Conjecture on primes (not mine=)

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)

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

CRGreathouse
Homework Helper
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)
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.

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

CRGreathouse
Homework Helper
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).

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.

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

CRGreathouse
Homework Helper
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!

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