Are there infinitely many Fermat primes?

[Guest]

"Hmm. Never underestimate the distribution of primes along the natural number line!"

In theory, is [tex]F_{n } = 2^{2^{n}} + 1[/tex] prime for infinitely many integers, n > 4?

Relevant Reference Links:

'Four Personalized Prime Number Formulae',

https://www.decodedscience.org/four-personalized-prime-number-formulae/;

'Fermat number',

https://en.wikipedia.org/wiki/Fermat_number.

[Guest]

The answer depends on the distribution of primes along the natural number line, and it also depends on the answer to the following question.

Does [tex]\prod_{n=5}^{\infty }x_{n } = 0[/tex] if [tex]x_{n } \rightarrow 1[/tex] as [tex]n \rightarrow \infty[/tex]?

[Guest]

The answer depends on the distribution of primes along the natural number line, and it also depends on the answer to the following question.

Does [tex]\prod_{n=5}^{\infty }x_{n } = 0[/tex] if [tex]x_{n } \rightarrow 1[/tex] as [tex]n \rightarrow \infty[/tex]?

If [tex]\prod_{n=5}^{\infty }x_{n } = 0[/tex] where [tex]x_{n } \rightarrow 1[/tex] as [tex]n \rightarrow \infty[/tex], then there are infinitely many Fermat primes.


If [tex]\prod_{n=5}^{\infty }x_{n } > 0[/tex] where [tex]x_{n } \rightarrow 1[/tex] as [tex]n \rightarrow \infty[/tex], then there is only a finite number of Fermat primes.

[Guest]

The answer depends on the distribution of primes along the natural number line, and it also depends on the answer to the following question.

Does [tex]\prod_{n=5}^{\infty }x_{n } = 0[/tex] if [tex]x_{n } \rightarrow 1[/tex] as [tex]n \rightarrow \infty[/tex]?

If [tex]\prod_{n=5}^{\infty }x_{n } = 0[/tex] where [tex]x_{n } \rightarrow 1[/tex] as [tex]n \rightarrow \infty[/tex], then there are infinitely many Fermat primes.


If [tex]\prod_{n=5}^{\infty }x_{n } > 0[/tex] where [tex]x_{n } \rightarrow 1[/tex] as [tex]n \rightarrow \infty[/tex], then there is only a finite number of Fermat primes.

Moreover,

[tex]x_{n } = \frac{\pi(\sqrt{F_{n}})}{\pi(\sqrt{F_{n}}) + 1}[/tex] where [tex]F_{n } = 2^{2^{n}} + 1[/tex].

Remark: [tex]\pi()[/tex] is the prime-counting function.

[Guest]

An Update: We replace "n = 4" with "n > 4" in two product expressions.

The answer depends on the distribution of primes along the natural number line, and it also depends on the answer to the following question.

Does [tex]\prod_{n > 4}^{\infty }x_{n } = 0[/tex] if [tex]x_{n } \rightarrow 1[/tex] as [tex]n \rightarrow \infty[/tex]?

If [tex]\prod_{n > 4}^{\infty }x_{n } = 0[/tex] where [tex]x_{n } \rightarrow 1[/tex] as [tex]n \rightarrow \infty[/tex], then there are infinitely many Fermat primes.


If [tex]\prod_{n > 4}^{\infty }x_{n } > 0[/tex] where [tex]x_{n } \rightarrow 1[/tex] as [tex]n \rightarrow \infty[/tex], then there is only a finite number of Fermat primes.

Moreover,

[tex]x_{n } = \frac{\pi(\sqrt{F_{n}})}{\pi(\sqrt{F_{n}}) + 1}[/tex] where [tex]F_{n } = 2^{2^{n}} + 1[/tex].

Remark: [tex]\pi()[/tex] is the prime-counting function

[Guest]

We strongly believe there is only a finite number of Fermat primes.

[Guest]

Hmm. Is your reasoning correct? I am not sure!

[Guest]

Hmm. Is your reasoning correct? I am not sure!

Predicting primes accurately is a very uncertain business. And therefore, the unexpected can happen unexpectedly for very large primes regardless of the form (Fermat, Mersenne, etc.) that they may take. So we encourage the author(s) of this post to review the work and check the calculations too. Good luck!

[Guest]

There may be infinitely many Fermat primes. But we cannot prove it.

[Guest]

There may be infinitely many Fermat primes. But we cannot prove it.

We know that the last digit (base-ten positional numeral system) of every odd prime number must be either 1, 3, 7, or 9.

And there are infinitely many prime numbers.

The last digit (base-ten positional numeral system) of every Fermat number ([tex]F_{n } = 2^{2^{n}} + 1[/tex]) must be either 1, 3, 7, or 9.

And therefore, there are infinitely many Fermat prime numbers.

[Guest]

There may be infinitely many Fermat primes. But we cannot prove it.

We know that the last digit (base-ten positional numeral system) of every odd prime number must be either 1, 3, 7, or 9.

And there are infinitely many prime numbers.

The last digit (base-ten positional numeral system) of every Fermat number ([tex]F_{n } = 2^{2^{n}} + 1[/tex]) must be either 1, 3, 7, or 9.

And therefore, there are infinitely many Fermat prime numbers.

Oops! The number, 5, is a Fermat prime.

Remark: The last digit, 5, of infinitely many Fermat numbers is extremely unlikely.

[Guest]

FYI: "With the exception of [tex]F_{0 }[/tex] and [tex]F_{1 }[/tex], the last digit of a Fermat number is 7."

Source Link:

https://en.wikipedia.org/wiki/Fermat_number.