Let $\Pi$ be the usual function counting prime numbers. Show that for any $n\in \mathbb{Z}$, exists $k\in \mathbb{Z}$ such that

$$\Pi((k+1)^2)-\Pi(k^2) \geq n$$

I tried to use the Prime Numbers Theorem to estimate how many prime numbers should exist between $(k+1)^2$ and $k^2$ in the form $$\lim_{n\to \infty} \frac{p_n}{n\log n} =1 $$ But that was not good to estimate the difference of the function between consecutive squares

  • $\begingroup$ And what question are you asking? $\endgroup$ – abiessu Feb 27 '16 at 3:33
  • $\begingroup$ See also Legendre's conjecture. $\endgroup$ – Lucian Feb 27 '16 at 10:52

If each prime less than $x$ is put into a group according to which squares it is surrounded by, there are $\lfloor \sqrt{x}\rfloor$ groups. Therefore at least one of those groups has more than $\frac{\pi(x)}{\sqrt{x}}$ primes, and by the prime number theorem this value tends to infinity.

  • $\begingroup$ Simple and elegant argument. Nice. Thanks $\endgroup$ – Terg Feb 28 '16 at 16:51

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