I was trying to answer the question at How to show that $\lim_{n \to \infty} \sin(n^2)$ does not exist?

I was able to show that $\sin(n^2)$ does not tend to a limit, but am still unable to show that $n^2$ itself is dense mod $\pi$. Can this be deduced in a straightforward way from the famous theorem that the integers are dense mod $\pi$?

Also, are all integer polynomials dense modulus an irrational number? I struggle to find the relevant literature online.

  • $\begingroup$ en.wikipedia.org/wiki/Weyl%27s_inequality $\endgroup$ Jul 3, 2016 at 12:39
  • 2
    $\begingroup$ expanding slightly on @JackD'Aurizio's answer: Let $f(x)$ be a nonconstant polynomial with integer coefficients. Weyl's inequality, applied to the sequence $\{f(n)/\pi\}$, shows that the sequence satisfies Weyl's criterion and is therefore equidistributed modulo 1. This is even a bit stronger than saying $\{f(n)\}$ is dense modulo $\pi$. $\endgroup$ Jul 3, 2016 at 19:42
  • 1
    $\begingroup$ so elementary methods aren't applicable? $\endgroup$ Jul 4, 2016 at 13:33

1 Answer 1


Yes, I think so. Answering loosely: say we want an integer $n$ so that $n^2$ is "close to" a certain value $r$ mod $\pi$. Let $s = \sqrt{r}$. Because the integers are dense mod $\pi$, there is an integer $k$ so that $k$ is "close to" $s$. Let $n = k$. $n^2 \mathrm{mod} \pi = k^2 \mathrm{mod} \pi = (k \mathrm{mod} \pi)^2$. So $n^2$ is "close to" $s^2 = r$ mod $\pi$.

  • 3
    $\begingroup$ This doesn't work; just because $a$ and $b$ are close mod $\pi$ does not mean that $a^2$ and $b^2$ are close mod $\pi$. $\endgroup$ Jul 3, 2016 at 19:24
  • $\begingroup$ No, you're right; on reflection, I was thinking too much in terms of integer modular arithmetic. Thanks for pointing that out. $\endgroup$ Jul 3, 2016 at 19:45

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .