Problem 1461

1. Given: $p$ is a prime number of the form $9k + 1$. This implies that $p \equiv 1 \pmod{9}$.

2. Objective: Show that there exists an integer $n$ such that $p \mid n^3 - 3n + 1$.

3. Step 1: Since $p \equiv 1 \pmod{9}$, we know that $9 \mid (p - 1)$. This implies that $p - 1 = 9m$ for some integer $m$.

4. Step 2: Consider the finite field $\mathbb{F}_{p^2}$. Since $p \equiv 1 \pmod{9}$, the multiplicative group $\mathbb{F}_{p^2}^*$ has order $p^2 - 1$.

5. Step 3: The order of $\mathbb{F}_{p^2}^*$ is $p^2 - 1$. Since $p \equiv 1 \pmod{9}$, we have $p^2 - 1 \equiv 0 \pmod{9}$. Therefore, $9 \mid (p^2 - 1)$.

6. Step 4: There exists an element $t$ in $\mathbb{F}_{p^2}$ with order 9. This is because the order of the multiplicative group $\mathbb{F}_{p^2}^*$ is divisible by 9.

7. Step 5: Consider the element $t + \frac{1}{t}$ in $\mathbb{F}_p$. We need to show that $t + \frac{1}{t}$ satisfies the equation $n^3 - 3n + 1 \equiv 0 \pmod{p}$.

8. Step 6: Compute $\left(t + \frac{1}{t}\right)^3 - 3\left(t + \frac{1}{t}\right) + 1$:
$$\left(t + \frac{1}{t}\right)^3 = t^3 + 3t + \frac{3}{t} + \frac{1}{t^3}$$
$$\left(t + \frac{1}{t}\right)^3 - 3\left(t + \frac{1}{t}\right) + 1 = t^3 + \frac{1}{t^3} + 3\left(t + \frac{1}{t}\right) - 3\left(t + \frac{1}{t}\right) + 1$$
$$= t^3 + \frac{1}{t^3} + 1$$

9. Step 7: Since $t$ has order 9, $t^9 = 1$. Therefore, $t^6 = \frac{1}{t^3}$. Substitute $t^6 = \frac{1}{t^3}$ into the equation:
$$t^3 + \frac{1}{t^3} + 1 = t^3 + t^6 + 1 = t^3 + \frac{1}{t^3} + 1 = 0$$

10. Step 8: Thus, $\left(t + \frac{1}{t}\right)^3 - 3\left(t + \frac{1}{t}\right) + 1 = 0$. Therefore, we can take $n \equiv t + \frac{1}{t} \pmod{p}$.

Conclusion:
$$\boxed{n \equiv t + \frac{1}{t} \pmod{p}}$$