Erdős Problem #997
Status Solved · erdosproblems.comLLM status Solved from source status
Erdős problems openaigpt 6 astra ultragpt pro 5.2 100% cataloguedSolvedSolvedStatement
A sequence $(x_m)_{m\ge 1}$ in $(0,1)$ is well-distributed (in the sense of Hlawka and Petersen) if for every $\epsilon>0$ there exists $k_0(\epsilon)$ such that for every integer $k\ge k_0$, every integer $n\ge 0$, and every interval $I\subseteq [0,1]$, \[ \left|\#\{n<m\le n+k : x_m\in I\} - k\,|I|\right| < \epsilon k. \] Let $p_n$ be the $n$-th prime and fix $\alpha\in(0,1)$. Consider the sequence \[ x_n := \{\alpha p_n\} \in [0,1). \] Question: Is it true that for every $\alpha\in(0,1)$ the sequence $(\{\alpha p_n\})_{n\ge 1}$ is not well-distributed?
Restatement from Math-6051-A2 r1, with the source manuscript’s qualifications.
Original problem on erdosproblems.com
Statement licence: CC-BY-4.0 · Source file
Contributions 2
GPT 6 Astra Ultra attempt on Math-6051
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.
The discrepancy obstruction
The fractional parts of these $m$ consecutive primes after multiplication by $\alpha$ all lie in a translate, on the circle, of the arc $(0,1/10)$. Its complement has length $9/10$ and meets $[0,1)$ in at most two intervals. One component has length at least $9/20$ and contains no point of the block. Shrinking it slightly gives a usual interval $I$ of length at least $2/5$ with zero block count. Its discrepancy is therefore at least $2m/5$. Since $m$ was arbitrary, the defining well-distribution inequality fails, for example at $\epsilon=1/3$. The interval may depend on the block, as allowed by the uniform quantifiers in that definition.Conclusion and verification scope
This proves the assertion for every real $\alpha$, using the explicit Maynard--Tao input. Ordinary equidistribution of irrational prime rotations is compatible with the failure of this stronger condition. Sources: https://www.erdosproblems.com/997 and https://arxiv.org/abs/1311.7003. No local Lean verification is claimed.COMPLETION ESTIMATE
COMPLETION: 100%gpt pro 5.2 attempt on Math-6051
1) FORMAL RESTATEMENT A sequence $(x_m)_{m\ge 1}$ in $(0,1)$ is well-distributed (in the sense of Hlawka and Petersen) if for every $\epsilon>0$ there exists $k_0(\epsilon)$ such that for every integer $k\ge k_0$, every integer $n\ge 0$, and every interval $I\subseteq [0,1]$, \[ \left|\#\{n<m\le n+k : x_m\in I\} - k\,|I|\right| < \epsilon k. \] Let $p_n$ be the $n$-th prime and fix $\alpha\in(0,1)$. Consider the sequence \[ x_n := \{\alpha p_n\} \in [0,1). \] Question: Is it true that for every $\alpha\in(0,1)$ the sequence $(\{\alpha p_n\})_{n\ge 1}$ is not well-distributed?
2) QUICK LITERATURE/CONTEXT CHECK The problem statement notes: Erdős proved non-well-distribution for lacunary $(n_k)$ and a.e. $\alpha$, and it records that there exists at least one irrational $\alpha$ with $(\{\alpha p_n\})$ not well-distributed. I do not use results beyond the definition.
3) ATTACK PLAN Prove necessary conditions for well-distribution, then show these fail at least for some classes of $\alpha$ (e.g. rational $\alpha$). For irrational $\alpha$, the problem appears to require deep information about primes in short intervals/arithmetic progressions to force irregularity in every block.
4) WORK
FAST REALITY CHECK (easy obstruction for rational $\alpha$). If $\alpha=a/q$ is rational in lowest terms, then $\{\alpha p_n\}\in\{0,1/q,2/q,\dots,(q-1)/q\}$ for all $n$, a finite set. A well-distributed sequence must be dense (Lemma 2 below), so this immediately rules out all rational $\alpha$.
Lemma 1 (Well-distributed $\Rightarrow$ uniformly distributed). If $(x_m)$ is well-distributed, then it is uniformly distributed in the usual sense: for every interval $I\subseteq[0,1]$, \[ \lim_{N\to\infty}\frac{1}{N}\#\{1\le m\le N: x_m\in I\} = |I|. \]
Proof. Fix an interval $I$ and $\epsilon>0$. Choose $k_0(\epsilon)$ from the definition. For $N\ge k_0$, write $N = qk + r$ with $k\ge k_0$ and $0\le r<k$. Partition $\{1,\dots,N\}$ into $q$ consecutive blocks of length $k$ plus a remainder of length $r$.
