Polynomials that pass through a lot of primes Whenever I mention a polynomial $p(x)$ passes through $n$, I mean there is an integer $z$ so that $p(z)=n$, also any polynomial I talk about is assumed to have integer coefficients.
It's a well known open to question to find a polynomial of degree 2 or more that passes through an infinite amount of primes.
Do we know of general polynomials of degree n that pass through at least f(n) primes where f isn't constant (so f could be like log(n) or 1.5*n)?
Additionally, do we know if the number of primes a quadratic pass through is unbounded (and similarly for other degrees)? Meaning if we look at $s(p(x))$ : the number of primes $p(x)$ passes through, where $p(x)$ is a quadratic, is $s(p(x))$ bounded?
 A: Your question is very difficult to answer in full.
For the quadratic case there are partial answers. I have kept, among others and for some time now, an article of Betty Garrison (San Diego State University. A.M.M. 97 (1990) p. 316-17) in which there is the following improvement of Sierpinski's result:
THEOREM. Let $k\ge 2$  be an integer and let $ M $ an arbitrarily large number. Then there exist positive integers $c,d$ such that  $x^k+c$, likewise $x^k-d$ is prime for more than $M$ positive integers $x$
The famous conjecture of Buniakowski (1854) say that for any irreducible polynomial $ f(x)\in \mathbb  Z [x]$ such that the set of values f (n) has no common divisor larger than $1$ ($f(x)=x^3+x+2$ is irreducible but $f(n)$ is always even) there are infinitely many primes $f(n)$. This conjecture is still one of the major unsolved problems in number theory when the degree of $f$ is greater than one. For degree 1 one has  Dirichlet’s theorem on primes in arithmetic progressions (obviously $ax+b; (a,b)=1$, is irreducible);  in other words, Buniakowski’s conjecture wants to generalize this celebrated theorem. 
A: Your last question is answered by a classical and beautiful theorem of Sierpinski:
For any integer $m \ge 1$, there exists a constant $k$ such that the quadratic $n^2 + k$ passes through at least $m$ primes.
Thus $s(p(x))$ is unbounded even when $p(x)$ is restricted to the simplest possible family of quadratic polynomials.
A: Getting an integer polynomial of degree $n$ where $p(0),p(1),\dots,p(n)$ are all prime is a little work, but relatively simple with a shout-out to Dirichlet's theorem on primes in arithmetic progressions.
Given a fixed $n$, we will find a sequence $p_k(z)$ of polynomials for $k=0,1,\dots,n$ such that $p_k$ is of degree at most $k$ and $p_k(i)$ is prime and bigger than $n$ for $0\leq i\leq k$. 
For $k=0$, we find a prime $q>n$, and define $p_0(z)=q$.
Then, given a $p_k(z)$ with $k<n$, we define $p_{k+1}(z)=p_k(z)+a_{k+1} z(z-1)\dots(z-k)$. We need to find $a_{k+1}$ so that $p_k(k+1)+a_{k+1} (k+1)!$ is a prime bigger than $n$.

Claim: $p_k(k+1)$ is relatively prime to $(k+1)!$. 
Proof: If $1\leq d\leq k+1$, then $$p_k(k+1)\equiv p_k(k+1-d)\pmod {d}.$$ Since $p_k(k+1-d)$ is a prime bigger than $n$, and hence bigger than $k$, then $p_k(k+1)$ is relatively prime to $d$.

So, by Dirichlet, we can pick an integer $a_k$ so that $p_k(k+1)+a_{k+1} (k+1)!$ is a prime.
Then $p_{k+1}(z)=p_k(z)+a_k z(z-1)(z-2)\dots(z-k)$ is of degree $k+1$ and has the property that $p_{k+1}(i)$ is prime for $i=0,1,2,\dots,k$.
