LemmaStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-28

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 a nonzero integer $a$ : $p^{v_{p} (a)} ∣ a$ and $p^{v_{p} (a) + 1} ∤ a$ ; $p^{k} ∣ a$ holds exactly for $k \leq v_{p} (a)$ ; $v_{p} (a) \geq 1$ exactly when $p ∣ a$ ; $v_{p} (1) = v_{p} (- 1) = 0$ ; and $v_{p} (p) = 1$

Statement

Let $p$ be a prime (Prime and composite integers: $p$ is prime when $p > 1$ and its only positive divisors are $1$ and $p$ ), let $a \in Z$ with $a \neq 0$ , and write $v : = v_{p} (a)$ (The $p$ -adic valuation $v_{p} (a)$ of a nonzero integer: the greatest $k \in N$ with $p^{k} ∣ a$ ). Powers are the natural powers of Powers $g^{n}$ : natural exponents in a monoid and integer exponents in a group, with $g^{0} = e$ in the commutative monoid $(Z, \cdot, 1)$ ( $(Z, \cdot, 1)$ is a commutative monoid whose group of units is ${1, - 1}$ ; equivalently $u ∣ 1$ holds exactly for $u = 1$ and $u = - 1$ ), and $k + 1$ means $σ (k)$ . Then:

$p^{k} \geq 1$ for every $k \in N$ ; in particular $p^{k} > 0$ and $p^{k} \neq 0$ ;
$p^{v} ∣ a$ and $p^{v + 1} ∤ a$ ;
for $k \in N$ : $p^{k} ∣ a$ if and only if $k \leq v$ ;
$a = p^{v} a^{'}$ for some $a^{'} \in Z$ with $a^{'} \neq 0$ and $p ∤ a^{'}$ ;
$v \geq 1$ if and only if $p ∣ a$ ;
$v_{p} (1) = v_{p} (- 1) = 0$ and $v_{p} (p) = 1$ .

Facts & Assumptions

Given: A prime $p$ , a nonzero integer $a$ , and $v : = v_{p} (a)$ .

[L1]

$v_{p} (a)$ is the greatest $k \in N$ with $p^{k} ∣ a$ ; in particular $p^{v} ∣ a$ , and $p^{k} ∣ a$ fails for every $k > v$ (The $p$ -adic valuation $v_{p} (a)$ of a nonzero integer: the greatest $k \in N$ with $p^{k} ∣ a$ ).

[L2]

$p^{0} = 1$ and $p^{σ (k)} = p^{k} \cdot p$ (Powers $g^{n}$ : natural exponents in a monoid and integer exponents in a group, with $g^{0} = e$ , Semigroup and monoid, $(Z, \cdot, 1)$ is a commutative monoid whose group of units is ${1, - 1}$ ; equivalently $u ∣ 1$ holds exactly for $u = 1$ and $u = - 1$ ).

[L3]

Exponent law in a monoid, for natural exponents: $g^{m + n} = g^{m} g^{n}$ (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).

[L4]

Divisibility is reflexive and transitive; $d ∣ u$ implies $d ∣ u c$ ; and $d ∣ u$ , $d ∣ - u$ , $- d ∣ u$ are equivalent (Divisibility is reflexive and transitive on $Z$ , and is linear: if $d ∣ a$ and $d ∣ b$ then $d ∣ a x + b y$ for all integers $x, y$ ; also $d ∣ a$ implies $d ∣ a c$ , $- d ∣ a$ and $d ∣ - a$ ).

[L5]

$d ∣ u$ means $u = d c$ for some $c \in Z$ (Divisibility in $Z$ : $d ∣ a$ when $a = d q$ for some integer $q$ ).

[L6]

$u ∣ 1$ exactly when $u = 1$ or $u = - 1$ ( $(Z, \cdot, 1)$ is a commutative monoid whose group of units is ${1, - 1}$ ; equivalently $u ∣ 1$ holds exactly for $u = 1$ and $u = - 1$ ).

[L7]

Every prime satisfies $p > 1$ (Prime and composite integers: $p$ is prime when $p > 1$ and its only positive divisors are $1$ and $p$ ).

