CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + claude-sonnet-5)audited 2026-08-17

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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

1.1assume-contraL2L3L5L6L7choose

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 $q \neq 2, 3$ .

2.1step 1.1L1L3L4L8discharge-contradiction∎

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.

Depends on

Used by

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Dependency tree · two levels

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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