A lot is known about the Fermat numbers $2^{2^k}+1$. For example, the first few
$$
2^1+1=3,\;2^2+1=5,\;2^4+1=17,\;2^8+1=257,\;2^{16}+1=65537
$$
are known to be prime, and Euler showed that the next ($2^{32}+1$) is not prime, being divisible by 641.


But what about the subset of special Fermat numbers formed by "tetration" or "power towers"?
$$
2+1 = 3\\
2^2+1 = 5\\
2^{2^2}+1 = 2^4+1=17\\
2^{2^{2^2}}+1 = 2^{16}+1=65537\\
2^{2^{2^{2^2}}}+1 = 2^{65536}+1
$$
I thought I had a vague memory of once reading that someone had conjectured that these are all prime (I guess people at one point conjectured that more generally *all* Fermat numbers were prime) and I *also* thought I had a vague memory of once reading that $2^{65536}+1$ had been proved composite.


However, I've been finding it difficult to get an answer to my question via a Google search. So, **is it known whether $2^{65536}+1$ is prime or composite?**



Primality of Fermat numbers associated with "tetration"
It's composite. It's divisible by 825753601.


Edit: It's also divisible by 188981757975021318420037633.