[L8]

If $x z = y z$ and $z \neq 0$ then $x = y$ ; and a product of two nonzero integers is nonzero (The integers have no zero divisors; multiplicative cancellation).

[L9]

On $N$ : $m \leq n$ means $m + c = n$ for some $c$ (Order on the natural numbers); $σ (k) = k + 1$ (Addition of natural numbers, The natural numbers $N$ (von Neumann)); $k < σ (k)$ and $m < σ (n)$ exactly when $m \leq n$ (On $N$ the order is membership: $m < n ⟺ m \in n$ ); $m < n$ exactly when $σ (m) \leq n$ (Discreteness: $σ (n)$ is the immediate successor); and $\leq$ is a linear order, so it is antisymmetric and total ( $\leq$ is a linear order on $N$ ).

[L10]

$Z$ is a commutative ring, its order is total, antisymmetric and transitive and compatible with addition, and $ι : N \to Z$ is injective, order preserving, with image the nonnegative integers and $ι (0) = 0$ , $ι (1) = 1$ (The integers form a commutative ring, Arithmetic on the integers, The integers as equivalence classes of pairs of naturals, The integers form a totally ordered ring, Order on the integers, The naturals embed in the integers).

[L11]

Induction on $N$ : a subset containing $0$ and closed under $σ$ is all of $N$ (The principle of mathematical induction).

Proof

technique · direct

1.1L2L10

$p^{1} = p^{σ (0)} = p^{0} \cdot p = 1 \cdot p = p$ .

1.2L7L10

$0 < 1 < p$ in $Z$ , since $1 = ι (1)$ is nonnegative and differs from $0$ ; hence $p \neq 0$ , $p \neq 1$ and $p \neq - 1$ .

1.3L1L9

Clause 1. $p^{v} ∣ a$ by [L1], and $v < σ (v) = v + 1$ , so $p^{v + 1} ∣ a$ would contradict the maximality of $v$ ; hence $p^{v + 1} ∤ a$ .

1.4L1L9

Clause 2, the other direction. If $p^{k} ∣ a$ then $k$ belongs to the set of which $v$ is the greatest element, so $k \leq v$ .

2.1step 1.1step 1.2L2L9L10L11

Clause 0: $p^{k} \geq 1$ for every $k \in N$ . The set of such $k$ contains $0$ , since $p^{0} = 1$ ; and if $p^{k} \geq 1$ then $p^{σ (k)} = p^{k} p$ has both factors $\geq 1 > 0$ , so the product is positive, hence $\geq 1$ by discreteness of the order on $Z$ (a positive integer is $ι (j)$ with $j \neq 0$ , so $1 = σ (0) \leq j$ and $1 \leq ι (j)$ ). By induction the set is all of $N$ , and in particular $p^{k} > 0$ and $p^{k} \neq 0$ .

2.2step 1.3L1L3L4L5L9

Clause 2, one direction. Let $k \leq v$ and write $v = k + c$ with $c \in N$ . Then $p^{v} = p^{k + c} = p^{k} p^{c}$ , so $p^{k} ∣ p^{v}$ ; with $p^{v} ∣ a$ and transitivity, $p^{k} ∣ a$ .

2.3step 1.3L1L2L5L9L10choose

Clause 3. By [L1] fix $a^{'}$ with $a = p^{v} a^{'}$ . Then $a^{'} \neq 0$ , since $a \neq 0$ . If $p ∣ a^{'}$ , say $a^{'} = p b$ , then $a = p^{v} (p b) = (p^{v} p) b = p^{v + 1} b$ , so $p^{v + 1} ∣ a$ , contradicting step 1.3. Hence $p ∤ a^{'}$ .

2.4step 1.1L3L5L9

If $k \geq 1$ then $p ∣ p^{k}$ : write $k = 1 + c$ , so $p^{k} = p^{1} p^{c} = p \cdot p^{c}$ .

