LemmaStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-07-31

How statement and proof provenance work

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.
AI-generated — a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

For a prime $p$ and $k\ge1$ , multiplication by $p$ bijects the standard representatives modulo $p^{k-1}$ with the representatives modulo $p^k$ divisible by $p$

Statement

Let $p$ be prime and let $k\in\mathbb N$ with $k\ge1$ . Write $k-1$ for the unique natural $j$ with $j+1=k$ . Multiplication by $p$ gives a bijection

$\{\,r\in\mathbb Z:0\le r<p^{k-1}\,\}\longrightarrow\{\,s\in\mathbb Z:0\le s<p^k,\ p\mid s\,\},\qquad r\longmapsto pr.$

Thus the standard representatives modulo $p^k$ divisible by $p$ are exactly $0,p,2p,\ldots,(p^{k-1}-1)p$ , and there are $p^{k-1}$ of them.

Facts & Assumptions

Given: A prime $p$ and a natural $k\ge1$ .

[F1]

Natural powers satisfy $p^0=1$ and $p^k=p^{k-1}p$ (Powers $g^{n}$ : natural exponents in a monoid and integer exponents in a group, with $g^{0} = e$ , Exponent laws in a group: $g^{m+n} = g^{m}g^{n}$ and $(g^{m})^{n} = g^{mn}$ for all $m, n \in \mathbb{Z}$ , and $(gh)^{n} = g^{n}h^{n}$ when $g$ and $h$ commute).

[F2]

For a positive modulus $m$ , the standard representatives are the integers $r$ with $0\le r<m$ , and there are $m$ of them (For $n\ge 1$ , every class in $\mathbb{Z}/n$ has one representative $r$ with $0\le r<n$ , so $\lvert\mathbb{Z}/n\rvert=n$ ; while $\mathbb{Z}/0$ is in bijection with $\mathbb{Z}$ , The cardinality $\lvert A\rvert$ of a finite set).

[F3]

A map is bijective when it is injective and surjective (Injection, surjection, bijection).

[L1]

If $px=py$ and $p\ne0$ , then $x=y$ (The integers have no zero divisors; multiplicative cancellation).

[F4]

A prime satisfies $p>1$ , hence $p>0$ and $p\ne0$ (Prime and composite integers: $p$ is prime when $p > 1$ and its only positive divisors are $1$ and $p$ ).

Proof

technique · direct

1.1

If $0\le r<p^{k-1}$ , then $0\le pr<p\cdot p^{k-1}=p^k$ , and $p\mid pr$ . Thus the displayed rule has values in the stated codomain.

F1F4algebra

1.2

The rule is injective: $pr=pr'$ implies $r=r'$ because $p\ne0$ .

L1F4

1.3

It is surjective: if $0\le s<p^k$ and $p\mid s$ , write $s=pr$ . Since $p>0$ , the inequalities give $0\le r<p^{k-1}$ after using $p^k=pp^{k-1}$ .

F1F4algebrachoose

2.1

Steps 1.1, 1.2 and 1.3 give a bijection, and [F2] transports the domain cardinality $p^{k-1}$ to the codomain.

step 1.1step 1.2step 1.3F2F3∎

Depends on

Used by

For a prime p and k≥1, φ(pᵏ)=pᵏ-pᵏ⁻¹ Theorem

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 80 results over 24 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

Euler's totient function (Wikipedia) (standard reference, not scraped)
Mathematics LibreTexts, Euler's phi Function (standard reference, not scraped)
K. Conrad, The Chinese Remainder Theorem (standard reference, not scraped)
Carnegie Mellon, number-theory lecture notes (standard reference, not scraped)

For a prime ppp and k≥1k\ge1k≥1, multiplication by ppp bijects the standard representatives modulo pk−1p^{k-1}pk−1 with the representatives modulo pkp^kpk divisible by ppp