False statementConstruction: Literature-sourcedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passverified 2026-08-03 (gpt-5.6-sol-codex-subscription)

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FALSE: for every finite list $p_0, \dots, p_{n-1}$ of distinct primes, $p_0 \cdots p_{n-1} + 1$ is prime

Statement

False claim: for every $n \in \mathbb{N}$ and every injective list $p : n \to \mathbb{Z}$ of primes (Prime and composite integers: $p$ is prime when $p > 1$ and its only positive divisors are $1$ and $p$ , Injection, surjection, bijection),

$\Bigl(\prod_{i<n} p_i\Bigr) + 1$

is prime, the product being that of 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 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 $u = -1$ .

The true statement is Euclid's theorem: for every $n \in \mathbb{N}$ and every list $p : n \to \mathbb{Z}$ of primes there is a prime not among $p_0, \dots, p_{n-1}$ ; consequently the set of primes is not finite, which concludes only that this integer has a prime divisor not on the list — never that it is itself prime.

Witness: $n = 6$ and $p = (2,3,5,7,11,13)$ . Here $2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13 = 30030$ and

$30031 \;=\; 59 \cdot 509 ,$

so $30031$ has the positive divisor $59$ , which is neither $1$ nor $30031$ : it is composite.

Numerals. For $k \in \mathbb{N}$ the symbol $k$ inside $\mathbb{Z}$ means $\iota(k)$ , the embedding of The naturals embed in the integers.

Facts & Assumptions

Given: The integers $2, 3, 5, 7, 11, 13, 59, 509, 30030, 30031$ .

[L1]

$p$ is prime when $p > 1$ and its only positive divisors are $1$ and $p$ ; an integer $> 1$ that is not prime is composite (Prime and composite integers: $p$ is prime when $p > 1$ and its only positive divisors are $1$ and $p$ ).

[L2]

$\prod_{i<0} g_i = e$ and $\prod_{i<\sigma(m)} g_i = \bigl(\prod_{i<m} g_i\bigr) g_m$ (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).

[L3]

Every integer $n > 1$ has a prime divisor, and the least divisor of $n$ exceeding $1$ is prime (Every integer $n > 1$ has a prime divisor; indeed the least divisor of $n$ that exceeds $1$ is prime).

[L4]

For $a \in \mathbb{Z}$ and $b > 0$ there is exactly one pair $(q,r)$ with $a = qb + r$ and $0 \le r < b$ , and $b \mid a$ exactly when $r = 0$ (Division with remainder in $\mathbb{Z}$ : for $a \in \mathbb{Z}$ and $b > 0$ there are unique $q, r \in \mathbb{Z}$ with $a = qb + r$ and $0 \le r < b$ ).

[L6]

$d \mid a$ means $a = dc$ for some $c$ ; divisibility is transitive (Divisibility in $\mathbb{Z}$ : $d \mid a$ when $a = dq$ for some integer $q$ , Divisibility is reflexive and transitive on $\mathbb{Z}$ , and is linear: if $d \mid a$ and $d \mid b$ then $d \mid ax + by$ for all integers $x, y$ ; also $d \mid a$ implies $d \mid ac$ , $-d \mid a$ and $d \mid -a$ ).

[L7]

$\mathbb{Z}$ is a commutative ring; its order is total, antisymmetric and transitive, is compatible with addition, and positives are closed under multiplication; a product of two nonzero integers is nonzero (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 integers have no zero divisors; multiplicative cancellation).

[L8]

$\iota$ is injective and order preserving with image the nonnegative integers, $\iota(0) = 0$ , $\iota(1) = 1$ (The naturals embed in the integers); $m < k$ exactly when $\sigma(m) \le k$ and $1 = \sigma(0)$ (Discreteness: $\sigma(n)$ is the immediate successor, The natural numbers $\mathbb{N}$ (von Neumann), Order on the natural numbers).

Refutation

technique · direct

1.1

$0 < 1$ , and every integer $y > 0$ satisfies $y \ge 1$ ; consequently $y > c$ implies $y \ge c + 1$ .

L7L8

1.2

$59 \cdot 509 = 30031$ : indeed $59 \cdot 500 = 29500$ and $59 \cdot 9 = 531$ , and $29500 + 531 = 30031$ . So $59 \mid 30031$ .