For each full block, the well-distribution inequality gives \[ \left|\#(\text{block}\cap I) - k|I|\right|<\epsilon k. \] Summing over the $q$ blocks gives \[ \left|\#\{1\le m\le qk: x_m\in I\} - qk|I|\right| < q\epsilon k=\epsilon qk. \] The remainder contributes at most $r\le k$ points, so \[ \left|\#\{1\le m\le N: x_m\in I\} - N|I|\right| \le \epsilon qk + k + r|I|. \] Divide by $N=qk+r\ge qk$ to get \[ \left|\frac{1}{N}\#\{1\le m\le N: x_m\in I\} - |I|\right| \le \epsilon + \frac{k+r|I|}{N} \le \epsilon + \frac{2k}{N}. \] Letting $N\to\infty$ (hence $k/N\to 0$ for fixed $k$) and then $\epsilon\to 0$ yields the claim. $\square$
Lemma 2 (Well-distributed $\Rightarrow$ dense and no long gaps). If $(x_m)$ is well-distributed, then for every interval $I\subseteq[0,1]$ with $|I|>0$ there exists $k_0$ such that every block of length $k\ge k_0$ contains at least one term in $I$. In particular, the set $\{x_m:m\ge 1\}$ is dense in $[0,1]$.
Proof. Fix an interval $I$ with $|I|>0$. Apply the definition with $\epsilon:=|I|/2$ to obtain $k_0$. For any $k\ge k_0$ and any $n\ge 0$, \[ \#\{n<m\le n+k: x_m\in I\} \ge k|I| - \epsilon k = \frac{|I|}{2}k. \] Since the right-hand side is $\ge 1$ for all $k\ge 2/|I|$, it follows that for all sufficiently large $k$ every such block contains at least one element of $I$. This implies density because every nonempty open interval eventually must contain some $x_m$. $\square$
Proposition 3 (Rational $\alpha$ gives non-well-distribution for primes). If $\alpha=a/q\in(0,1)$ is rational, then $(\{\alpha p_n\})$ is not well-distributed.
Proof. As noted in the FAST REALITY CHECK, the values lie in the finite set $\{0,1/q,\dots,(q-1)/q\}$. Choose an interval $I\subseteq[0,1]$ of length $|I|>0$ that avoids this finite set (e.g. a small open interval between two consecutive multiples of $1/q$). Then $\#\{n<m\le n+k: x_m\in I\}=0$ for all $n,k$, but $k|I|$ grows linearly in $k$, contradicting the definition for any $\epsilon<|I|$. Hence the sequence is not well-distributed. $\square$
5) VERIFICATION - Lemma 1 uses only the block definition of well-distribution and basic counting; no hidden number theory. - Lemma 2 uses the well-distribution inequality with $\epsilon=|I|/2$ to force a positive lower bound. - Proposition 3 is immediate once Lemma 2 is established.
6) FINAL UNRESOLVED (i) Strongest proved partial result: For every rational $\alpha\in(0,1)$, the prime sequence $(\{\alpha p_n\})$ is not well-distributed (Proposition 3). More generally, any well-distributed sequence must be dense (Lemma 2) and uniformly distributed (Lemma 1). (ii) First gap: Handle irrational $\alpha$. Prove that for every irrational $\alpha$ there exist intervals $I$ and arbitrarily large blocks of consecutive primes indices for which the count in $I$ deviates by $\gg k$ from $k|I|$. (iii) Top 3 next moves: 1. Reduce well-distribution failure to a concrete prime-in-arithmetic-progressions-in-short-index-intervals statement and try to prove that statement (even with weak parameters). 2. For fixed irrational $\alpha$, search computationally for large $k$ and $n$ where the block discrepancy for $x_m=\{\alpha p_m\}$ is unusually large, to guess a mechanism (e.g. clustering near rationals). 3. Attempt a "Diophantine approximation + primes in progressions" strategy: approximate $\alpha$ by $a/q$ and compare $\{\alpha p_n\}$ to $\{a p_n/q\}$; prove that short blocks of primes have biased residue distribution mod $q$ strongly enough to break well-distribution. (iv) Minimal counterexample structure (if the statement is false): a counterexample would be an irrational $\alpha$ such that $(\{\alpha p_n\})$ is well-distributed, hence dense and uniformly distributed with uniformly small discrepancy on every consecutive block. Such an $\alpha$ would force extremely strong "local" equidistribution of primes through the map $p\mapsto \{\alpha p\}$, far beyond global equidistribution.