How can I know if $2^{2^{2^{2^{2}}}}+1=?$ is prime?

Question

I could calculate the following prime numbers $$2+1=3$$ $$2^{2}+1=5$$ $$2^{2^{2}}+1=17$$ $$2^{2^{2^{2}}}+1=65537$$

Are the following numbers prime??? $$2^{2^{2^{2^{2}}}}+1=?$$ $$2^{2^{2^{2^{2^{2}}}}}+1=??$$

Answer 1

Note that $$ 2^{2^{2^{2^2}}}+1=2^{2^{16}}+1=F_{16}, $$ the $16$-th Fermat number, and $$ 2^{2^{2^{2^{2^2}}}}+1=2^{2^{65536}}+1=F_{65536}. $$ All Fermat numbers $F_n$ for $5\le n \le 32$ are known to be composite (Wikipedia). Summaries of factoring status are given here. Specifically, $F_{16}$ is known to be divisible by $$ 1575\cdot 2^{19} + 1 = 825753601 $$ and $$ 180227048850079840107 \cdot 2^{20} + 1 = 188981757975021318420037633. $$ As far as I can tell, nothing is known about the primality of $F_{65536}$.