CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-16

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A coprime exponent gives a unique nonzero $k$ -th root modulo a prime

Statement

Let $p$ be prime, let $k \geq 1$ , and let $a \in Z$ satisfy $p ∤ a$ and $\gcd (k, p - 1) = 1$ . Then $x^{k} \equiv a (m o d p)$ has a unique nonzero solution class. If $ℓ \geq 0$ and

$k ℓ \equiv 1 (m o d p - 1),$

then that class is $[a^{ℓ}]_{p}$ ; the formula is independent of the chosen nonnegative representative $ℓ$ of the inverse class.

Facts & Assumptions

Given: A prime $p$ , an integer $k \geq 1$ , and $a \in Z$ with $p ∤ a$ and $\gcd (k, p - 1) = 1$ .

[L1]

If $n \geq 1$ admits a primitive root, $\gcd (a, n) = 1$ , $m \geq 1$ , and $d = \gcd (m, φ (n))$ , then $x^{m} \equiv a (m o d n)$ is solvable if and only if $a^{φ (n) / d} \equiv 1 (m o d n)$ (Euler's criterion: if $n$ has a primitive root, $\gcd (a, n) = 1$ , and $m \geq 1$ , then $x^{m} \equiv a (m o d n)$ is solvable if and only if $a^{φ (n) / \gcd (φ (n), m)} \equiv 1 (m o d n)$ ).

[L2]

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

[L3]

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

[L4]

Under the hypotheses of [L1], a soluble congruence $x^{m} \equiv a (m o d n)$ 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]

For $u, v \in Z$ and $n \geq 1$ , the congruence $u y \equiv v (m o d n)$ is soluble exactly when $\gcd (u, n)$ divides $v$ , and then it has exactly $\gcd (u, n)$ 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 $n$ ).

[L6]

If $p$ is prime and $p ∤ a$ , then $a^{p - 1} \equiv 1 (m o d p)$ (Fermat's little theorem: for prime $p$ , $p ∤ a$ implies $a^{p - 1} \equiv 1 (m o d p)$ , and always $a^{p} \equiv a (m o d p)$ ).

[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

1.1L1L2L3L6L7given

By [L7], $\gcd (a, p) = 1$ . Facts [L2] and [L3] specialise [L1] to modulus $p$ , while $\gcd (k, p - 1) = 1$ reduces its test to $a^{p - 1} \equiv 1 (m o d p)$ , which holds by [L6]. Thus the root congruence is soluble.

2.1L2L3L4step 1.1given

Applying [L4] with $n = p$ and $m = k$ gives exactly $\gcd (p - 1, k) = 1$ solution class, so the root is unique. It is nonzero because a zero root would give $x^{k} \equiv 0 (m o d p)$ , contrary to $p ∤ a$ .

3.1L5L6L7step 2.1algebra∎

By [L5], the congruence $k ℓ \equiv 1 (m o d p - 1)$ has one solution class and has nonnegative representatives. If $p = 2$ , every such $a^{ℓ}$ is odd, so it represents the sole unit class and hence the unique root from step 2.1. Suppose $p$ is odd. For a nonnegative representative $ℓ$ , the congruence forces $ℓ > 0$ , so $k ℓ = 1 + t (p - 1)$ with $t \geq 0$ ; then [L6] gives $(a^{ℓ})^{k} = a^{1 + t (p - 1)} \equiv a (m o d p)$ . Two nonnegative representatives differ by a multiple of $p - 1$ ; ordering them and applying [L6] to the nonnegative difference shows that their powers of $a$ represent the same class.

Depends on

Used by

The unique cube root of 2 modulo 29 is -3 Example

Dependency tree · two levels

40 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

N. Mascot, Legendre Symbols and Quadratic Reciprocity, Chapter 3 (standard reference, not scraped)

A coprime exponent gives a unique nonzero kk-th root modulo a prime