Is 641 the Smallest Factor of any Composite Fermat Number?

Question

Consider the sequence $a_n = 2^{2^n}+1$ of so-called Fermat numbers. It's well known that $a_5$ isn't prime ($a_5 = 641 \cdot 6700417$, this is due to Euler). What I want to know about this sequence is the smallest factor of any of its composite terms. Is it simply 641, or does a later composite term in the sequence have a smaller factor?

I originally wanted to ask if there had been any effort to solve similar problems for other sequences as well, but that's not something I feel MSE would want. If someone has any info on this, however, they should feel free to comment or add it to their answer!

Answer 1

If $q$ is a prime factor of $a_n=2^{2^n}+1$, then $q\equiv1\pmod{2^{n+1}}$ and hence $$q>2^{n+1}\ .$$ So if we are looking for $q<641$ then $n\le8$; if we want $a_n$ composite then $n\ge5$. Thus $n=5,6,7,8$, and there is no prime factor $q$ less than $641$, because all these numbers have been completely factorised.

Answer 2

There is a deterministic answer to this, easy computer program. For each prime $p$ of interest, calculate $$ 2,4,16,256,\ldots, 2^{\left( 2^n \right)}, \ldots \pmod p $$ until you find repetition. For the prime being worked on, this tells you whether $2^{\left( 2^n \right)}$ can ever be congruent to $-1 \pmod p.$