2

In this page (http://mathworld.wolfram.com/FermatPrime.html) we have the following result:

$2^{2^n}+1$ is a Fermat prime if and only if the period length of $1/(2^{2^n}+1)$ is equal to $2^{2^n}$. In other words, Fermat primes are full reptend primes.

Thus, we get:

$2^{2^n}+1$ is not a Fermat prime if and only if the period length of $1/(2^{2^n}+1)$ is strictly less then $2^{2^n}$. See: Period of the decimal expression for the rational number $\frac{1}{n}$ is at most $n-1$

My question is: Can we deduce that there is infinitely many indices $n$ such that the period length of $1/(2^{2^n}+1)$ is strictly less then $2^{2^n}$.

Safwane
  • 3,840
  • @Sil: From this can we say that this problem is undecidable! – Safwane Dec 22 '19 at 13:59
  • 1
    I think it is just undecided, undecidable would mean that someone actually proved that it cannot be answered, but the fact is that we just don't know. – Sil Dec 22 '19 at 14:01
  • I've converted my comment to an answer and cleaned the comments. – Sil Dec 22 '19 at 14:07
  • Probably, there is no Fermat prime beyond $\ n=4\ $ , but in fact , as mentioned in the answer, we do not even know whether infinite many Fermat numbers are composite. The first unknown case is $\ n=33\ $ – Peter Dec 22 '19 at 14:14
  • 1
    @Helena Considering Goedel's results , we do not even know whether a statement like "There are infinite many composite Fermat numbers" can be proven or disproven. It might be independent of ZFC like the continuum hypothesis. – Peter Dec 22 '19 at 14:22
  • @Peter: Yes. I just read about this. Thanks. – Safwane Dec 22 '19 at 14:23

1 Answers1

2

Showing existence of infinitely many such $n$'s would imply infinitely many $n$'s for which Fermat numbers are composite, which is an open problem (and as suggested in comments, period of $1/n$ cannot be larger than $n-1$). See also Composite Fermat's numbers and Period of the decimal expression for the rational number $\frac{1}{n}$ is at most $n-1$.

Sil
  • 16,612