$14 \cdot 29$ Find all prime numbers of the form $\left[\frac{n^{2}}{3}\right]$, where $n$ is a natural number.
(9th All-Russian Mathematical Olympiad, 1983)
［Solution］Let $p=\left[\frac{n^{2}}{3}\right]$.
(1) When $n=3 k(k \in N)$,
$$
p=\left[\frac{9 k^{2}}{3}\right]=3 k^{2} \text {. }
$$

Only when $k=1$, $p$ is a prime number, $p=3$.
(2) When $n=3 k+1(k \in N \cup\{0\})$,
$$
p=\left[\frac{n^{2}}{3}\right]=\left[\frac{(3 k+1)^{2}}{3}\right]=3 k^{2}+2 k=k(3 k+2) \text {. }
$$

Only when $k=1$, $p$ is a prime number, $p=5$.
(3) When $n=3 k+2 \quad(k \in N \cup\{0\}$),
$$
p=\left[\frac{n^{2}}{3}\right]=\left[\frac{(3 k+2)^{2}}{3}\right]=3 k^{2}+4 k+1=(k+1)(3 k+1) \text {. }
$$

When $k=0$, $p=1$, when $k>0$, $p$ is a composite number. Therefore, when $n=3 k+2$, $p$ is not a prime number.
In summary, $p=3$ and 5.