# Primality of Fermat numbers associated with “tetration”

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?

• I would not say that "a lot is known" about Fermat numbers. As far as I know, the existence of other Fermat primes has neither been proven nor ruled out. – Sylvain JULIEN Aug 19 at 19:33

• Factors of $2^{2^{16}}+1$ must be of the form $k\cdot2^{18}+1$. Here $k=3150$ which was found with a short computer search. – Thomas Browning Aug 19 at 19:38