GPT 6 Astra Ultra attempt on Math-6051
mathematicsErdős problemscontributionManuscript
Contents
Erdős Problem 997: prime rotations are never well-distributed
Statement and source
For every real $\alpha$, we prove that $(\{\alpha p_j\})$ is not well-distributed: there are arbitrarily long consecutive blocks and intervals whose counting error is a fixed positive proportion of the block length. The known solution is credited on the problem page to an internal OpenAI model and uses Banks--Freiberg--Turnage-Butterbaugh, Consecutive primes in tuples, arXiv:1311.7003v3. We reconstruct the admissible tuple and the consecutive-prime reduction. The deep external input is the Maynard--Tao theorem for admissible affine forms, stated in Section 1 of that paper. No novelty is claimed.External prime input
For every $m\ge2$ there is $k_m$ such that an admissible family of $k\ge k_m$ distinct affine forms $a_jn+b_j$, with positive leading coefficients and nonproportional pairs, takes at least $m$ prime values simultaneously for infinitely many $n$. Admissibility means that no prime divides the product of all forms for every residue class of $n$.An admissible tuple whose phases cluster
Rational $\alpha$ is immediate: all fractional parts lie in a finite grid, so some fixed interval of positive length contains none of them. Now let $\alpha$ be irrational, fix any $m\ge2$, and choose $k\ge k_m$. Let $Q$ be the product of the primes at most $k$. Since $\alpha Q$ is irrational, the fractional parts of its positive multiples are dense. Choose distinct positive integers \[ h_1<\cdots<h_k,\qquad Q\mid h_j,\qquad \{\alpha h_j\}\in(0,1/10). \] The forms $n+h_j$ are admissible. If a prime $\ell\le k$, all shifts are zero modulo $\ell$, so a nonzero residue of $n$ avoids divisibility. If $\ell>k$, at most $k<\ell$ forbidden residues occur.Forcing actual consecutive primes
This is the elementary CRT step in Banks--Freiberg--Turnage-Butterbaugh. For each integer $t\in[h_1,h_k]\setminus\{h_1,\ldots,h_k\}$ choose a different prime $q_t>h_k+1$. Set $W=\prod_tq_t$ and choose $a$ by the Chinese remainder theorem with $a+t\equiv0\pmod{q_t}$. Consider \[ L_j(u)=Wu+a+h_j. \] For primes dividing $W$, no $L_j$ vanishes modulo that prime, because $h_j-t$ is nonzero and has modulus less than $q_t$. For primes not dividing $W$, multiplication by $W$ is invertible, and admissibility follows from the original tuple. The forms have positive leading coefficient and are pairwise nonproportional.By the stated theorem, infinitely many sufficiently large $u$ produce at least $m$ primes among the $L_j(u)$. All other integers in the interval $[Wu+a+h_1,Wu+a+h_k]$ are composite: each has a prescribed divisor $q_t$, and for large $u$ exceeds that divisor. Thus all primes in this interval belong to the tuple. Any $m$ successive primes in the interval are also $m$ consecutive primes in the global prime sequence.