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The discriminant counts roots of $A x^{2} + B x + C \equiv 0 (m o d p)$ for odd prime $p ∤ A$

Statement

Let $p$ be an odd prime, and let $A, B, C \in Z$ with $p ∤ A$ . Put $Δ = B^{2} - 4 A C$ . Then

$A x^{2} + B x + C \equiv 0 (m o d p)$

has exactly

$1 + (\frac{Δ}{p})$

solution classes modulo $p$ .

Facts & Assumptions

Given: An odd prime $p$ and integers $A, B, C$ with $p ∤ A$ ; write $Δ = B^{2} - 4 A C$ .

[L1]

Addition and multiplication make $Z / p$ a commutative ring (For every natural $n$ , $(Z / n, +)$ is an abelian group, multiplication is a commutative monoid operation, and both distributive laws hold).

[L2]

A class $[u]_{p}$ is a unit exactly when $\gcd (u, p) = 1$ (For $n \geq 1$ , $[a]_{n}$ is a unit if and only if $\gcd (a, n) = 1$ ).

[L3]

If $u, v \in Z$ , $n \geq 1$ , and $d = \gcd (u, n)$ , then $u x \equiv v (m o d n)$ is soluble exactly when $d ∣ v$ , and when soluble it has exactly $d$ solution classes (For $n \geq 1$ , $a x \equiv b (m o d n)$ is solvable exactly when $\gcd (a, n) ∣ b$ , and then has exactly $\gcd (a, n)$ solution classes modulo $n$ ).

[L4]

For every integer $d$ and odd prime $p$ , the congruence $y^{2} \equiv d (m o d p)$ has exactly $1 + (d / p)$ solution classes ( $x^{2} \equiv a (m o d p)$ has exactly $1 + (a / p)$ solution classes).

Proof

technique · direct

1.1

In the commutative ring [L1], the identity $(2 A x + B)^{2} - Δ = 4 A (A x^{2} + B x + C)$ holds.

L1givenalgebra

2.1

Since $p$ is odd and $p ∤ A$ , the prime $p$ divides neither $2 A$ nor $4 A$ , so [L2] makes both classes units. Fact [L3] then says that $x \mapsto 2 A x + B$ is a bijection of the residue classes, and cancellation of the unit $4 A$ in step 1.1 shows that the original congruence is equivalent to $(2 A x + B)^{2} \equiv Δ (m o d p)$ .

L2L3step 1.1

3.1

The bijection in step 2.1 preserves the number of solutions, and [L4] gives exactly $1 + (Δ / p)$ solutions to the square congruence.

L4step 1.1step 2.1∎

Depends on

Used by

A nonmonic quadratic congruence solved through its discriminant Example

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 80 results over 19 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

A. Gorodnik, Number Theory, Lecture 9, Section 1 (standard reference, not scraped)

The discriminant counts roots of Ax2+Bx+C≡0(modp)Ax2+Bx+C≡0(modp) for odd prime p∤Ap∤A