200-n)+2 n+1$, that is, $d_{n} \mid 401$. Consequently, $1 \leq d_{n} \leq 401$ for all positive integers $n$.
Now when $n=200$, we have $a_{n}=a_{200}=200^{2}+100=401 \times 100$ and $a_{n+1}=a_{201}=201^{2}+100=401 \times 101$. Therefore $d_{200}=\operatorname{gcd}\left(a_{200}, a_{201}\right)=401$. Hence it follows that the maximum value of $d_{n}$ when $n$ ranges over all positive integers is 401 , which is attained at $n=200$.