Is the sequence $\sqrt{p}-\lfloor\sqrt{p}\rfloor$ , $p$ running over the primes , dense in $[0,1]$? Another formulation of the title : 

If we have real numbers $a,b$ with $0<a<b<1$, can we always find a prime $p$ such that $a<\sqrt{p}-\lfloor\sqrt{p}\rfloor<b$ holds ?

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 $a<\sqrt{n}-\lfloor\sqrt{n}\rfloor<b$ for $a,b$ as above, but what is the situation if we restrict to the primes ?
 A: 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.
A: 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 $u_n-\lfloor u_n \rfloor <a<u_{n+1} -\lfloor u_{n+1} \rfloor$. for these n, $u_{n+1}-u_n> b-a$, so $u_{n+1}-u_n$ does not converge to zero.
We just need to prove that $\sqrt(p_{n+1})-\sqrt(p_n) \to 0$
