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It is conjectured that for every intever $n\geq1$ there is a prime $p$ with $n^2<p<(n+1)^2$. Show that if this conjecture is true then $\pi(x)\geq\lfloor\sqrt{x}\rfloor$ for all $x\geq2$.

I understand that the conjecture is true since in every interval there must be a prime in between. And I understand that $\pi(x)$ is the number of primes less than $x$ with $x\geq2$. I am just very confused on how to setup the proof to get the desired result. Can anyone give me a hint into the direction??

Andrew Chin
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user287133
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  • What's the source of this question? – Andrew Chin Jul 06 '20 at 20:39
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    The conjecture is open. The exercise is to show that it implies the given inequality. – Peter Jul 06 '20 at 20:40
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    Your reasoning does not show that the conjecture must be true. Not every interval in the natural numbers contains a prime. – D. Brogan Jul 06 '20 at 20:40
  • Sorry I am not focusing on the conjecture, more on how to use the conjecture to prove the statement. I have spent 4 days on this proof and cannot get the desired result and was simply hoping someone could point me in the right direction – user287133 Jul 06 '20 at 20:42
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    Assuming the conjecture, there's a prime in the interval $(1,4)$ another in the interval $(4,9)$, yet another in the interval $(9,16)$, and so on. How many such intervals are there that are $\leq x$? – saulspatz Jul 06 '20 at 20:45
  • I just wanted to point out that the conjecture is not obvious which you seem to believe , but remained unproven until today. Of course, for the exercise we do not need that the conjecture is actually true. – Peter Jul 06 '20 at 20:46
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    @Peter: It was proved today?! – Brian Tung Jul 06 '20 at 20:48
  • @BrianTung Oops, I used an unlucky formulation :) – Peter Jul 06 '20 at 20:49
  • Ahh, yes, that is a common English bugaboo: "...but remains unproved to the present day" or something like that is a better formulation. :-) – Brian Tung Jul 06 '20 at 20:50
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    That'd be pretty big news. – Brian Tung Jul 06 '20 at 20:51
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    Some of the comments above and Robert Israel's answer seem to construe everything about the posted question literally, despite evidence of some confusion's in the poster's mind. – Michael Hardy Jul 06 '20 at 21:08

3 Answers3

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Hint: At least one prime from $1^2$ to $2^2$, at least one from $2^2$ to $3^2$, ..., at least one from $(m-1)^2$ to $m^2$. How many is that?

Robert Israel
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4 primes between 1 and 9 plus at least one prime between every $t^2$ and $(t+1)^2$ for every $3\le t\le \lfloor \sqrt{x}\rfloor-1$. Altogether gives you $>\lfloor \sqrt{x}\rfloor$ primes smaller than $x$.

markvs
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  • Can you elaborate on the inequality? – user287133 Jul 06 '20 at 20:53
  • Which inequality? – markvs Jul 06 '20 at 20:54
  • I fixed the answer replacing 4 by 9. – markvs Jul 06 '20 at 20:55
  • Why are you comparing between 1 and 9 since it should only be between n and n+1 which would actually be 1 and 4 for the first interval correct? And I am confused how you set up the inequality... – user287133 Jul 06 '20 at 20:58
  • If you start with $1^2$ and not $3^2$ you will get fewer than needed primes. What inequality did I set up? – markvs Jul 06 '20 at 21:02
  • 3 <= t <= floor(sqrt(x)) - 1 – user287133 Jul 06 '20 at 21:05
  • I start applying the fact from your question with $n=3$ (before ]$3^2$ there are 4 primes), hence $3\le t$. I stop counting when $(t+1)^2>x$, hence the upper bound. The number of applications of the fact is $\lfloor\sqrt{x}\rfloor - 1-2$. So altogether the number of primes $\le x$ is at least $4+\lfloor\sqrt{x}\rfloor - 1-2=\lfloor\sqrt{x}\rfloor +1$. – markvs Jul 06 '20 at 21:11
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Just a general remark, in fact $\pi(x)>\frac{x}{\ln{x}}>\sqrt{x}$ holds without assuming the conjecture. But let's use it, i.e. Legendre's conjecture to prove the inequality.

P1. Legendre's conjecture $\iff \pi\left((n+1)^2\right)-\pi\left(n^2\right)\geq 1$ for any integer $n\geq 1$

It's obvious.

If $\pi\left((n+1)^2\right)-\pi\left(n^2\right)\geq 1$, then $\{1,2,...,(n+1)^2\}$ contains more primes than $\{1,2,...,n^2\}$. Thus, there is at least one prime between $n^2$ and $(n+1)^2$.

If there is at least one prime between $n^2$ and $(n+1)^2$, then $\{1,2,...,(n+1)^2\}$ contains more primes than $\{1,2,...,n^2\}$. Thus $\pi\left((n+1)^2\right)-\pi\left(n^2\right)\geq 1$.

P2. $\pi(n^2)\geq n$, for any integer $n\geq 2$.

By induction:

  • it's true for $\pi(2^2)=2\geq 2$.
  • from the induction hypotheses $\pi(n^2)\geq n$ we have $$\pi\left((n+1)^2\right)=\pi\left((n+1)^2\right)-\pi\left(n^2\right)+\pi\left(n^2\right)\overset{P1}{\geq} 1+\pi\left(n^2\right)\geq 1+n$$

Finally for all $x\geq2$ $$\pi\left(x\right)\geq \pi\left(\lfloor\sqrt{x}\rfloor^2\right)\overset{P2}{\geq}\lfloor\sqrt{x}\rfloor$$

simply because

  • $\pi(x)$ is ascending ($x\geq y \Rightarrow \pi(x)\geq \pi(y)$) and
  • $x\geq \lfloor\sqrt{x}\rfloor^2$ for $x\geq 0$, from $$\sqrt{x}=\lfloor\sqrt{x}\rfloor + \{x\}\Rightarrow x = \lfloor\sqrt{x}\rfloor^2 + 2 \lfloor\sqrt{x}\rfloor \{x\} +\{x\}^2 \geq \lfloor\sqrt{x}\rfloor^2$$
rtybase
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