$3 \cdot 2$ Find all prime numbers $p$ such that $4 p^{2}+1$ and $6 p^{2}+1$ are also prime.
(Polish Mathematical Competition, 1964)
【Sol】Let $u=4 p^{2}+1, v=6 p^{2}+1$.
And let $p=5 k+r, \quad k \in \mathbb{Z}, \quad r \in\{0,1,2,3,4\}$.
Then
$$
\begin{array}{l}
u=100 k^{2}+40 k r+4 r^{2}+1, \\
v=150 k^{2}+60 k r+6 r^{2}+1 .
\end{array}
$$

If $p \equiv 0 \quad(\bmod 5)$, then
$$
u \equiv 1, \quad v \equiv 1 \quad(\bmod 5) \text {; }
$$

If $p \equiv 1(\bmod 5)$, then
$$
u \equiv 0, \quad v \equiv 2(\bmod 5)
$$

If $p \equiv 2(\bmod 5)$, then
$$
u \equiv 2, \quad v \equiv 0 \quad(\bmod 5)
$$

If $p \equiv 3 \quad(\bmod 5)$, then
$$
u \equiv 2, \quad v \equiv 0 \quad(\bmod 5) \text {; }
$$

If $p \equiv 4 \quad(\bmod 5)$, then
$$
u \equiv 0, \quad v \equiv 2(\bmod 5) .
$$

Therefore, for any integer $p$, among the three numbers $p, u, v$, there is one and only one that can be divided by 5.

Since when $p$ cannot be divided by 5, one of $u$ and $v$ can be divided by 5, i.e., one of $u$ and $v$ is not a prime number, so if $u$ and $v$ are both prime numbers, it is only possible when $p$ can be divided by 5. Since $p$ is a prime number, $p=5$.

It is not difficult to verify that when $p=5$, $u=4 \cdot 5^{2}+1=101, \quad v=6 \cdot 5^{2}+1=151$ are both prime numbers.
Therefore, this problem has only one unique solution.