CorollaryStatement: Literature-sourcedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passverified 2026-08-03 (gpt-5.6-sol-codex-subscription)

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Every nonzero integer $n$ is $u \prod_{i<r} p_i$ with $u \in \{1,-1\}$ and every $p_i$ prime; $u$ and $r$ are determined by $n$ , and the list is determined up to a permutation

Statement

Let $n \in \mathbb{Z}$ with $n \ne 0$ , and take finite products in the commutative monoid $(\mathbb{Z},\cdot,1)$ of $(\mathbb{Z}, \cdot, 1)$ is a commutative monoid whose group of units is $\{1, -1\}$ ; equivalently $u \mid 1$ holds exactly for $u = 1$ and $u = -1$ , as in The product $g_0 g_1 \cdots g_{n-1}$ of a finite list in a monoid, by recursion, with the empty product ( $n = 0$ ) equal to the identity.

Existence. There are $u \in \{1,-1\}$ , $r \in \mathbb{N}$ and a list $p : r \to \mathbb{Z}$ of primes (Prime and composite integers: $p$ is prime when $p > 1$ and its only positive divisors are $1$ and $p$ ) with

$n \;=\; u \prod_{i<r} p_i .$
Uniqueness. If also $n = u' \prod_{j<s} q_j$ with $u' \in \{1,-1\}$ and $q : s \to \mathbb{Z}$ a list of primes, then $u = u'$ , $r = s$ , and $q_i = p_{\pi(i)}$ for every $i < r$ , for some $\pi \in \operatorname{Sym}(r)$ (The symmetric group $\operatorname{Sym}(X)$ : the bijections of a set $X$ under composition).

Facts & Assumptions

Given: A nonzero integer $n$ .

[L1]

Every integer $m \ge 1$ is $\prod_{i<r} p_i$ for some $r \in \mathbb{N}$ and some list $p : r \to \mathbb{Z}$ of primes; and conversely every such product is $\ge 1$ (Every integer $n \ge 1$ is a finite product of primes: there are $r \in \mathbb{N}$ and a list $p : r \to \mathbb{Z}$ of primes with $n = \prod_{i<r} p_i$ , the case $n = 1$ being the empty product, The product $g_0 g_1 \cdots g_{n-1}$ of a finite list in a monoid, by recursion, with the empty product ( $n = 0$ ) equal to the identity, $(\mathbb{Z}, \cdot, 1)$ is a commutative monoid whose group of units is $\{1, -1\}$ ; equivalently $u \mid 1$ holds exactly for $u = 1$ and $u = -1$ , Semigroup and monoid).

[L2]

If $\prod_{i<r} p_i = \prod_{j<s} q_j$ for lists of primes, then $r = s$ and $q_i = p_{\pi(i)}$ for all $i < r$ , for some $\pi \in \operatorname{Sym}(r)$ (The fundamental theorem of arithmetic: every integer $n \ge 1$ is a product of primes, and the factorisation is unique up to order — if $\prod_{i<r} p_i = \prod_{j<s} q_j$ with every $p_i$ and $q_j$ prime, then $r = s$ and $q_i = p_{\pi(i)}$ for some $\pi \in \operatorname{Sym}(r)$ , The symmetric group $\operatorname{Sym}(X)$ : the bijections of a set $X$ under composition).

[L3]

$|x| = x$ when $x \ge 0$ and $|x| = -x$ when $x < 0$ (The absolute value $|a|$ of an integer); $|x| \ge 0$ , $|x| = 0$ exactly when $x = 0$ , and $|xy| = |x|\,|y|$ (Absolute value in $\mathbb{Z}$ : $|a| \ge 0$ ; $|a| = 0$ exactly when $a = 0$ ; $|-a| = |a|$ ; $|ab| = |a|\,|b|$ ; $-|a| \le a \le |a|$ ; and $|a| \le c$ exactly when $-c \le a \le c$ ).

[L4]

If $xz = yz$ and $z \ne 0$ then $x = y$ (The integers have no zero divisors; multiplicative cancellation).

[L5]

$\mathbb{Z}$ is a commutative ring: multiplication is associative and commutative, $x \cdot 1 = x$ , $x \cdot (-1) = -x$ , and every $x$ has an additive inverse, with $-(-x) = x$ (The integers form a commutative ring, Arithmetic on the integers, The integers as equivalence classes of pairs of naturals).

[L6]

The order on $\mathbb{Z}$ is total, antisymmetric and transitive and is compatible with addition (The integers form a totally ordered ring, Order on the integers).

