3. Let $p$ and $5 p^{2}-2$ both be prime numbers: Find the value of $p$.
(2012, National Junior High School Mathematics Competition, Tianjin Preliminary Round)
It is easy to prove that when $3 \times p$, $3 \mid \left(5 p^{2}-2\right)$.
Since $5 p^{2}-2>3$, thus $5 p^{2}-2$ is not a prime number, which contradicts the given condition.
Therefore, $3 \mid p$. Then $p=3$.