$1 \cdot 33$ Find all natural numbers $n \in[1000,2000]$, such that $a_{n}=\sqrt{57121+35 n}$ is a natural number.
[Sol] According to the condition $1000 \leqslant n \leqslant 2000$, we get

and $a_{n} \leqslant \sqrt{57121+35 \times 2000}=356.54032 \cdots$,
which means $304 \leqslant a_{n} \leqslant 357$.
Also, since $57121=1632 \times 35+1$,
we have $a_{n}=\sqrt{35(n+1632)+1}$.
Thus, $a_{n}^{2}-1=\left(a_{n}-1\right)\left(a_{n}+1\right)$ can be divided by $35=5 \times 7$.
Therefore, $7 \mid \left(a_{n}-1\right)$ or $7 \mid \left(a_{n}+1\right)$.
This means $a_{n}=7 k+1$ or $a_{n}=7 k-1 .(k \in \mathbb{Z})$
In the first case, $304 \leqslant 7 k+1 \leqslant 357$, which means $44 \leqslant k \leqslant 50$. Using a calculator to compute $a_{n}^{2}-1=(7 k+1)^{2}-1$ for $k=44,45, \cdots, 50$, and selecting those divisible by 5, we get four values of $n=\left[\frac{a_{n}^{2}-1}{35}\right]-1632$: $1096\left(k=44, a_{n}=309\right)$, $1221\left(k=45, a_{n}=316\right), 1749\left(k=49, a_{n}=344\right)$, and $1888(k=50$, $a_{n}=351$ ).

In the second case, $a_{n}=7 k-1$ and $44 \leqslant k \leqslant 51$. Similarly, we get four values of $n$: $1185\left(k=45, a_{n}=314\right), 1312\left(k=46, a_{n}=321\right), 1848\left(k=50, a_{n}=\right.$ 349) and 1989(k=51, $\left.a_{n}=356\right)$.