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Another formulation of the title :

If we have real numbers $a,b$ with $0

The fractional parts of the numbers $\sqrt{2}\cdot n$ , $n$ running over the positive integers, are known to be even equidistributed modulo $1$, so we can always find a natural number $n$ with $a<\sqrt{n}-\lfloor\sqrt{n}\rfloor

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    It wouldn't surprise me if this was unsolved. The obvious way to prove it would be to show that there are lots of primes between any two sufficiently large consecutive perfect squares, but it's not known that there is even one. So any proof of this would have to involve choosing $\lfloor \sqrt{p}\rfloor$ carefully based on $a$ and $b$; just taking any sufficiently large $\lfloor \sqrt{p}\rfloor$ isn't going to fly. – 2017-01-09
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    If it were known to be false this would also be very surprising, since it would mean our current thinking about large prime gaps is vastly inaccurate: there would have to exist arbitrarily large prime gaps of size $\Omega(\sqrt{p})$ rather than the conjectured maximum $O((\log p)^{2 + \epsilon})$. – 2017-01-09

2 Answers 2

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The question is whether the fractional parts $\{\sqrt{p}\}$, with $p$ running over primes, are dense in $[0,1]$. One can show the stronger result that the fractional parts are equidistributed modulo 1.

By Weyl's criterion and elementary estimates, it suffices to show that $$\sum_{n \leq x} \Lambda(n) e(h \sqrt{n}) =o(x)$$ for each fixed integer $h \neq 0$, the implied constant possibly depending on $h$. Here $\Lambda$ is the von Mangoldt function and $e(x) = e^{2\pi i x}$. This exponential sum can be bounded using Vaughan's identity and some straightforward exponential sum estimates.

Bounding this exponential sum actually shows up as an exercise in chapter 13 of Iwaniec and Kowalski's book on analytic number theory. Unfortunately, I do not have a copy handy at the moment. Maybe someone here can get the reference for us. Another place to start might be Xiumin Ren's paper "Vinogradov's exponential sum over primes," which can be found here.

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    [there](http://www.paris8.free.fr/Ivaniec%20H.,%20Kowalski%20E.%20Analytic%20number%20theory%20(2004)(KA)(T)(610s)_MT_.pdf) p.347 – 2017-05-31
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we will use the lemma:

Let $u_n$ a increasing sequence such that $u_n \to \infty$ and $u_{n+1}-u_n \to 0$. Then $u_n-\lfloor u_n \rfloor $ is dense in $[0,1]$

Proof: By contradiction: take $a,b$ so that no $u_{n+1}-u_n$ is between $a$ and $b$:

Then there is an infinite number of terms such that $u_n-\lfloor u_n \rfloor b-a$, so $u_{n+1}-u_n$ does not converge to zero.

We just need to prove that $\sqrt(p_{n+1})-\sqrt(p_n) \to 0$

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    The last statement which needs to be proven is a famous open problem. But this is a good way to show that $\{\sqrt{n}\}$ is dense without needing to use equidistribution theory. – 2017-01-09
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    The best result in this direction is Baker-Harman-Pintz, that $\sqrt{p_{n+1}} - \sqrt{p_n} \ll p_n^{0.025}$, of course it is true that $\sqrt{p_{n+1}} - \sqrt{p_n}$ is very small for most values of $n$, we just don't have good control over the exceptional $n$. – 2017-01-09
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    It would suffice if you could find a subsequence with this property. – 2017-01-09
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    @erick-wong $0.025$, eh, that's so close! – 2017-01-09
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    @Baconaro Yes, under Riemann Hypothesis we can replace the $0.025$ with $\epsilon$, but we're pretty far from being close to that :). – 2017-01-09
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    @ReneSchipperus Which property are you referring to? Certainly no subsequence of an increasing sequence can satisfy $u_{n+1}-u_n \to 0$ unless the whole sequence does. – 2017-01-09
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    @Baconaro The given double inequality cannot hold for infinite many $n$, if the sequence $u$ tends to $\infty$ – 2017-01-09
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    How do you conclude that there will be *an infinite number of terms such that $\left \lfloor u_n \right \rfloor$u_n \rightarrow \infty$ assumption straightaway? – 2017-01-10