Is $n^2+n+41$ prime for all whole numbers $n$?
Furthermore, how can we prove/disprove this?
Oh, sorry, I meant 41...
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Since the question has been modified but the answer is still "no", let me answer the question raised in the comments. (What follows is all standard, I am just writing it to try and put the question to bed.)
EDIT as Gerry Myerson points out, the original argument I gave is insufficient when $c=1$. I think the following patch should work.
Let $p$ be a polynomial, not identically zero, with non-negative integer coefficients. If $p(0)=0$ then $p(4)$ is strictly positive and divisible by $4$, hence composite. Otherwise, we note that since $p(0)\geq 1$, $p(1)$ is strictly greater than $1$. Set $m=p(1)$ and note that $p(1+m)-p(1)$ is strictly positive and divisible by $m$ (because any positive power of $1+m$ is congruent to $1$ modulo $m$), so that $p(1+m)=km$ for some integer $k\geq 2$.
If one considers polynomials $p$ with arbitrary integer coefficients: my suspicion is that one can still always find a positive integer $n$ such that $p(n)$ is composite and non-zero, but I have not thought through the details properly.
** FURTHER EDIT** Thanks to Gerry Myerson again, this time for pointing out that in between my initial wrong proposal and my edited version, André sketched a better version of this approach in comments to the original question. I would encourage people to vote up his comment instead of this answer.