Is there a polynomial $p(x)$ with real coefitients and degree at least one that $[p(n)]$ for everey natural number like $n$ be a prime?


If yes, what is such a polynomial $p(x)$ and if no, how to prove.(I have asked this problem in math.stackexchange.com in [this question](https://math.stackexchange.com/questions/1142240/generating-primes-with-floor-of-a-polynomial-pn) but has'nt given any solution to the problem yet.)



Generating primes with floor of a polynomial $[p(n)]$
With Vesselin's idea in the comments proof is ready as below:


If $p(x)-p(0) \in \mathbb{Q}[x]$ then the problem isn't so hard.


If $p(x)-p(0) \not \in \mathbb{Q}[x]$ then there is an irrational coefficient for a term of degree bigger than or equal one. There is a problem in ergodic theory that says that the sequence $p(n) \text{ mod 1}$ is equidistributed in $[0,1)$. It can prove similarly that the sequence $p(n) \text{ mod 2}$ is equidistributed in $[0,2)$, and with this observation we find that there are a lot of $n \in \mathbb{N}$ that for them $[p(n)]$ is even and thus isn't prime.


Also with the equidistributivity of $p(n) \text{ mod m}$ in $[0,m)$ for all $m \in \mathbb{N}$ we find that natural density of 


$$\{n\ |\ m\ |[p(n)]\}$$


is $\frac{1}{m}$.


