TITLE: Generalization of Dirichlet's theorem
QUESTION [9 upvotes]: Is there a generalization of Dirichlet's theorem along these lines?
If $p(n)$ is a polynomial of degree $k$ with positive integers as coefficients, such that the coefficients are relatively prime, then the sequence $p(1),p(2),p(3),\ldots$ has infinitely many primes?
$k=1$ is the good old Dirichlet's theorem on arithmetic progressions.
If the above is false, is there a trivial example for some degree '$k$'?

REPLY [14 votes]: This is false as stated. Take, for example, $p(n) = n^2 + n$. The correct condition is that $p$ is irreducible and there exists no $d > 1$ such that $d | p(n)$ for all $n$. With this condition this is a big open problem, the Bunyakovsky conjecture, and it is open for any such polynomial $p$ of degree greater than $1$.

REPLY [12 votes]: In 1918 Stackel published the following simple observation:
THEOREM  If $ p(x)$ is a composite integer coefficient polynomial
then  $ p(n) $ is  composite for all $|n| > B $, for some bound $B$,
in fact $ p(n) $ has at most $ 2d $ prime values, where  $ d = {\rm deg}(p)$.
The simple proof can be found online in Mott & Rose [3], p. 8.
I highly recommend this delightful and stimulating 27 page paper
which discusses prime-producing polynomials and related topics.
Contrapositively, $ p(x) $ is prime (irreducible) if it assumes a prime value
for large enough $ |x| $. Conversely Bouniakowski conjectured (1857)
that prime $ p(x) $ assume infinitely many prime values (except in trivial
cases where the values of $p$ have an obvious common divisor, e.g. $ 2 | x(x+1)+2$ ).
As an example, Polya-Szego popularized A. Cohn's irreduciblity test, which
states that $ p(x) \in {\mathbb Z}[x]$ is prime if  $ p(b) $
yields a prime in radix $b$ representation (so necessarily $0 \le p_i < b$).
For example $f(x) = x^4 + 6 x^2 + 1 \pmod p$ factors for all primes $p$,
yet  $f(x)$ is prime since  $f(8) = 10601$ octal $= 4481$ is prime.
Note: Cohn's test fails if, in radix $b$, negative digits are allowed, e.g.
$f(x) = x^3 - 9 x^2 + x-9 = (x-9)(x^2 + 1)$ but $f(10) = 101$ is prime.
For further discussion see my prior post [1], along with Murty's online paper [2].
[1] Dubuque, sci.math 2002-11-12, On prime producing polynomials
http://groups.google.com/groups?selm=y8zvg4m9yhm.fsf%40nestle.ai.mit.edu 
[2] Murty, Ram.  Prime numbers and irreducible polynomials.
Amer. Math. Monthly, Vol. 109 (2002), no. 5, 452-458.
http://www.mast.queensu.ca/~murty/polya4.dvi 
[3] Mott, Joe L.; Rose, Kermit.
Prime producing cubic polynomials
Ideal theoretic methods in commutative algebra, 281-317.
Lecture Notes in Pure and Appl. Math., 220, Dekker, New York, 2001.
http://web.math.fsu.edu/~aluffi/archive/paper134.ps

REPLY [4 votes]: It is conjectured there are infinitely many primes of the form $n^2 + 1$ but not proven.