[Paper Review] Constructing the Primitive Roots of Prime Powers

Nathan Jolly|ArXiv.org|Sep 12, 2008

History and Theory of Mathematics7 references3 citations

TL;DR

This paper presents a constructive method to generate all primitive roots of $p^{k+1}$ from the primitive roots of $p^k$, where $p$ is an odd prime and $k \geq 2$, using only addition and multiplication modulo $p^{k+1}$. The key result shows that for each primitive root $g$ of $p^k$, the $p$ values $g + tp^k$ for $0 \leq t \leq p-1$ are all primitive roots of $p^{k+1}$, and collectively generate all $\varphi(\varphi(p^{k+1}))$ primitive roots without repetition.

ABSTRACT

We use only addition and multiplication to construct the primitive roots of $p^{k+1}$ from the primitive roots of $p^{k}$, where $p$ is an odd prime and $k$ is at least 2.

Motivation & Objective

• To provide an explicit, elementary construction of all primitive roots of $p^{k+1}$ using only addition and multiplication, avoiding exponentiation.
• To extend Niven et al.'s construction for $p^2$ to higher prime powers $p^{k+1}$ with $k \geq 2$.
• To demonstrate that all $\varphi(\varphi(p^{k+1}))$ primitive roots of $p^{k+1}$ can be generated from the primitive roots of $p^k$ via linear lifts $g + tp^k$.
• To establish that this construction yields exactly $p \cdot \varphi(\varphi(p^k))$ distinct primitive roots, matching the known count $\varphi(\varphi(p^{k+1}))$.

Proposed method

• Lift each primitive root $g$ of $p^k$ to $p$ candidates $g + tp^k$ modulo $p^{k+1}$ for $0 \leq t \leq p-1$.
• Use Hensel's Lemma to analyze the lifting behavior of solutions to $x^{\varphi(p^k)} \equiv 1 \pmod{p^{k+1}}$, showing that $g$ does not lift to solutions when $k \geq 2$.
• Prove that $\mathrm{ord}_{p^{k+1}}(g + tp^k) = \varphi(p^{k+1})$ by showing the order cannot be $\varphi(p^k)$, relying on the derivative condition $f'(g) \equiv 0 \pmod{p}$.
• Verify that the constructed values $g + tp^k$ are all distinct modulo $p^{k+1}$ and cover all primitive roots without duplication.
• Confirm that the total number of constructed roots $p \cdot \varphi(\varphi(p^k))$ equals $\varphi(\varphi(p^{k+1}))$, ensuring completeness.
• Use the fact that $g$ is a primitive root of $p^k$ and $k \geq 2$ to ensure $g^{\varphi(p^k)} \not\equiv 1 \pmod{p^{k+1}}$, which rules out order $\varphi(p^k)$.

Experimental results

Research questions

• RQ1 Can all primitive roots of $p^{k+1}$ be constructed from the primitive roots of $p^k$ using only addition and multiplication, without exponentiation?
• RQ2 For $k \geq 2$, does every primitive root $g$ of $p^k$ generate $p$ distinct primitive roots of $p^{k+1}$ via the form $g + tp^k$?
• RQ3 Why does the lifting $g + tp^k$ avoid producing elements of order $\varphi(p^k)$ modulo $p^{k+1}$, and how does this ensure primitivity?
• RQ4 Is the number of such constructed roots equal to the known count $\varphi(\varphi(p^{k+1}))$?
• RQ5 What role does the condition $k \geq 2$ play in ensuring the validity of the construction, particularly in relation to Hensel's Lemma?

Key findings

• For each primitive root $g$ of $p^k$ with $k \geq 2$, all $p$ values $g + tp^k$ for $0 \leq t \leq p-1$ are primitive roots of $p^{k+1}$, and no others are generated.
• The construction produces exactly $p \cdot \varphi(\varphi(p^k))$ distinct primitive roots of $p^{k+1}$, which matches the known count $\varphi(\varphi(p^{k+1}))$.
• The order of $g + tp^k$ modulo $p^{k+1}$ is never $\varphi(p^k)$, which is essential for ensuring primitivity, and this is proven via Hensel's Lemma and the vanishing derivative $f'(g) \equiv 0 \pmod{p}$.
• The construction fails for $k=1$ due to the derivative condition not being sufficient to block lifting, which justifies the requirement $k \geq 2$.
• The method is fully constructive and elementary: it uses only addition and multiplication, avoiding exponentiation, making it suitable for hand computation.
• The result generalizes Niven et al.'s construction for $p^2$ to all $p^{k+1}$ with $k \geq 2$, providing a uniform lifting mechanism.