What is required for large integers to be prime?

[Guest]

Large primes are a very rare breed over all integers. Indeed! But there are infinitely many primes... Go figure!

We shall uncover some important details (...) about primes.

Relevant Reference Link:

https://en.m.wikipedia.org/wiki/Prime_number

[Guest]

Hmm. Let's begin with an important mathematical paradox which is a central focus in prime number theory, the wonderful Harmonic Series, [tex]H_{k} = \sum_{k=1}^{n }\frac{1}{k}[/tex]. As [tex]n \rightarrow \infty[/tex], [tex]\frac{1}{k} \rightarrow 0[/tex] and [tex]H_{k} \rightarrow \infty[/tex]. What a paradox! Indeed!

Relevant Reference Links:

https://en.m.wikipedia.org/wiki/Harmonic_series_(mathematics);

https://en.m.wikipedia.org/wiki/Prime_number_theorem.

[Guest]

Fundamentally, the larger the odd integer, [tex]N[/tex], is (excluding all multiples of 5), the more the number of primes, [tex]\pi(\sqrt{N})[/tex] is required to determine if N is prime.

And according to the Prime Number Theorem, as [tex]N \rightarrow \infty[/tex], the average chance of very large [tex]N[/tex] being prime is [tex]\frac{\pi(N)}{N} = \frac{1}{log(N)} \rightarrow 0[/tex] relative to almost all odd integers excluding all multiples of 5.

[Guest]

Now let's consider the Fermat number, [tex]N = F_{k} = 2^{2^{k}} + 1[/tex] for large k. The average chance of very large [tex]N[/tex] being prime is again almost zero, but there are infinitely many primes. And therefore, [tex]N = F_{k} = 2^{2^{k}} + 1[/tex] would be prime for infinitely many values of k > 4.

The extreme rarity of Fermat primes over all k > 4 and the infinitely many Fermat primes are indeed paradoxical.

Relevant Reference Link:

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

[Guest]

Known Fermat primes exists for k = 0, 1, 2, 3, and 4. And they would reappear over large k.

Now let's consider the infinite ordered sequence of Fermat numbers, {[tex]F_{k }[/tex] over all k > 4}. The chance that the sequence is devoid of primes is

[tex]\prod_{k > 4}^{\infty }\frac{\pi(\sqrt{F_{k }})}{\pi(\sqrt{F_{k }} + 1)} = 0[/tex].

And [tex]\frac{\pi(\sqrt{F_{k }})}{\pi(\sqrt{F_{k }} + 1)} \rightarrow 1[/tex] as [tex]k \rightarrow \infty[/tex].

What a paradox! Indeed!

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

[Guest]

Moreover, the expectation, [tex]E[/tex], that [tex]N = F_{k } = 2^{2^{k}} + 1[/tex] is prime for large k is

[tex]E(N = F_{k }) = \frac{F_{k }}{\pi(\sqrt{F_{k }} + 1)} \rightarrow F_{k }[/tex] as [tex]k \rightarrow \infty[/tex].

The greater the expectation value, the more likely [tex]N = F_{k }[/tex] is prime.

Remark: We shall refute, if possible, our reasoning here and elsewhere.

Relevant Reference Links:

'Are there infinitely many Fermat primes?'

https://www.math10.com/forum/viewtopic.php?f=63&t=8803&sid=90227135cd46b7cec448b7e2c669985d;

'Are there infinitely many Mersenne primes?'

https://www.math10.com/forum/viewtopic.php?f=63&t=8804.

[Guest]

An Update:

Known Fermat primes exists for k = 0, 1, 2, 3, and 4. And they would reappear over large k.

Now let's consider the infinite ordered sequence of Fermat numbers, {[tex]F_{k }[/tex] over all k > 4}. The chance that the sequence is devoid of primes is

[tex]\prod_{k > 4}^{\infty }\frac{\pi(\sqrt{F_{k }})}{\pi(\sqrt{F_{k }}) + 1} = 0[/tex].

And [tex]\frac{\pi(\sqrt{F_{k }})}{\pi(\sqrt{F_{k }})+ 1} \rightarrow 1[/tex] as [tex]k \rightarrow \infty[/tex].

What a paradox! Indeed!

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

[Guest]

An Update:

Moreover, the expectation, [tex]E[/tex], that [tex]N = F_{k } = 2^{2^{k}} + 1[/tex] is prime for large k is

[tex]E(N = F_{k }) = \frac{F_{k }}{\pi(\sqrt{F_{k }}) + 1} \rightarrow F_{k }[/tex] as [tex]k \rightarrow \infty[/tex].

The greater the expectation value, the more likely [tex]N = F_{k }[/tex] is prime.

Remark: We shall refute, if possible, our reasoning here and elsewhere.

Relevant Reference Links:

'Are there infinitely many Fermat primes?'

https://www.math10.com/forum/viewtopic.php?f=63&t=8803&sid=90227135cd46b7cec448b7e2c669985d;

'Are there infinitely many Mersenne primes?'

https://www.math10.com/forum/viewtopic.php?f=63&t=8804.

