TITLE: Can second degree polynomials generate as many as we wish prime numbers in the way described?
QUESTION [15 upvotes]: While I was getting in my pyjamas, a few minutes ago, the Euler polynomial $n^2+n+41$ came into my mind. As you know, this polynomial is famous because the set $\{f(0),f(1),...f(39)\}$ consists of prime numbers only, so this polynomial takes for the first $40$ of its values on the set $\mathbb N_0$ only prime numbers.
So the natural question to ask is:

Is it true that for every $m \in \mathbb N$ there exists second degree real polynomial of a real variable $P$ and a number $k(m) \in \mathbb N$ such that all numbers in the set $\{P(k(m)), P(k(m)+1),...,P(k(m)+m-1)\}$ are prime numbers?

REPLY [2 votes]: If you allow non-monic polynomials the question is not very arithmetic in nature.    
Any class of integer functions closed under the operations $f(n) \to Af(n)+B$ with integer $A,B$, is guaranteed to have this property for any set that contains long arithmetic progressions (such as the primes,  by the Green/Tao theorem as people have mentioned).  This includes degree $k$ polynomials for any $k$.  
For monic non-linear polynomials, as long as $f(n)$ does not assume all values mod $p$ for some prime, then for all $k$ there is an integer $A$ such that $(f(x) + A)$ has no congruence obstruction to having a string of $k$ consecutive prime values. According to the the Hypothesis H conjectures (Schinzel, Bateman-Horn) such strings then occur infinitely often.