[ Decimal numeral system ]

Find all prime numbers of the form $P^{P}+1$ ( $P$ - natural), containing no more than 19 digits.
If $P=m q$, where $q$ is odd, then $P^{P}+1$ is divisible by $P^{m}+1$. It remains to check the cases when $P$ is a power of two.

$1^{1}+1=2,2^{2}+1=5,4^{4}+1=257.8^{8}+1=2^{24}+1$ is divisible by $2^{8}+1.16^{16}=2^{64}>10 \cdot 2^{60}=10 \cdot 1024^{6}>10^{19}$, which means it contains more than 19 digits.

## Answer

$1,2,4$.