The 17th question: Find all positive integers $n$ such that $\left(n^{2}+11 n-4\right) \cdot n!+33 \times 13^{n}+4$ is a perfect square.
The 17th question:
Answer: When $n \geq 4$, $8 \mid n!$, so $\left(n^{2}+11 n-4\right) \cdot n!+33 \times 13^{n}+4 \equiv$ 
\[
\left\{\begin{array}{l}
1(\bmod 8), \text { when } n \text { is odd } \\ 
5(\bmod 8), \text { when } n \text { is even }
\end{array}\right.
\]
While the remainder of a perfect square modulo 8 can only be $0, 1$, or 4, hence when $n \geq 4$, $n$ can only be odd.

When $n \geq 7$, $7 \mid n!$, so $\left(n^{2}+11 n-4\right) \cdot n!+33 \times 13^{n}+4 \equiv$
\[
\left\{\begin{array}{l}
6(\bmod 7), \text { when } n \text { is odd } \\ 
2(\bmod 7), \text { when } n \text { is even }
\end{array}\right.
\]
While the remainder of a perfect square modulo 7 can only be $0, 1, 2$, or 4, hence when $n \geq 7$, $n$ can only be even.

Combining the above two aspects, when $n \geq 7$, there are no solutions that satisfy the conditions. And when $n \geq 4$, $n$ can only be odd, so the possible values of $n$ that satisfy the conditions can only be $1, 2, 3, 5$.

When $n=5$, $\left(n^{2}+11 n-4\right) \cdot n!+33 \times 13^{n}+4 \equiv 3(\bmod 5)$, so $\left(n^{2}+11 n-4\right) \cdot n!+33 \times 13^{n}+4$ cannot be a perfect square;

When $n=3$, $\left(n^{2}+11 n-4\right) \cdot n!+33 \times 13^{n}+4 \equiv 3(\bmod 5)$, so $\left(n^{2}+11 n-4\right) \cdot n!+33 \times 13^{n}+4$ cannot be a perfect square.
When $n=2$, $\left(n^{2}+11 n-4\right) \cdot n!+33 \times 13^{n}+4=5625=75^{2}$;
When $n=1$, $\left(n^{2}+11 n-4\right) \cdot n!+33 \times 13^{n}+4=441=21^{2}$;
In summary, the values of $n$ that satisfy the conditions are $n=1, 2$.