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?

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    $\begingroup$ 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. $\endgroup$ – Sylvain JULIEN Aug 19 at 19:33

It's composite. It's divisible by 825753601.

Edit: It's also divisible by 188981757975021318420037633.

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    $\begingroup$ 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. $\endgroup$ – Thomas Browning Aug 19 at 19:38
  • $\begingroup$ ...And the number 825753601 is very searchable by Google. It looks like this may have been first noticed by Selfridge in 1953? In any case, thanks for your reply! $\endgroup$ – idmercer Aug 19 at 19:41
  • $\begingroup$ See also factordb.com/index.php?query=2%5E65536%2B1 $\endgroup$ – Max Alekseyev Aug 19 at 22:31
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    $\begingroup$ That 27-digit factor was found by Brent, Crandall, and Dilcher in 1997, maths-people.anu.edu.au/~brent/pd/rpb175tr.pdf $\endgroup$ – Gerry Myerson Aug 20 at 0:57

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