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?