Conjecture on primes (not mine=)

**zetafunction** · May11-09, 01:18 PM

i saw this conjecture on the web but do not know if is true

the number of primes between the expressions

and

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)

**Hurkyl** · May11-09, 03:45 PM

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

**CRGreathouse** · May11-09, 04:59 PM

Originally Posted by zetafunction

i saw this conjecture on the web but do not know if is true

the number of primes between the expressions and

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.

**robert Ihnot** · May11-09, 06:54 PM

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%27s_conjecture

**chhitiz** · May12-09, 04:58 AM

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

**CRGreathouse** · May12-09, 08:48 AM

Originally Posted by chhitiz

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).

**chhitiz** · May13-09, 03:11 AM

bertrand's postulate can show that there is a prime between (p+1)²/2 and (p+1)². p² is greater than (p+1)²/2 after 2.

**chhitiz** · May14-09, 09:07 AM

well, somebody please tell me if i was correct. can bertrand's postulate be used to prove legendre's conjecture?

**CRGreathouse** · May14-09, 09:59 AM

Originally Posted by chhitiz

well, somebody please tell me if i was correct. can bertrand's postulate be used to prove legendre's conjecture?

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

**camilus** · Jun9-09, 05:07 PM

Originally Posted by chhitiz

well, somebody please tell me if i was correct. can bertrand's postulate be used to prove legendre's conjecture?

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 wont 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).

**CRGreathouse** · Jun10-09, 12:26 AM

Originally Posted by camilus

Although saying the RH is too weak is a bold statement. I think with the RH proved, Legendre's wont put up much of a fight.

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

**camilus** · Jun11-09, 09:59 PM

Originally Posted by CRGreathouse

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

I dont even think its necessary, but how would you say the prime counting function grows, linear, logarithmic, exponential..?