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2020-10-06, 18:20   #3
jshort

"James Short"
Mar 2019
I didn't think a counter example could be found for such small $n$. My mistake was in assuming that the non-primitive part of $2^{n} + 1$ must contain many factors from smaller $2^{m} + 1$ where $m < n$ is another highly composite integer. However just because $m$ is highly composite, doesn't mean that $m | n$.
In this case, even considering only the primitive part of $2^{24} + 1$ which is $65281 = 97 \cdot 673$ we still get a composite integer, which is kind of surprising because we can also prove that any prime factor of the primitive part of $2^{24} + 1$ must be of the form $1 + 4 \cdot 24k = 1 + 96k$, which greatly restricts the possible number of "eligible" prime factors to nearly 1% of all primes that the are in the range from $(2, \sqrt{65281})$. However when $k=1$, we get $97$ which is a factor lol!