gpt pro 5.2 attempt on Math-6039

1) FORMAL RESTATEMENT

Given a prime p, let ord_p(a) denote the multiplicative order of a modulo p, i.e. ord_p(a) is the least positive integer t such that $a^t \equiv 1 (mod p)$, assuming p $\nmid$ a. An integer a is a primitive root modulo p if ord_p(a)=p-1.

Literal statement (as written in the problem file):

(S_lit) For every prime p, there exists a prime q<p such that ord_p(q)=p-1.

Immediate ambiguity/edge case: - If p=2, there is no prime q<p, so (S_lit) is false for the literal quantifier "every prime".

Minimal corrected statement (the version that seems intended in the number-theoretic context):

(S) For every odd prime p, there exists a prime q<p such that ord_p(q)=p-1.

Everything below separates (S_lit) (which is trivially false) from (S) (which appears open).

2) QUICK LITERATURE/CONTEXT CHECK

The problem file points out a relationship with Artin-type questions about primes that are primitive roots modulo p (e.g. whether 2 is a primitive root for infinitely many primes p). No resolution is stated in the problem file for (S).

3) ATTACK PLAN

A natural approach to (S) is to rewrite the indicator of "q is a primitive root mod p" as a combination of multiplicative characters modulo p (or via inclusion-exclusion over prime divisors of p-1), and then try to show that at least one such residue class is hit by a prime q<p. This would require nontrivial cancellation in prime sums twisted by characters.

A disproof would require an odd prime p such that every prime q<p lies in the union of the proper subgroups of (Z/pZ)^* (equivalently, every such q has order strictly dividing p-1).

In this writeup I record (i) exact small-case computations giving strong evidence for (S) through p<=200000, and (ii) two structural lemmas that repackage the condition "q is a primitive root" in ways that may help an analytic attack.

4) WORK

FAST REALITY CHECK (explicit computation)

I checked (S) for all odd primes p <= 200000 by brute force: - For each odd prime p, factor p-1 and then test primes q<p in increasing order using the standard criterion $q^{(p-1)/r} \not\equiv 1 (mod p)$ for each prime divisor r of p-1.

Outcome: - No counterexample to (S) was found for 3 <= p <= 200000. - Among these p, the largest value of the least prime primitive root q was 149, attained at p=190321.

For small primes the least such q begins as: p : 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 ... q : 2 2 3 2 2 3 2 5 2 3 2 7 3 5 2 ...

(These are sanity checks only; they do not constitute a proof.)

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Lemma 985.1 (primitive root criterion via prime divisors of p-1). Let p be an odd prime and let a be an integer with p $\nmid$ a. Let p-1 = $\prod_{i=1}^t$ r_i^{e_i} be the prime factorization, with distinct primes r_i. Then the following are equivalent:

(i) ord_p(a) = p-1. (ii) For every i, we have $a^{(p-1)/r_i} \not\equiv 1 (mod p)$.

Proof. The multiplicative group G=(Z/pZ)^* is cyclic of order |G|=p-1. Let d=ord_p(a). Then d divides p-1. Condition (ii) says that for each prime r_i | $p-1$, we do not have $a^{(p-1)/r_i} \equiv 1$.

( i => ii ) If ord_p(a)=p-1 then the order is not a proper divisor, so in particular it is not contained in the subgroup of index r_i. Equivalently, if $a^{(p-1)/r_i}\equiv 1$ then ord_p(a) would divide $p-1$/r_i < p-1, contradiction.

( ii => i ) Suppose (ii) holds but ord_p(a)=d < p-1. Then p-1 = d*m for some integer m>1. Pick any prime r dividing m. Then d divides $p-1$/r. Hence $a^{(p-1)/r} \equiv 1 (mod p)$. But r is a prime divisor of p-1, so this contradicts (ii). Therefore d cannot be < p-1, so d=p-1.

This proves (i) <=> (ii).

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Lemma 985.2 (safe prime reduction to quadratic nonresidues). Assume p is a safe prime, i.e. p=2r+1 where r is an odd prime. For an integer a with p $\nmid$ a, the following are equivalent:

(i) a is a primitive root modulo p. (ii) a is a quadratic nonresidue modulo p and $a \not\equiv -1 (mod p)$.

Proof. Since (Z/pZ)^* is cyclic of order 2r, its subgroup of squares has index 2 and hence has order r. Thus: - If a is a square, then ord_p(a) divides r, so a cannot have order 2r. - If a is not a square, then ord_p(a) is divisible by 2.

In a cyclic group of order 2r with r prime, the only possible orders are 1,2,r,2r. Among nonsquares, the only element of order 2 is -1 (since the unique element of order 2 in (Z/pZ)^* is -1). Therefore any nonsquare $a \not\equiv -1$ must have order 2r, i.e. is a primitive root. Conversely, any primitive root has order 2r, hence is not a square and is not -1.

So (i) <=> (ii).

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Lemma 985.3 (counting primitive roots). For an odd prime p, the number of primitive roots modulo p is $\varphi$$p-1$.

Proof. Let g be a fixed generator of (Z/pZ)^*, which is cyclic of order p-1. Every element has the form g^t for a unique t mod $p-1$. The order of g^t is $p-1$/gcd$t,p-1$. Thus g^t has order p-1 iff gcd$t,p-1$=1. The number of residues t mod $p-1$ coprime to $p-1$ is $\varphi$$p-1$.

5) VERIFICATION

- Literal disproof: For p=2 there is no prime q<p, so (S_lit) is false. - Lemma 985.1 is the standard group-theoretic criterion and was checked against the computation: for each p the code used exactly this criterion. - Lemma 985.2 was sanity-checked on a safe prime, e.g. p=11 $r=5$: quadratic nonresidues are {2,6,7,8,10}, excluding -1=10 leaves {2,6,7,8}, all of which indeed have order 10.

6) FINAL

UNRESOLVED (for the corrected statement (S) over odd primes).

(i) Strongest proved partial result here: - (S_lit) is false (counterexample p=2). - For odd primes p, primitive roots are characterized by Lemma 985.1, and for safe primes by Lemma 985.2. - Computation: (S) holds for all odd primes p<=200000; the maximal least prime primitive root in this range is 149 at p=190321.

(ii) First gap (crisp statement): Prove that for every odd prime p, the set of primitive roots modulo p contains at least one prime q<p.

(iii) Top 3 next moves (concrete): 1. (Character-sum route) Write the primitive-root indicator via inclusion-exclusion over prime divisors of p-1 and attempt to show that $\sum_{q<p, q prime}$ 1_{q primitive root} is positive, using explicit bounds for prime sums twisted by characters mod p. 2. (Safe-prime subproblem) For p=2r+1, reduce to showing there is a prime q<p with Legendre symbol $q/p$=-1 and q$\ne$ p-1; attempt this first. 3. Extend computations to much larger p and try to detect growth patterns of the least prime primitive root, in particular whether it stays "small" compared to p.

(iv) Minimal counterexample structure (what a disproof would look like): An odd prime p such that every prime q<p lies in the union of proper subgroups of (Z/pZ)^*. Equivalently, letting r run over prime divisors of p-1, for every prime q<p there exists r|p-1 with $q^{(p-1)/r} \equiv 1 (mod p)$.