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Source. GPT 5.6 Sol Pro, Coprime Power Differences, public manuscript shared by Liam Price in a proof claim on erdosproblems.com, 16 July 2026 (Overleaf snapshot accessed 5 September 2026), Theorem 1.1, p. 1; proof on pp. 2–3. Provenance is on the source card.

Statement

For n≥2n\ge2, the manuscript (p. 1) defines

K(n)=min⁡{k≥2:gcd⁡(kn−1,2n−1)=1}.K(n)=\min\{k\ge2:\gcd(k^n-1,2^n-1)=1\}.

This is the H1(n)H_1(n) of Erdős (1974). Its existence, the inequality K(n)>2K(n)>2, and H(n)≤K(n)H(n)\le K(n) are noted on p. 1 and established in the threshold comparison. Let τ(n)\tau(n) count the positive divisors of nn.

Theorem 1.1 (p. 1). There is an absolute constant C>0C>0 such that, for every n≥2n\ge2,

log⁡K(n)≤Cτ(n)(log⁡(n+2))2.\log K(n)\le C\tau(n)(\log(n+2))^2.

Proof sketch

Fix nn and write A=2n−1A=2^n-1. For a prime p∣Ap\mid A, the nn-th roots of unity modulo pp number gcd⁡(n,p−1)\gcd(n,p-1), by cyclicity of Fp×\mathbb F_p^\times. Two kinds of prime divisor arise.

  • When p−1∣np-1\mid n, every unit modulo pp is an nn-th root of unity, so an admissible base must be divisible by pp. These primes are injectively indexed by divisors of nn and are at most n+1n+1, so their product MM has log⁡M≤τ(n)log⁡(n+1)\log M\le\tau(n)\log(n+1) (display (2), p. 3).
  • For each other prime, writing the base as MtMt forbids exactly gcd⁡(n,p−1)\gcd(n,p-1) residues of tt, at most half of all residues. The ratio (p−1)/gcd⁡(n,p−1)(p-1)/\gcd(n,p-1) takes each value for at most τ(n)\tau(n) primes, and fewer than nn such primes divide AA. This bounds the total forbidden density by τ(n)(1+log⁡(n+1))\tau(n)(1+\log(n+1)) and the total number of forbidden residues by n2n^2.

Lemma 2.1 then gives an admissible tt with log⁡t≪τ(n)(log⁡(n+2))2\log t\ll\tau(n)(\log(n+2))^2. Then k=Mtk=Mt satisfies gcd⁡(kn−1,A)=1\gcd(k^n-1,A)=1 and k≥2k\ge2, so log⁡K(n)≤log⁡M+log⁡t\log K(n)\le\log M+\log t.

Reading note. The manuscript's chain bounding the forbidden density (p. 3) opens with a strict inequality, which fails when no prime of the second kind exists; the weak inequality holds in every case and suffices. The bound itself is unaffected.

Dependencies. Lemma 2.1, the elementary threshold comparison, Fermat's theorem and the cyclicity of the multiplicative group of a finite field. No prime-distribution theorem is used.

Scope. The manuscript asserts the existence of an absolute constant and gives no value. The separately linked Lean source states a numerical constant 500500; this page does not certify that value. The source and formal-evidence distinctions are recorded on the card.

Bears on. #820, through Corollary 1.2, which deduces from this theorem the eventual upper bound asked for in the problem's questions on H(n)H(n) and on the least k≥2k\ge2 with gcd⁡(kn−1,2n−1)=1\gcd(k^n-1,2^n-1)=1. The theorem says nothing on whether gcd⁡(2n−1,3n−1)=1\gcd(2^n-1,3^n-1)=1 infinitely often.