For any positive number $c$ and positive integer $k$, show that there are infinitely many positive integer $n$ that satisfy
$$|\sin(n^k)| < c$$
I proved the case when $k=1$, but I can't solve when $k\geq2$.
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You could try 2 dimensional proof by induction. – Leonhard Euler Feb 01 '18 at 10:07
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Could this question be helpful? – Watson Feb 03 '18 at 13:50
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This is a direct consequence of the fact that ${n^a }$ is equidistributed modulo $2\pi$ for any $a>0$. – pisco Feb 03 '18 at 13:50
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Due to Weyl's inequality and the equidistribution theorem, the irrationality of $\pi$ implies that $e^{i f(n)}$ is equidistributed (hence dense) on the unit circle for any $f(x)\in\mathbb{Z}[x]$. Since the map $z\to \text{Im}(z)$ is continuous it preserves density and the claim is a straightforward consequence.
Jack D'Aurizio
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