19. Let $p=a^{b}+b^{a}$. If $a, b$ and $p$ are all prime, what is the value of $p$ ?
19. Answer: 17
If both $a, b$ are odd prime numbers, then $p$ is even and $p \geq 4$, contradicting the condition that $p$ is prime. Thus $a=2$ or $b=2$.
Assume that $a=2$. Then $b \neq 2$; otherwise, $p=8$ which is not prime.
Thus $b$ is an odd prime number. Let $b=2 k+1$, where $k$ is an integer greater than 1 . Thus
$$
p=2^{2 k+1}+(2 k+1)^{2}=2 \times 4^{k}+(2 k+1)^{2} .
$$

We shall show that $k=1$.
Suppose that $k \geq 2$. If $k \equiv 1(\bmod 3)$, then $b>3$ and
$$
b=2 k+1 \equiv 0(\bmod 3),
$$
contradicting the condition that $b$ is prime. If $k \not 11(\bmod 3)$, then
$$
p=2 \times 4^{k}+(2 k+1)^{2} \equiv 2+4 k^{2}+4 k+1 \equiv 4 k(k+1) \equiv 0(\bmod 3),
$$
a contradiction too. Thus $k=1$ and $b=3$. Therefore
$$
p=2^{3}+3^{2}=8+9=17
$$