The 52nd question: Given a prime number $p>3$, define the sequence $\left\{a_{n}\right\}$ as follows: $a_{n}=\left\{\begin{array}{cc}n, & \text { when } n \in\{0,1, \ldots, p-1\} \\ a_{n-1}+a_{n-p}, & \text { when } n \geq p\end{array}\right.$, try to find the remainder of $\mathrm{a}_{\mathrm{p} 3}$ divided by $\mathrm{p}$.
The 52nd question:
Solution: Clearly, the sequence is a $p$-order linear recursive integer sequence, and the characteristic equation corresponding to the sequence is $f(x)=x^{p}-x^{p-1}-1$. Note that $f'(x)=p x^{p-1}-(p-1) x^{p-2}$ has no common roots with $f(x)$, so the equation has exactly $p$ distinct roots $x_{1}, x_{2}, \ldots, x_{p}$. Therefore, there exist constants $A_{1}, A_{2}, \ldots, A_{p}$ such that
$$
a_{n}=\sum_{k=1}^{p} A_{k} x_{k}^{n}
$$

According to the binomial theorem, we know:
$$
x_{k}^{p^{2}}=\left(x_{k} p^{2}-1+1\right)^{p}=\sum_{j=0}^{p} C_{p}^{j} x_{k}^{j(p-1)}
$$

Thus, according to (1), we have:
$$
\begin{array}{l}
a_{n+p^{2}}=\sum_{k=1}^{p} A_{k} x_{k}^{n}+p^{2}=\sum_{k=1}^{p} A_{k} x_{k}^{n}\left(\sum_{j=0}^{p} C_{p}^{j} x_{k}^{j(p-1)}\right) \\
=\sum_{j=0}^{p} C_{p}^{j} \sum_{k=1}^{p} A_{k} x_{k} x^{n+j(p-1)} \\
=\sum_{j=0}^{p} C_{p}^{j} a_{n+f(p-1)} \\
\equiv a_{n}+a_{n+p(p-1)}(\bmod p)
\end{array}
$$

According to the given condition, $a_{n+p^{2}}=a_{n+p(p-1)}+a_{n+p^{2}-1}$, combining with (2), we know:
$$
a_{n+p^{2}-1} \equiv a_{n}(\bmod p)
$$

Since $p^{3}=p(p^{2}-1)+p$, according to (3), we know:
$$
a_{p^{3}} \equiv a_{p}=a_{p-1}+a_{0}=p-1(\bmod p)
$$

In summary, the remainder of $a_{p}^{3}$ divided by $p$ is $p-1$.