ExampleConstruction: AI-adaptedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-16

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$2$ is a primitive root modulo $13$ by testing the prime divisors of $12$

Example

The class of $2$ is a primitive root modulo $13$ .

Facts & Assumptions

Given: The prime modulus $13$ and the unit class of $2$ .

[L1]

A unit is a primitive root modulo $n$ when its order is $φ (n)$ (Primitive roots modulo $n$ ).

[L2]

$φ (13) = 12$ ( $φ (1) = 1$ , and $φ (p) = p - 1$ for every prime $p$ ).

[L3]

If an element has finite order $r$ , its powers equal the identity exactly at exponents divisible by $r$ (If $ord (g) = n$ then $g^{k} = e$ iff $k$ is an integer multiple of $n$ , the powers $g^{0}, \dots, g^{n - 1}$ are distinct, and $⟨ g ⟩$ has exactly $n$ elements; if $g$ has infinite order then $g^{j} = g^{k}$ only for $j = k$ ).

Verification

technique · direct

1.1

Direct calculation gives $2^{4} \equiv 3 \equiv̸ 1$ and $2^{6} \equiv - 1 \equiv̸ 1 (m o d 13)$ ; in particular $2^{12} \equiv 1 (m o d 13)$ .

algebra

2.1

By step 1.1 and [L3], the order divides $12$ . Every proper divisor of $12$ divides either $12 / 2 = 6$ or $12 / 3 = 4$ , so step 1.1 and [L3] exclude every proper divisor. The order is therefore $12 = φ (13)$ by [L2], and [L1] makes $2$ primitive.

step 1.1L1L2L3∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 84 results over 22 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

William Stein, Elementary Number Theory, Example 2.5.9 (standard reference, not scraped)

22 is a primitive root modulo 1313 by testing the prime divisors of 1212