On the Number of Primes between n^2 and (n+1)^2

[Guest]

Please prove the following conjecture:

For any positive integer, n ≥ 2, there are at least two prime numbers between [tex]n^{2}[/tex] and[tex](n+1)^{2}[/tex] .

https://www.quora.com/Can-we-disprove-that-for-any-positive-integer-n-there-are-at-least-two-prime-numbers-between-n-2-and-n-1-2/answer/David-Cole-146

[Guest]

The question is stronger than the related Legendre's Conjecture (https://en.wikipedia.org/wiki/Legendre%27s_conjecture).

And the key result is:

[tex]\pi((n+1)^{2})[/tex] - [tex]\pi(n^{2})[/tex] [tex]\rightarrow[/tex] [tex]\pi(n)[/tex] = n/log(n) as [tex]n \rightarrow \infty.[/tex]

Example: [tex]\pi(75,001^{2})[/tex] - [tex]\pi(75,000^{2})[/tex] - [tex]\pi(75,000)[/tex]

= 262,872,577 - 262,865,922 - 7493 = -738 (almost 10% error).

That excellent result corresponds very well with our theory!

Dave the antworker

[Guest]

The question is stronger than the related Legendre's Conjecture (https://en.wikipedia.org/wiki/Legendre%27s_conjecture).

And the key result is:

[tex]\pi((n+1)^{2})[/tex] - [tex]\pi(n^{2})[/tex] [tex]\rightarrow[/tex] [tex]\pi(n)[/tex] = n/log(n) as [tex]n \rightarrow \infty.[/tex]

Example: [tex]\pi(75,001^{2})[/tex] - [tex]\pi(75,000^{2})[/tex] - [tex]\pi(75,000)[/tex]

= (262,872,577 - 262,865,922) - 7493

= 6755 - 7493 = -738 (almost 11% error);

6755 - 75,000/log(75,000) = 6755 - 6681

= 74 (almost 1% error).

Those excellent results correspond very well with our theory!

Dave the antworker