2
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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
3
$\begingroup$

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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.