Is it true that, for every prime , there is a prime which is a primitive root modulo ?

claimed — no verifier run, no signed judgmentunreviewedOpen. Worked here; no verified result yet.

number theory · open · formalized (Lean) · 0 attempts

machinery: artin-primitive-root,least-primitive-root,GRH-conditional,character-sum-Burgess,prime-distribution,sieve/Brun-Titchmarsh,Hooley-Heath-Brown-density

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vela registry pull vfr_37aec80d874a0239
vela reproduce examples/erdos-problems

Evidence

unverified AI candidates (2)

gpt-erdos · GPT-5.2 Pro + Deep Research · unverified

Taken literally, the statement is **false** because of the trivial case $p=2$: there is no prime (q<p).

candidate solution ↗

llm-hunter · gpt pro 5.2 · unverified

1 LLM attack(s) recorded (gpt pro 5.2); unverified.

candidate solution ↗

Formal proof

AMS 11 · open (literature)

theorem erdos_985 : answer(sorry) ↔ ∀ᵉ (p : ℕ) (hp_prime : p.Prime) (hp_nontrivial : p ≠ 2),
    ∃ q, q.Prime ∧ q < p ∧ orderOf (q : ZMod p) = p - 1
formal-conjectures/985.lean ↗

OEIS3

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  • packet.json · sha256 6febe9fda89f83be77eafa8d33dee91c8fb5c4cd273651f14d695d545dee7d48

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