19. Proof: Let $N$ be any given positive integer, and let $p_{1}, \ldots, p_{s}$ be all the primes of the form $8k+3$ that do not exceed $N$. Define
$$q=\left(p_{1} p_{2} \ldots p_{s}\right)^{2}+2$$

Let $p_{j}=2 m_{j}+1$. It is easy to see that
$$p_{j}^{2}=\left(2 m_{j}+1\right)^{2}=8 \cdot \frac{m_{j}\left(m_{j}+1\right)}{2}+1 \equiv 1(\bmod 8)$$

Thus, $q \equiv 1^{2}+2=3(\bmod 8)$. If $q$ itself is a prime, then $q > N$, and the problem is already proved. If $q$ is not a prime, let $p$ be any prime factor of $q$. Then we have
$$\left(p_{1} p_{2} \ldots p_{s}\right)^{2} \equiv-2(\bmod p)$$

Hence, $-2$ is a quadratic residue modulo $p$. Since
$$\left(\frac{-2}{p}\right)=\left(\frac{-1}{p}\right)\left(\frac{2}{p}\right)=(-1)^{\frac{p-1}{2}+\frac{p^{2}-1}{8}},$$

and
$$(-1)^{\frac{p-1}{2}}=\left\{\begin{array}{c} 1, p \equiv 1(\bmod 4), \\ -1, p \equiv-1(\bmod 4), \end{array},(-1)^{\frac{p^{2}-1}{8}}=\left\{\begin{array}{c} 1, p \equiv \pm 1(\bmod 8), \\ -1, p \equiv \pm 3(\bmod 8), \end{array}\right.\right.$$

Thus, $\left(\frac{-2}{p}\right)=1$ is the solution set of the following system of congruences:
$$\left\{\begin{array}{l} p \equiv 1(\bmod 4) \\ p \equiv 1(\bmod 8) \end{array},\left\{\begin{array}{l} p \equiv 1(\bmod 4), \\ p \equiv-1(\bmod 8) \end{array},\left\{\begin{array}{l} p \equiv-1(\bmod 4), \\ p \equiv 3(\bmod 8) \end{array},\left\{\begin{array}{l} p \equiv-1(\bmod 4), \\ p \equiv-3(\bmod 8) \end{array}\right.\right.\right.\right.$$

The second and fourth groups have no solutions. From the first and third groups, we get $p \equiv 1$ and $p \equiv 3(\bmod 8)$, respectively. However, the prime factors of $q$ cannot all be of the form $8k+1$, otherwise we would have $q \equiv 1(\bmod 8)$, which contradicts the earlier result that $q \equiv 3(\bmod 8)$. Let $p$ be a prime factor of $q$ of the form $8k+3$. It is easy to see that $p \neq 2, p_{1}, p_{2}, \cdots, p_{s}$, so $p > N$. Therefore, for any given $N$, there exists a prime of the form $8k+3$ that is greater than $N$. Q.E.D.