CorollaryStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + claude-sonnet-5)audited 2026-08-17

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.

There are infinitely many primes congruent to $1$ modulo $3$

Statement

The set of primes $p$ satisfying $p \equiv 1 (m o d 3)$ is infinite (Finite, countably infinite, countable, uncountable).

Facts & Assumptions

Given: The set $S = {p : p is prime and p \equiv 1 (m o d 3)}$ .

[L1]

For every odd prime $p \neq 3$ , the congruence $x^{2} \equiv - 3 (m o d p)$ is soluble if and only if $p \equiv 1 (m o d 3)$ (Odd primes represented by a divisor of $x^{2} + 3$ ).

[L2]

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

[L3]

A finite product has empty-product value $1$ and satisfies $\prod_{i < k + 1} g_{i} = (\prod_{i < k} g_{i}) g_{k}$ (The product $g_{0} g_{1} \dots g_{n - 1}$ of a finite list in a monoid, by recursion, with the empty product ( $n = 0$ ) equal to the identity).

[L4]

In a commutative monoid, permuting the entries of a finite product does not change its value (Generalised associativity: in a monoid the product of a finite list does not depend on the bracketing, and in a commutative monoid it does not depend on the order of the factors either).

[L5]

A set $A$ is finite if $A \approx n$ for some $n \in N$ (Finite, countably infinite, countable, uncountable).

[L6]

The relation $A \approx B$ means that there exists a bijection $A \to B$ (Equinumerous sets, $A \approx B$ and $A ⪯ B$ ).

[L7]

A bijection is, in particular, surjective (Injection, surjection, bijection).

[L8]

If $d ∣ u$ and $d ∣ v$ , then $d ∣ u x + v y$ for all integers $x, y$ (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 $d ∣ - a$ ).

Proof

technique · contradiction

1.1

Suppose, for contradiction, that $S$ is finite. By [L5] and [L6], choose a bijection from some natural number $r$ onto $S$ and write its prime values as $p_{0}, \dots, p_{r - 1}$ ; surjectivity is [L7]. Let $P = \prod_{i < r} p_{i}$ , using the value $P = 1$ if $r = 0$ from [L3], and set $N = 12 P^{2} + 1 > 1$ . By [L2], choose a prime $q ∣ N$ . Since $N$ is odd and $N \equiv 1 (m o d 3)$ , one has $q \neq 2, 3$ .

assume-contraL2L3L5L6L7choose

2.1

From $q ∣ N$ and $(6 P)^{2} + 3 = 3 N$ , the class of $6 P$ solves $x^{2} \equiv - 3 (m o d q)$ , so [L1] gives $q \equiv 1 (m o d 3)$ . Thus $q \in S$ and equals some $p_{i}$ ; by [L4] move that factor to the end of the finite product, and then [L3] gives $q ∣ P$ , hence $q ∣ 12 P^{2}$ . Together with $q ∣ 12 P^{2} + 1$ , [L8] gives $q ∣ 1$ , impossible for a prime. Therefore $S$ is not finite.

step 1.1L1L3L4L8discharge-contradiction∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 82 results over 27 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

A. Gorodnik, Number Theory, Lecture 9, consequence after Example 2.4 (standard reference, not scraped)

There are infinitely many primes congruent to 11 modulo 33