Question 13. The general term of the sequence $101, 104, 109, 116, \cdots$ is $a_{n}=100+n^{2}$. Here $n=1,2,3$, $\cdots$. For each $n$, let $d_{n}$ denote the greatest common divisor of $a_{n}$ and $a_{n+1}$. Find the maximum value of $\mathrm{d}_{\mathrm{n}}$, where $n$ takes all positive integers.
We can prove more generally: If $a$ is a positive integer, and $d_{n}$ is the greatest common divisor of $a+n^{2}$ and $a+(n+1)^{2}$, then when $n=2a$, $d_{n}$ reaches its maximum value of $4a+1$. From this, the answer is $4(100)+1 = 401$.

To prove the above proposition, first note that if $d_{n}$ divides both $a+(n+1)^{2}$ and $a+n^{2}$, then it also divides their difference, i.e., $d_{n} \mid (2n+1)$.
Since $2(a+n^{2}) = n(2n+1) + (2a-n)$, from (1) we know $d_{n} \mid (2a-n)$, (2) and from (1) and (2) we have
$$
d_{n} \mid [(2n+1) + 2(2a-n)]
$$

i.e., $d_{n} \mid (4a+1)$.
Therefore, $1 \leqslant d_{n} \leqslant 4a+1$, so $d_{n}$ cannot exceed $4a+1$. Moreover, $d_{n}$ can indeed reach this value, for example, when $n=2a$, we have
$$
\begin{array}{l}
a+n^{2} = a + (2a)^{2} = a(4a+1) \\
a+(n+1)^{2} = a + (2a+1)^{2} \\
= (a+1) \quad (4a+1).
\end{array}
$$

Thus, $d_{n}$ reaches its maximum value of $4a+1$. In particular, when $a=100$, the maximum value of $d_{n}$ is 401.