4. Given four numbers: $3^{147}-5^{21}, 2^{2^{4}}+1,147^{4}-$ $121^{2}, 13^{4}-11$ where one of them is a prime number. Then this prime number is $\qquad$ .
$4 \cdot 2^{2^{4}}+1$.
$3^{147}-5^{21}=\left(3^{21}\right)^{7}-\left(5^{3}\right)^{7}$, it has a factor $3^{21}-5^{3}$, $147^{4}-121^{2}$ can be divided by $147^{2}-121$, $13^{4}$ has the last digit 1, so $13^{4}-11$ can be divided by 10. $\therefore 2^{2^{4}}+1$ is a prime number.