L6L7algebra

2.1

A composite integer $n$ has a prime divisor $q$ with $q \cdot q \le n$ . Let $q$ be the least divisor of $n$ exceeding $1$ , which is prime by [L3], and write $n = qm$ . Then $m > 0$ , since $n > 0$ and $q > 0$ ; and $m \ne 1$ , since $m = 1$ would make $n = q$ prime. So $m > 1$ , and $m \mid n$ , so $m$ is a divisor of $n$ exceeding $1$ and minimality gives $q \le m$ ; multiplying by $q > 0$ gives $q \cdot q \le qm = n$ .

step 1.1L1L3L6L7

2.2

$59 \ne 1$ and $59 \ne 30031$ , and $59 > 0$ ; also $30031 > 1$ . So $30031$ has a positive divisor other than $1$ and itself, hence is not prime, and being greater than $1$ it is composite.

step 1.1step 1.2L1L7

3.1

$2$ , $3$ , $5$ , $7$ , $11$ and $13$ are prime. Each exceeds $1$ , so by step 2.1 it suffices to check the primes $q$ with $q \cdot q$ at most the number. For $2$ and $3$ there is none, since the least prime is $2$ and $2 \cdot 2 = 4 > 3$ . For $5$ and $7$ only $q = 2$ qualifies, and $5 = 2 \cdot 2 + 1$ , $7 = 3 \cdot 2 + 1$ . For $11$ and $13$ only $q = 2$ and $q = 3$ qualify, since $4 \cdot 4 = 16 > 13$ , and $11 = 5 \cdot 2 + 1$ , $11 = 3 \cdot 3 + 2$ , $13 = 6 \cdot 2 + 1$ , $13 = 4 \cdot 3 + 1$ . In every case no such divisor exists, so none of the six is composite, and each is therefore prime.

step 2.1L1L4L7algebra

4.1

The six are pairwise distinct, and the list $p = (2,3,5,7,11,13)$ is therefore an injective list of primes of length $6$ .

step 3.1L7L8

5.1

$\prod_{i<6} p_i = 30030$ : applying [L2] six times, $1 \cdot 2 = 2$ , $2 \cdot 3 = 6$ , $6 \cdot 5 = 30$ , $30 \cdot 7 = 210$ , $210 \cdot 11 = 2310$ and $2310 \cdot 13 = 30030$ . Hence the integer named by the claim is $30030 + 1 = 30031$ .

step 4.1L2L7algebra

6.1

Steps 4.1, 5.1 and 2.2 exhibit an injective list of primes whose product plus $1$ is composite: the claim is false.

step 4.1step 5.1step 2.2∎

Remarks

Re-reading the true theorem shows it never says otherwise. Euclid's theorem: for every $n \in \mathbb{N}$ and every list $p : n \to \mathbb{Z}$ of primes there is a prime not among $p_0, \dots, p_{n-1}$ ; consequently the set of primes is not finite takes $N = \prod_{i<n} p_i + 1$ , applies Every integer $n > 1$ has a prime divisor; indeed the least divisor of $n$ that exceeds $1$ is prime to get a prime divisor $q$ of $N$ , and then shows $q$ is not among the $p_i$ . Primality of $N$ is never claimed and is never used.
The promise the theorem does make is kept here. $59$ is prime — by step 2.1 the only candidates are the primes $q$ with $q \cdot q \le 59$ , that is $q \in \{2,3,5,7\}$ since $8 \cdot 8 = 64 > 59$ , and $59 = 29 \cdot 2 + 1$ , $59 = 19 \cdot 3 + 2$ , $59 = 11 \cdot 5 + 4$ , $59 = 8 \cdot 7 + 3$ all have nonzero remainder — and $59$ is not among $2,3,5,7,11,13$ , exactly as Euclid's theorem: for every $n \in \mathbb{N}$ and every list $p : n \to \mathbb{Z}$ of primes there is a prime not among $p_0, \dots, p_{n-1}$ ; consequently the set of primes is not finite promises. Nothing is claimed here about $509$ .
Smaller lists do not refute the claim. For $n = 0,1,2,3,4,5$ with the first primes in order, the six values $2$ , $3$ , $7$ , $31$ , $211$ and $2311$ are classically known to be prime; this item does not verify that, and it is beside the point — one witness suffices, and no amount of small cases could establish a universal claim.

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Sources

Euclid number (Wikipedia) (standard reference, not scraped)
Euclid's theorem (Wikipedia) (standard reference, not scraped)
Discrete mathematics notes: Prime numbers and Euclid's argument (standard reference, not scraped)

FALSE: for every finite list p0,…,pn−1p_0, \dots, p_{n-1}p0​,…,pn−1​ of distinct primes, p0⋯pn−1+1p_0 \cdots p_{n-1} + 1p0​⋯pn−1​+1 is prime