[Guest]

An Update:

Moreover, the expectation, [tex]E[/tex], that [tex]N = F_{k } = 2^{2^{k}} + 1[/tex] is prime for large k is

[tex]E(N = F_{k }) = \frac{F_{k }}{\pi(\sqrt{F_{k }}) + 1} \rightarrow \sqrt{F_{k }} * log(\sqrt{F_{k }} )[/tex] as [tex]k \rightarrow \infty[/tex].

The greater the expectation value, the more likely [tex]N = F_{k }[/tex] is prime.

Remark: We shall refute, if possible, our reasoning here and elsewhere.

Relevant Reference Links:

'Are there infinitely many Fermat primes?'

https://www.math10.com/forum/viewtopic.php?f=63&t=8803&sid=90227135cd46b7cec448b7e2c669985d;

'Are there infinitely many Mersenne primes?'

https://www.math10.com/forum/viewtopic.php?f=63&t=8804.

[Guest]

An Update:

Moreover, the expectation, [tex]E[/tex], that [tex]N = F_{k } = 2^{2^{k}} + 1[/tex] is prime for large k is

[tex]E(N = F_{k }) = \frac{F_{k }}{\pi(\sqrt{F_{k }}) + 1} \rightarrow \sqrt{F_{k }} * log(\sqrt{F_{k }} )[/tex] as [tex]k \rightarrow \infty[/tex].

The greater the expectation value, the more likely [tex]N = F_{k }[/tex] is prime.

Remark: We shall refute, if possible, our reasoning here and elsewhere.


Relevant Reference Links:

'Are there infinitely many Fermat primes?'

https://www.math10.com/forum/viewtopic.php?f=63&t=8803&sid=90227135cd46b7cec448b7e2c669985d;

'Are there infinitely many Mersenne primes?'

https://www.math10.com/forum/viewtopic.php?f=63&t=8804.

An Example:

For k = 8, [tex]\frac{1}{E(F_{8 })} \approx 5.215 * 10^{-37}[/tex]. And therefore, we strongly expect [tex]F_{8}[/tex] is not prime.

Hmm. We have doubts galore! ...

[Guest]

Hmm. Almost all Fermat numbers are not primes in theory. But there are still infinitely many Fermat primes!

What a paradox! Indeed, we are dumbfounded!

When we ponder prime number theory, we should consider the very unexpected predictions of quantum physics!

"When you have eliminated the impossible, whatever remains, however improbable, must be the truth!." -- Sherlock Holmes.

[Guest]

Hmm. Almost all Fermat numbers are not primes in theory. But there are still infinitely many Fermat primes!

What a paradox! Indeed, we are dumbfounded!

When we ponder prime number theory, we should consider the very unexpected predictions of quantum physics!

"When you have eliminated the impossible, whatever remains, however improbable, must be the truth!." -- Sherlock Holmes.

A Remark: Our Central Idea on infinitely many primes among the much more abundant non-prime odd integers (excluding all multiples of 5) as discussed here and in previous posts is

[tex]10^{-k} * \infty = \infty[/tex] where [tex]10^{-k}[/tex] is almost zero/approaching zero, and it also represents the prime density among all odd integers excluding all multiples of 5.

Go figure!

[Guest]

Hah! We propose a counter-argument to our reasoning and conclusion on infinitely many Fermat primes.

We should consider the ordered sequence of Fermat numbers, {[tex]F_{k}[/tex] for all [tex]k > 4[/tex]}, with zero or a finite number of exceptions (Fermat primes) is an ordered sequence of multiples of distinct primes.

[Guest]

Hah! We propose a counter-argument to our reasoning and conclusion on infinitely many Fermat primes.

We should consider the ordered sequence of Fermat numbers, {[tex]F_{k}[/tex] for all [tex]k > 4[/tex]}, with zero or a finite number of exceptions (Fermat primes) is an ordered sequence of multiples of distinct primes.

Furthermore, for integers,

[tex]k > 4[/tex], [tex]i \ge 1[/tex], and [tex]j \ge 1[/tex],

[tex]F_{k } = 2^{2^{k}} + 1 = n_{ij }*p_{i }[/tex]

with integers, [tex]n_{ij} \ge 1[/tex],

and with distinct odd primes, [tex]p_{i}[/tex].

Remark: [tex]n_{ij} = 1[/tex] occurs finitely.

[Guest]

Hah! We propose a counter-argument to our reasoning and conclusion on infinitely many Fermat primes.

We should consider the ordered sequence of Fermat numbers, {[tex]F_{k}[/tex] for all [tex]k > 4[/tex]}, with zero or a finite number of exceptions (Fermat primes) is an ordered sequence of multiples of distinct primes.

Furthermore, for integers,

[tex]k > 4[/tex], [tex]i \ge 1[/tex], and [tex]j \ge 1[/tex],

[tex]F_{k } = 2^{2^{k}} + 1 = n_{ij }*p_{i }[/tex]

with integers, [tex]n_{ij} \ge 1[/tex],

and with distinct odd primes, [tex]p_{i}[/tex].

