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$x^{2} \equiv a (m o d p)$ has exactly $1 + (a / p)$ solution classes

Statement

For every integer $a$ and odd prime $p$ , the congruence

$x^{2} \equiv a (m o d p)$

has exactly $1 + (a / p)$ solution classes modulo $p$ .

Facts & Assumptions

Given: An integer $a$ and an odd prime $p$ .

[L1]

The Legendre symbol is $0$ when $p ∣ a$ , $1$ when $a$ is a quadratic residue modulo $p$ , and $- 1$ when $a$ is a quadratic nonresidue (The Legendre symbol, including its zero value).

[L2]

The quotient $Z / p$ is a field (For every prime $p$ , the two operations on $Z / p$ make it a field).

[L3]

Every field is an integral domain (Every field is a commutative ring with $1 \neq 0$ ; it is an integral domain, and it is a commutative division ring).

[L4]

Under its primitive-root, coprimality, positivity, and solubility hypotheses, $x^{m} \equiv a (m o d n)$ has exactly $\gcd (φ (n), m)$ solution classes (If $n$ has a primitive root, $\gcd (a, n) = 1$ , $m \geq 1$ , and $x^{m} \equiv a (m o d n)$ is solvable, then it has exactly $\gcd (φ (n), m)$ solution classes modulo $n$ ).

[L5]

Every prime admits a primitive root modulo that prime (Every prime modulus admits a primitive root).

[L6]

For every prime $p$ , $φ (p) = p - 1$ ( $φ (1) = 1$ , and $φ (p) = p - 1$ for every prime $p$ ).

[L7]

If $p ∤ a$ , then $p$ and $a$ are coprime (For a prime $p$ and any integer $a$ , $\gcd (p, a)$ is $p$ when $p ∣ a$ and $1$ otherwise; so $p ∤ a$ makes $p$ and $a$ coprime).

Proof

technique · direct

1.1L1L2L3given

If $p ∣ a$ , the equation in the field [L2] is $[x]_{p}^{2} = [0]_{p}$ . Since [L3] gives no zero divisors, $[x]_{p} = [0]_{p}$ is the unique solution. This count is $1 + 0$ by [L1].

1.2L1L4L5L6L7algebra

Suppose $p ∤ a$ . By [L7], $\gcd (a, p) = 1$ . If $(a / p) = 1$ , the congruence is soluble by [L1], and [L4], [L5], and [L6] give exactly $\gcd (p - 1, 2) = 2$ roots. If $(a / p) = - 1$ , [L1] says that no root exists.

2.1L1step 1.1step 1.2∎

The three possible symbol values $0, 1, - 1$ therefore give respectively one, two, and zero solution classes, which in every case equals $1 + (a / p)$ .

Depends on

Used by

The discriminant counts roots of Ax²+Bx+C≡0 (mod p) for odd prime p∤ A Corollary
A soluble square congruence need not define a quadratic residue Counterexample
Unit square criterion and root count modulo odd prime powers Theorem

Dependency tree · two levels

46 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

W. Stein, Elementary Number Theory, Corollary 4.2.3 (standard reference, not scraped)
A. Gorodnik, Number Theory, Lecture 9, Section 1 (standard reference, not scraped)

x2≡a(modp)x2≡a(modp) has exactly 1+(a/p)1+(a/p) solution classes