LemmaStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-16

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For an odd prime $p$ and a primitive root $g$ modulo $p$ , at least one of $g$ and $g + p$ is primitive modulo $p^{2}$

Statement

Let $p$ be an odd prime and let the integer $g$ represent a primitive root modulo $p$ . Then at least one of $g$ and $g + p$ represents a primitive root modulo $p^{2}$ .

Facts & Assumptions

Given: An odd prime $p$ and a primitive root $g$ modulo $p$ .

[L1]

A class is a primitive root when its order equals the totient of the modulus (Primitive roots modulo $n$ ), and it is a unit exactly when its representative is coprime to the modulus (For $n \geq 1$ , $[a]_{n}$ is a unit if and only if $\gcd (a, n) = 1$ ).

[L2]

The order of an element of a finite group divides the group order (The order of every element of a finite group divides the order of the group), and $φ (p^{2}) = p (p - 1)$ (For a prime $p$ and $k \geq 1$ , $φ (p^{k}) = p^{k} - p^{k - 1}$ ).

[L3]

If an element has finite order $r$ , a power is the identity exactly when its exponent is 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 $j = k$ ).

[L4]

Group powers satisfy the usual addition and iteration laws (Exponent laws in a group: $g^{m + n} = g^{m} g^{n}$ and $(g^{m})^{n} = g^{m n}$ for all $m, n \in Z$ , and $(g h)^{n} = g^{n} h^{n}$ when $g$ and $h$ commute).

[L5]

Congruence modulo $p^{2}$ means divisibility of the difference by $p^{2}$ (Congruence modulo an integer: $a \equiv b (m o d n)$ when $n ∣ (a - b)$ , including the moduli $0$ and $1$ ).

[L6]

Mathematical induction holds on $N$ (The principle of mathematical induction).

Proof

technique · cases

1.1

Both $g$ and $g + p$ are units modulo $p^{2}$ by [L1] and reduce to the same primitive root modulo $p$ . If either has order $r$ modulo $p^{2}$ , reduction modulo $p$ makes its $r$ th power $1$ , so $p - 1 ∣ r$ by [L1] and [L3]. By [L2], $r$ also divides $p (p - 1)$ , and hence $r$ is either $p - 1$ or $p (p - 1)$ .

givenL1L2L3algebra

1.2

For every $r \geq 1$ , induction using the product law gives $(g + p)^{r} \equiv g^{r} + r p g^{r - 1} (m o d p^{2})$ .

L4L5L6

2.1

Assume first that $g^{p - 1} \equiv̸ 1 (m o d p^{2})$ . Then step 1.1 excludes order $p - 1$ , so $g$ has order $p (p - 1)$ and is primitive modulo $p^{2}$ .

assume-case firststep 1.1

2.2

Assume instead that $g^{p - 1} \equiv 1 (m o d p^{2})$ . Step 1.2 gives $(g + p)^{p - 1} \equiv 1 + p (p - 1) g^{p - 2} (m o d p^{2})$ ; the second term is not divisible by $p^{2}$ , since neither $p - 1$ nor $g$ is divisible by $p$ . Thus $(g + p)^{p - 1} \equiv̸ 1 (m o d p^{2})$ , and step 1.1 makes $g + p$ primitive.

assume-case secondstep 1.1step 1.2L1L5

3.1

The two cases are exhaustive, and in each one of the two representatives is primitive modulo $p^{2}$ .

step 2.1step 2.2cases-exhaustive∎

Depends on

Used by

For every odd prime p and k≥1, (ℤ/pᵏℤ)^× is cyclic of order pᵏ⁻¹(p-1) Theorem

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 121 results over 28 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

Peter Hackman, Elementary Number Theory, Lemma C.IV.1 (standard reference, not scraped)

For an odd prime pp and a primitive root gg modulo pp, at least one of gg and g+pg+p is primitive modulo p2p2