Primes modulo which a given quadratic equation has roots Given a quadratic polynomial $ax^2 + bx + c$, with $a$, $b$ and $c$ being integers, is there a characterization of all primes $p$ for which the equation
$$ax^2 + bx + c \equiv 0 \pmod p$$
has solutions?
I have seen it mentioned that it follows from quadratic reciprocity that the set is precisely the primes in some arithmetic progression, but the statement may require some tweaking. The set of primes modulo which $1 + \lambda = \lambda^2$ has solutions seems to be $$5, 11, 19, 29, 31, 41, 59, 61, 71, 79, 89, 101, 109, 131, 139, 149, 151, 179, 181, 191, 199, \dots$$ which are ($5$ and) the primes that are $1$ or $9$ modulo $10$.
(Can the question also be answered for equations of higher degree?)

I never noticed this one before.
$$  x^3 - x - 1 \equiv 0 \pmod p  $$
has one root for odd primes $p$ with $(-23|p) = -1.$
$$  x^3 - x - 1 \equiv 0 \pmod p  $$
has three distinct roots for odd $p$ with $(-23|p) = 1$ and $p = u^2 + 23 v^2 $ in integers.
$$  x^3 - x - 1 \equiv 0 \pmod p  $$
has no roots for odd $p$ with $(-23|p) = 1$ and $p = 3u^2 + 2 u v + 8 v^2 $ in integers (not necessarily positive integers).
Here we go, no roots $\pmod 2,$ but a doubled root and a single $\pmod {23},$ as
$$  x^3 - x - 1 \equiv (x - 3)(x-10)^2 \pmod {23}. $$
Strange but true. Easy to confirm by computer for primes up to 1000, say. 
The example you can see completely proved in books, Ireland and Rosen for example, is $x^3 - 2,$ often with the phrase "the cubic character of 2" and the topic "cubic reciprocity." $2$ is a cube for primes $p=2,3$ and any prime $p \equiv 2 \pmod 3.$ Also, $2$ is a cube for primes $p \equiv 1 \pmod 3$ and $p = x^2 + 27 y^2$ in integers. However, $2$ is not a cube for primes $p \equiv 1 \pmod 3$ and $p = 4x^2 +2 x y + 7 y^2$ in integers. (Gauss)
$3$ is a cube for primes $p=2,3$ and any prime $p \equiv 2 \pmod 3.$ Also, $3$ is a cube for primes $p \equiv 1 \pmod 3$ and $p = x^2 + x y + 61 y^2$ in integers. However, $3$ is not a cube for primes $p \equiv 1 \pmod 3$ and $p = 7x^2 +3 x y + 9 y^2$ in integers. (Jacobi)
