Prime number logic 
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??
 A: 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?
A: 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 $x$.
A: 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 $$\sqrt{x}=\lfloor\sqrt{x}\rfloor + \{x\}\Rightarrow
x = \lfloor\sqrt{x}\rfloor^2 + 2 \lfloor\sqrt{x}\rfloor \{x\} +\{x\}^2 \geq \lfloor\sqrt{x}\rfloor^2$$
