# Conjecture on primes (not mine=)

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)

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

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.

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

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

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).

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.

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

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!

camilus
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 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).

Homework Helper
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.

camilus
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..?