Alphabeta Math
ExampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passaudited 2026-08-16
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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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2 is primitive modulo every power of 5

Example

For every k≥1, the class of 2 is a primitive root modulo 5k.

Facts & Assumptions

Given: The integer 2 and powers of the odd prime 5.

[L1]

For an odd prime p, an integer u with p∤u, and k≥1, the class of 1+pu has order pk−1 modulo pk (For odd prime p, p∤u, and k≥1, the class of 1+pu has order pk−1 modulo pk).

[L2]

If an element has order r, then its fourth power has order r/gcd⁡(r,4) (In a cyclic group of order m, ga has order m/gcd⁡(a,m)).

[L5]

A class modulo n is a unit exactly when its representative is coprime to n, and a primitive root is a unit of order φ(n) (For n≥1, [a]n is a unit if and only if gcd⁡(a,n)=1, Primitive roots modulo n).

Verification

technique · direct
1.1L3L4L5algebra

Modulo 5, 21,22,23,24 are 2,4,3,1, so 2 has order 4. This settles k=1.

1.2L1L2

Let k≥2 and let r be the order of 2 modulo 5k. Since 24=1+5⋅3, [L1] says that 24 has order 5k−1. Hence [L2] gives r/gcd⁡(r,4)=5k−1.

2.1step 1.1step 1.2L3L4L5∎

Reduction modulo 5 and step 1.1 show that 4∣r by [L3]. Therefore gcd⁡(r,4)=4, and step 1.2 yields r=4⋅5k−1=φ(5k) by [L4]. Since 2 is a unit by [L5], it is primitive by the definition in [L5].

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources