Is it possible for a polynomial to be integer-valued only for prime inputs? Could one write a polynomial that produces an integer output only for prime number inputs? 
I know that if a polynomial of degree $n$ takes integer values at $n+1$ consecutive integer arguments, then it takes integer values at all integer arguments. However, since the prime number inputs aren't consecutive, then how would one go about proving it would be possible to write?
 A: No. Suppose there is such a polynomial $f(x)$ of degree $k$, say. First off, this polynomial has to have rational coefficients (by Lagrange interpolation, say). Let $N$ be the least common denominator of the coefficients. Then $g(x)=Nf(x)$ is a polynomial with integer coefficients and $g(2)$ is divisible by $N$. But $g(2-N)\equiv g(2)\equiv 0\pmod N$, so $g(2-N)$ is divisible by $N$, hence $f(2-N)$ is an integer. But clearly $2-N$ is not a prime, so we get a contradiction.
Edit: for the sake of clarifying, here is how I have interpreted the question: is there a polynomial $f$ such that, for $n\in\mathbb Z$, $f(n)$ is an integer iff $n$ is prime? I show that there is no such polynomial.
Edit 2: (in reply to Yves Daoust) Suppose that $f$ is a polynomial of degree $k$ which takes integer value at every prime. Pick any primes $p_1,\dots,p_{k+1}$. Consider polynomial $L(x)$ defined here for $x_i=p_i,y_i=f(p_i)$. Since every $\ell_i$ is a polynomial with rational coefficients of degree at most $k$, the same holds for $L$.
Now note $f-L$ is a polynomial of degree at most $k$ which is zero at at least $k+1$ points $p_1,\dots,p_{k+1}$, hence it must be a zero polynomial. Thus $f=L$ has rational coefficients.
A: Yes.
All polynomials of the form
$$Q(x)+\sqrt2\prod_{k=1}^n(x-p_k)$$ where $Q$ is a polynomial of integer coefficients, are integer only at the primes $p_k$.

Addendum:
The interpretation of the question as "the only integer values of the polynomial over $\mathbb R$ occur at primes" wouldn't make sense as for sufficiently large $n$, the slope of any polynomial exceeds $1$ and the integer values occur at arguments closer than $1$.