3.1step 1.1step 2.2step 1.4

Clause 4. If $v \geq 1$ then $p = p^{1} ∣ a$ by step 2.2. Conversely if $p ∣ a$ then $p^{1} ∣ a$ , so $1 \leq v$ by step 1.4.

3.2step 1.2step 2.4L1L2L4L6L9

$v_{p} (1) = 0$ . If $p^{k} ∣ 1$ with $k \geq 1$ , then $p ∣ p^{k} ∣ 1$ by step 2.4 and transitivity, so $p = 1$ or $p = - 1$ , contradicting step 1.2. So no $k \geq 1$ satisfies $p^{k} ∣ 1$ , while $p^{0} = 1 ∣ 1$ ; the greatest such $k$ is $0$ .

3.3step 1.1step 1.2step 2.2step 1.4L2L5L8L9

$v_{p} (p) = 1$ . First $p^{1} = p ∣ p$ , so $1 \leq v_{p} (p)$ by step 1.4 applied to $a : = p$ . Next $p^{2} ∤ p$ , where $2 = σ (1)$ : otherwise $p = p^{2} c = p (p c)$ for some $c$ , and cancelling $p \neq 0$ gives $1 = p c$ , so $p ∣ 1$ , contradicting step 1.2. Hence $v_{p} (p) < 2$ by step 2.2, that is $v_{p} (p) \leq 1$ , and antisymmetry gives $v_{p} (p) = 1$ .

4.1step 3.2L1L4

$v_{p} (- 1) = 0$ , because $p^{k} ∣ - 1$ holds exactly when $p^{k} ∣ 1$ , so the two sets of exponents coincide.

5.1step 1.3step 2.1step 2.2step 1.4step 2.3step 3.1step 3.2step 4.1step 3.3∎

Clauses 0 to 5 are established.

Remarks

Clause 3 is the working form. Every later proof on this page writes a nonzero integer as $a = p^{v_{p} (a)} a^{'}$ with $p ∤ a^{'}$ and then argues about $a^{'}$ ; that is how $v_{p} (a b) = v_{p} (a) + v_{p} (b)$ for nonzero integers $a, b$ , and $v_{p} (a + b) \geq \min {v_{p} (a), v_{p} (b)}$ whenever $a$ , $b$ and $a + b$ are all nonzero gets additivity, and it is where Euclid's lemma enters.
The hypothesis $a \neq 0$ is carried everywhere because $v_{p} (0)$ is not defined: every power of $p$ divides $0$ , so there is no greatest exponent (The $p$ -adic valuation $v_{p} (a)$ of a nonzero integer: the greatest $k \in N$ with $p^{k} ∣ a$ ).
Clause 2 says the exponents form an initial segment, which is what makes "the greatest $k$ with $p^{k} ∣ a$ " the same as "the number of times $p$ divides $a$ ". Without it the maximum could in principle skip values.

For a prime $p$ and a nonzero integer $a$ : $p^{v_{p} (a)} ∣ a$ and $p^{v_{p} (a) + 1} ∤ a$ ; $p^{k} ∣ a$ holds exactly for $k \leq v_{p} (a)$ ; $v_{p} (a) \geq 1$ exactly when $p ∣ a$ ; $v_{p} (1) = v_{p} (- 1) = 0$ ; and $v_{p} (p) = 1$

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

For a prime pp and a nonzero integer aa: pvp(a)∣apvp​(a)∣a and pvp(a)+1∤apvp​(a)+1∤a; pk∣apk∣a holds exactly for k≤vp(a)k≤vp​(a); vp(a)≥1vp​(a)≥1 exactly when p∣ap∣a; vp(1)=vp(−1)=0vp​(1)=vp​(−1)=0; and vp(p)=1vp​(p)=1

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

For a prime $p$ and a nonzero integer $a$ : $p^{v_{p} (a)} ∣ a$ and $p^{v_{p} (a) + 1} ∤ a$ ; $p^{k} ∣ a$ holds exactly for $k \leq v_{p} (a)$ ; $v_{p} (a) \geq 1$ exactly when $p ∣ a$ ; $v_{p} (1) = v_{p} (- 1) = 0$ ; and $v_{p} (p) = 1$