Is there a uniform family of polynomials $f_p(x) =x^2 + a(p)x + b(p)$ such that $f_p(x)\in \mathbb{Z}[x]$ is irreducible and irreducible mod $p$?

Question

Let $p\in\mathbb{Z}$ be a positive prime number.

Is there a "uniform" family of polynomials $f_p(x) =x^2 + a(p)x + b(p)$ of degree two such that $f_p(x)\in \mathbb{Z}[x]$ is both irreducible and irreducible mod $p$ ?

If $p=4k+3$ for some positive integer $k$ then $f_p(x):=x^2+1$ does the trick and this does not even depend on $p$.

By the Minkowski bound (see,e.g., https://en.wikipedia.org/wiki/Minkowski%27s_bound), it is not possible to find such a polynomial which stays irreducible mod $p$ for all prime numbers $p$.

What happens if $p=4k+1$ ?

Doe we even have a "uniform" family of polynomials $f_p(x) =x^2 + a(p)x + b(p)$ which works for all (odd) primes $p$ ?

Edit: "uniform" should be interpreted as follows: are there $a(t),b(t)\in\mathbb{Z}[t]$ so that the statement is true?

Thank you for the help.

Answer 1

No, this can't happen: For any polynomials $a,b \in \mathbb Z[t]$, there are infinitely many primes $p$ such that $f_p(x) := x^2 + a(p)x + b(p)$ is reducible modulo $p$.

Let $a_0 := a(0)$ and $b_0 := b(0)$. Then we have $f_p(x) \equiv x^2 + a_0x + b_0 \pmod p$, which is reducible (for odd $p$) if and only if $\Delta := a_0^2 - 4b_0$ is a square modulo $p$. But any integer $\Delta$ is a quadratic residue for infinitely many primes $p$, so $f_p$ will be reducible modulo infinitely many primes $p$.