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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 but has'nt given any solution to the problem yet.)

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    $\begingroup$ Because of large gaps, the degree can't be one. Based on results for integer coefficient polynomials, my guess is the answer is no. $\endgroup$
    – The Masked Avenger
    Commented Feb 10, 2015 at 20:21
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    $\begingroup$ No. Here is a hint (or rather a suggestion): Assuming $p(x) - p(0) \notin \mathbb{Q}[x]$ (otherwise your problem is easy), prove that there are arbitrarily large $n \in \mathbb{N}$ for which $\lfloor p(n) \rfloor \equiv 0 \mod{2}$ is even (and arbtrarily large $n$ for which it is odd). Start with $p(n) = \alpha n$ to get an idea; in this basic case it will suffice to use the density of fractional parts $\{\alpha n\}$, for $\alpha$ irrational. Can you then do the case $p(n) = \alpha n^2$? $\endgroup$
    – Vesselin Dimitrov
    Commented Feb 10, 2015 at 21:01
  • $\begingroup$ Thankd @VesselinDimitrov! your idea completes the proof. $\endgroup$
    – Mohammad Ghiasi
    Commented Feb 10, 2015 at 22:25

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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 $\frac{1}{m}$.

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    $\begingroup$ Wasn't it an idea by @VesselinDimitrov ? (I didn't see you mentioning any references). $\endgroup$
    – Włodzimierz Holsztyński
    Commented Feb 11, 2015 at 6:33
  • $\begingroup$ @WłodzimierzHolsztyński yes this idea is from him. how I can mention the refrence? $\endgroup$
    – Mohammad Ghiasi
    Commented Feb 11, 2015 at 6:41
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    $\begingroup$ Just, first thing, state in your Answer that this is Vesselin's idea. $\endgroup$
    – Włodzimierz Holsztyński
    Commented Feb 11, 2015 at 6:52
  • $\begingroup$ aha! thanks @WłodzimierzHolsztyński $\endgroup$
    – Mohammad Ghiasi
    Commented Feb 11, 2015 at 6:53

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