Remark: [tex]n_{ij} = 1[/tex] occurs finitely.

The size of the sample space for primes, [tex]p_{i}[/tex], is [tex]\pi(\sqrt{F_{k }} )[/tex] where [tex]\pi()[/tex] is the exact odd prime-counting function.

Now let's consider the infinite ordered sequence of Fermat numbers, {[tex]F_{k }[/tex] over all k > 4}. The chance that the sequence is devoid of Fermat primes is again,

[tex]\prod_{k > 4}^{\infty }\frac{\pi(\sqrt{F_{k }})}{\pi(\sqrt{F_{k }}) + 1} = 0[/tex].

And [tex]\frac{\pi(\sqrt{F_{k }})}{\pi(\sqrt{F_{k }})+ 1} \rightarrow 1[/tex] as [tex]k \rightarrow \infty[/tex].

Hmm. We have a contradiction! And therefore, there are infinitely many Fermat primes!

[Guest]

The size of the sample space for primes, [tex]p_{i}[/tex], is [tex]\pi(\sqrt{F_{k }} )[/tex] where [tex]\pi()[/tex] is the exact odd prime-counting function.

Now let's consider the infinite ordered sequence of Fermat numbers, {[tex]F_{k }[/tex] over all k > 4}. The chance that the sequence is devoid of Fermat primes is again,

[tex]\prod_{k > 4}^{\infty }\frac{\pi(\sqrt{F_{k }})}{\pi(\sqrt{F_{k }}) + 1} = 0[/tex].

And [tex]\frac{\pi(\sqrt{F_{k }})}{\pi(\sqrt{F_{k }})+ 1} \rightarrow 1[/tex] as [tex]k \rightarrow \infty[/tex].

Hmm. We have a contradiction! And therefore, there are infinitely many Fermat primes!

Remark: While there may be many primes [tex]p_{i} \le \sqrt{F_{k }}[/tex], that divide [tex]F_{k }[/tex], only one is required.

[Guest]

The size of the sample space for primes, [tex]p_{i}[/tex], is [tex]\pi(\sqrt{F_{k }} )[/tex] where [tex]\pi()[/tex] is the exact odd prime-counting function.

Now let's consider the infinite ordered sequence of Fermat numbers, {[tex]F_{k }[/tex] over all k > 4}. The chance that the sequence is devoid of Fermat primes is again,

[tex]\prod_{k > 4}^{\infty }\frac{\pi(\sqrt{F_{k }})}{\pi(\sqrt{F_{k }}) + 1} = 0[/tex].

And [tex]\frac{\pi(\sqrt{F_{k }})}{\pi(\sqrt{F_{k }})+ 1} \rightarrow 1[/tex] as [tex]k \rightarrow \infty[/tex].

Hmm. We have a contradiction! And therefore, there are infinitely many Fermat primes!

Remark: While there may be many primes [tex]p_{i} \le \sqrt{F_{k }}[/tex], that divide [tex]F_{k }[/tex], only one is required.

Remark: Further investigation of the Fermat numbers is warranted. We still have some lingering doubts since the nature of the beast (Fermat numbers) is not fully understood.

[Guest]

The size of the sample space for primes, [tex]p_{i}[/tex], is [tex]\pi(\sqrt{F_{k }} )[/tex] where [tex]\pi()[/tex] is the exact odd prime-counting function.

Now let's consider the infinite ordered sequence of Fermat numbers, {[tex]F_{k }[/tex] over all k > 4}. The chance that the sequence is devoid of Fermat primes is again,

[tex]\prod_{k > 4}^{\infty }\frac{\pi(\sqrt{F_{k }})}{\pi(\sqrt{F_{k }}) + 1} = 0[/tex].

And [tex]\frac{\pi(\sqrt{F_{k }})}{\pi(\sqrt{F_{k }})+ 1} \rightarrow 1[/tex] as [tex]k \rightarrow \infty[/tex].

Hmm. We have a contradiction! And therefore, there are infinitely many Fermat primes!

Remark: While there may be many primes [tex]p_{i} \le \sqrt{F_{k }}[/tex], that divide [tex]F_{k }[/tex], only one is required.

Remark: Further investigation of the Fermat numbers is warranted. We still have some lingering doubts since the nature of the beast (Fermat numbers) is not fully understood.

Relevant Reference Links:

'Integer factorization',

https://en.wikipedia.org/wiki/Integer_factorization;


'A faster method for multiplying very big numbers',

https://phys.org/news/2019-04-faster-method-big.html.

[Guest]

Remark: We believe an efficient algorithm for factoring very large Fermat numbers is required.

[Guest]

Remark: We believe an efficient algorithm for factoring very large Fermat numbers is required.

Relevant Reference Link:

'Number Theory and Cryptography',

https://www.math10.com/forum/viewtopic.php?f=63&t=8634.