From the properties of Pythagorean triples $(a, b, c)$, we know that
$a=2pq, b=p^2-q^2, c=p^2+q^2$,
where $p$ and $q$ (with $q < p$) are coprime and one of them is even, the other is odd. Given that $2 \times 2000^2$,
we have $2000 \sqrt{2} \approx \frac{2000}{51}$, so $q \geqslant 20$.
Thus, $p^2 + 20^2 \leqslant p^2 + q^2 < 3000$, which means $p^2 < 2600$.
Therefore, $p \leqslant 50$.
Hence, $p = 50$.
Then, $50^2 + q^2 < 3000$, which means $q^2 < 500$.
Therefore, $q \leqslant 22$.
Since $p$ is even, $q$ must be odd.
Thus, $q = 21$.
It is easy to see that $a = 2100, b = 2059, c = 2941$ is the unique Pythagorean triple that meets the requirements.
Example 6 (2006 French National Team Selection Exam) Find all integer triples $(a, b, c)$ such that $a^{2}+b^{2}=c^{2}, \operatorname{gcd}(a, b, c)=1,2000 \leqslant a, b, c \leqslant 3000$