- #1

zetafunction

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the number of primes between the expressions [tex] x^2 [/tex] and [tex] (x+1)^2 [/tex]

for every x or at least for x bigger than 100

is equal to the Number of primes less than 2x+1 (the x are the same)

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- Thread starter zetafunction
- Start date

- #1

zetafunction

- 391

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the number of primes between the expressions [tex] x^2 [/tex] and [tex] (x+1)^2 [/tex]

for every x or at least for x bigger than 100

is equal to the Number of primes less than 2x+1 (the x are the same)

- #2

Hurkyl

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Isn't this "obviously" false? The two intervals you consider are the same length...

- #3

CRGreathouse

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the number of primes between the expressions [tex] x^2 [/tex] and [tex] (x+1)^2 [/tex]

for every x or at least for x bigger than 100

is equal to the Number of primes less than 2x+1 (the x are the same)

That conjecture is false (counterexamples: 101, 102, 103, ..., 10000, ...). Perhaps you mean

"the number of primes between x^2 and (x+1)^2 is at most the number of primes below 2x+1"

which is a special case of a conjecture of Hardy and Littlewood. Of course this conjecture is widely believed to be false, because it is incompatible with the prime tuple conjecture. I don't know if this special case is possible under the prime tuple conjecture.

- #4

robert Ihnot

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Zetafunction may have confused Legendre's Conjecture, which states there is a prime number between n^2 and (n+1)^2. **This remains unproven as of 2009. **

They are conjectured tighter bounds, but this indicates just how little is know of this problem. http://en.wikipedia.org/wiki/Legendre's_conjecture

They are conjectured tighter bounds, but this indicates just how little is know of this problem. http://en.wikipedia.org/wiki/Legendre's_conjecture

Last edited:

- #5

chhitiz

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can't Legendre's Conjecture be proven using bertrand's postulate?

- #6

CRGreathouse

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can't Legendre's Conjecture be proven using bertrand's postulate?

Bertrand's postulate can be used to show that there is a prime between p^2 and 2p^2. But (p+1)^2 = p^2 + 2p + 1 is smaller than 2p^2 (for p prime).

- #7

chhitiz

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- #8

chhitiz

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- #9

CRGreathouse

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No. Bertrand's postulate isn't nearly strong enough. Even the Riemann hypothesis is too weak!

- #10

camilus

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What CR said is correct. How do I know? Trust me I've tried it.

Although saying the RH is too weak is a bold statement. I think with the RH proved, Legendre's won't put up much of a fight.

The thing about conjectures such as Legendre's is that they are similar to FLT, nearly an unlimited amount of conjectures similar to it can be made: Just from Legendre's conjecture I can make a bunch of other conjectures similar to it without any proofs (as of yet, especially without the RH).

- #11

CRGreathouse

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Although saying the RH is too weak is a bold statement. I think with the RH proved, Legendre's won't put up much of a fight.

I challenge you to write a proof of Legendre's conjecture conditional on the RH.

- #12

camilus

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I challenge you to write a proof of Legendre's conjecture conditional on the RH.

I don't even think its necessary, but how would you say the prime counting function grows, linear, logarithmic, exponential..?

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