Conjecture on primes (not mine=)

  • #1
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i saw this conjecture on the web but do not know if is true

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)
 

Answers and Replies

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

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
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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
 
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  • #5
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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
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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
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well, somebody please tell me if i was correct. can bertrand's postulate be used to prove legendre's conjecture?
 
  • #9
CRGreathouse
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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!
 
  • #10
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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).
 
  • #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 wont put up much of a fight.
I challenge you to write a proof of Legendre's conjecture conditional on the RH.
 
  • #12
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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..?
 

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