[L7]

$\iota : \mathbb{N} \to \mathbb{Z}$ is injective, preserves the order, and has as image exactly the nonnegative integers, with $\iota(0) = 0$ and $\iota(1) = 1$ (The naturals embed in the integers).

[L8]

On $\mathbb{N}$ : $0 \le k$ for every $k$ (Order on the natural numbers); $m < k$ exactly when $\sigma(m) \le k$ (Discreteness: $\sigma(n)$ is the immediate successor); $1 = \sigma(0)$ (The natural numbers $\mathbb{N}$ (von Neumann)).

Proof

technique · direct

1.1

$0 < 1$ , since $1 = \iota(1)$ is nonnegative and differs from $0 = \iota(0)$ by injectivity; and if $0 < x$ then $1 \le x$ , because $x = \iota(k)$ with $k \ne 0$ , so $1 = \sigma(0) \le k$ and $\iota$ preserves the order.

L7L8

2.1

$|n| \ge 0$ and $|n| \ne 0$ , so $|n| > 0$ and hence $|n| \ge 1$ .

step 1.1L3L6

2.2

For uniqueness, suppose $n = u P = u' P'$ where $P := \prod_{i<r} p_i$ and $P' := \prod_{j<s} q_j$ and $u, u' \in \{1,-1\}$ . By [L1] both $P \ge 1$ and $P' \ge 1$ , so both are positive and $|P| = P$ , $|P'| = P'$ .

step 1.1L1L3L6

3.1

By [L1] there are $r \in \mathbb{N}$ and a list $p$ of primes of length $r$ with $|n| = \prod_{i<r} p_i$ .

step 2.1L1choose

3.2

Taking absolute values, $|n| = |u|\,|P| = P$ and likewise $|n| = |u'|\,|P'| = P'$ , since $|1| = 1$ and $|-1| = 1$ . Hence $P = P'$ .

step 2.2L3L5

4.1

The order is total and $n \ne 0$ , so $n > 0$ or $n < 0$ . If $n > 0$ then $|n| = n$ and $n = 1 \cdot \prod_{i<r} p_i$ ; if $n < 0$ then $|n| = -n$ , so $n = -(-n) = -|n| = (-1) \prod_{i<r} p_i$ . In both cases clause 1 holds, with $u = 1$ and $u = -1$ respectively.

step 3.1L3L5L6

4.2

By [L2] applied to $P = P'$ we get $r = s$ and a permutation $\pi \in \operatorname{Sym}(r)$ with $q_i = p_{\pi(i)}$ for every $i < r$ .

step 3.2L2

4.3

And $u P = u' P$ with $P \ne 0$ , since $P \ge 1 > 0$ ; cancellation gives $u = u'$ .

step 2.2step 3.2L4L6

5.1

Clause 1 is step 4.1 and clause 2 is steps 4.2 and 4.3.

step 4.1step 4.2step 4.3∎

Remarks

This is why Prime and composite integers: $p$ is prime when $p > 1$ and its only positive divisors are $1$ and $p$ can insist on $p > 1$ without loss. The sign of $n$ is carried by the unit $u$ , not by the primes, so admitting negative primes would buy nothing and would break uniqueness, since $(-2)(-3)$ and $2 \cdot 3$ would be different lists with the same product.
The uniqueness of $u$ needs $P \ne 0$ , and that is why the empty product is harmless. At $r = 0$ the statement reads $n = u$ , so the nonzero integers with an empty prime list are exactly $1$ and $-1$ — which is $(\mathbb{Z}, \cdot, 1)$ is a commutative monoid whose group of units is $\{1, -1\}$ ; equivalently $u \mid 1$ holds exactly for $u = 1$ and $u = -1$ again, from a different direction.
Nothing is claimed at $n = 0$ . Zero is divisible by every prime and is not a product of primes times a unit at all, since every such product is $1$ or $-1$ times a positive integer. It is excluded by hypothesis, not overlooked.

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Sources

Fundamental theorem of arithmetic (Wikipedia) (standard reference, not scraped)
Inquiry into Advanced Algebra: Division, primes, and factorisation (standard reference, not scraped)

Every nonzero integer nnn is u∏i<rpiu \prod_{i<r} p_iu∏i<r​pi​ with u∈{1,−1}u \in \{1,-1\}u∈{1,−1} and every pip_ipi​ prime; uuu and rrr are determined by nnn, and the list is determined up to a permutation