You're given some integer-valued irreducible quadratic function $f(x):=ax^2+bx+c$ with positive lead coefficient $a$, but you don't know anything else about it. However, you're able to get the divisor count for as many values of $x$ as you'd like with $\sigma_0(f(x))$.
I'm curious whether it is possible to identify the specific function you've been given using finitely many of these divisor counts, but its usage is otherwise unrestricted. If it's relevant, assume the function meets the criteria to be able to yield primes.
It seems to me that whether this task is possible may end up depending on the infinitude of polynomial primes, but I worry I'm overlooking more obvious solutions, or properties that make it trivially impossible.
Lets try one out: $$f(x)=23x^2+3x+3$$ f(1) is prime so: $$\sigma_0(f(1))=2$$ to find out which primes are possibly f(1) we can test values that are 1 more than primes until we hit one with more than 2 (guaranteed to happen as $f(f(1)+1)$ has the same remainder on division by $f(1)$ as $f(1)$ does by polynomial remainder theorem). This at least gives us a minimum value of $a+b+c$. So lets try that:
$$\sigma_0(f(3))=4\\\sigma_0(f(4))=2\\\sigma_0(f(6))=4\\\sigma_0(f(8))=2\\\sigma_0(f(12))=4\\\sigma_0(f(14))=4\\\sigma_0(f(18))=4\\\sigma_0(f(20))=4\\\sigma_0(f(24))=4\\\sigma_0(f(30))=8\\\vdots$$
Okay not so useful it seems, f(2) can be used showing $3a+b$ to be a difference of primes. This is only even ( the most likely case unless one of the primes is 2) if $a$ and $b$ are of the same parity. As 2 is even f(2) has the same parity as $c$. $f(4)$,$f(8)$, and $f(38)$ Also create differences in primes etc.