# Conjecture on primes (not mine=)

1. May 11, 2009

### zetafunction

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

the number of primes between the expressions $$x^2$$ and $$(x+1)^2$$

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. May 11, 2009

### Hurkyl

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

3. May 11, 2009

### CRGreathouse

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. May 11, 2009

### robert Ihnot

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

Last edited: May 11, 2009
5. May 12, 2009

### chhitiz

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

6. May 12, 2009

### CRGreathouse

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. May 13, 2009

### chhitiz

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.

8. May 14, 2009

### chhitiz

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

9. May 14, 2009

### CRGreathouse

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

10. Jun 9, 2009

### camilus

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).

11. Jun 9, 2009

### CRGreathouse

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

12. Jun 11, 2009

### camilus

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