Reused writeup on Math-1007 (gpt pro 5.2)

Erdős Problem 972 --- collection record

PROVENANCE AND STATUS

Official problem: https://www.erdosproblems.com/972

Saved collaborative-database status: open. Snapshot checked: 2026-09-23T17:22:24Z. Latest status and formalization links: https://teorth.github.io/erdosproblems/

This is an attributed collection of existing writeups. The model-folder names below identify their original attribution; placement in GPT_6_Astra_Ultra does not attribute their mathematical work to that model. All locally available source versions selected for this problem are reproduced below, without editing their text.

Proof claims, computations, literature statements, and completion estimates in the reused text have not been independently verified in this collection pass. A source claiming a solution does not change the saved upstream status.

Primary source (latest numbered version in the preferred model folder): attacks/erdos/gpt_pro_5.2/972.tex.

REUSED SOURCE 1

Original attribution: gpt_pro_5.2. Original file: attacks/erdos/gpt_pro_5.2/972.tex.

Let $\alpha>1$ be irrational. Are there infinitely many primes $p$ such that $\lfloor p\alpha\rfloor$ is also prime? Vinogradov [Vi48] proved that the sequence $\{p\alpha\}$ is uniformly distributed for every irrational $\alpha$, and hence there are infinitely many primes $p$ of the shape $p=\lfloor n\alpha\rfloor$ for every irrational $\alpha>1$. Indeed, this occurs if and only if\[\frac{p}{\alpha}\leq n<\frac{p+1}{\alpha},\]which is true if and only if $\{p\alpha^{-1}\}>1-\alpha^{-1}$, which happens infinitely often by the uniform distribution of $\{p\alpha^{-1}\}$. References [Vi48] Vinogradov, I. M., On an estimate of trigonometric sums with prime numbers . Izv. Akad. Nauk SSSR Ser. Mat. $1948$, 225--248.

1) FORMAL RESTATEMENT
Fix an irrational real number $\alpha>1$. The question is whether \[ \#\{p\text{ prime}: \lfloor \alpha p\rfloor\text{ is prime}\}=\infty. \] Equivalently, do there exist infinitely many prime pairs $(p,q)$ such that $q$ is prime and \[ q=\lfloor \alpha p\rfloor \quad\Longleftrightarrow\quad q\le \alpha p<q+1. \]

2) QUICK LITERATURE/CONTEXT CHECK
The problem statement recalls Vinogradov's uniform distribution result for $\{p\alpha\}$ (and $\{p\alpha^{-1}\}$) and explains how it implies infinitely many primes of the form $\lfloor n\alpha\rfloor$ for irrational $\alpha>1$. The present problem asks for simultaneous primality of $p$ and $\lfloor \alpha p\rfloor$, which is not resolved by that uniform distribution statement alone.

3) ATTACK PLAN
Proof track: reinterpret the condition as a thin set of prime pairs with a linear constraint and attempt a sieve for the pair $(p,\lfloor\alpha p\rfloor)$.

Disproof track: attempt to find an $\alpha$ for which $\lfloor\alpha p\rfloor$ is forced to have a small prime factor for all large $p$ (e.g. near an integer); however $\alpha$ is irrational so only approximate congruence obstructions are possible.

I reach only equivalences and computational evidence.

4) WORK
FAST REALITY CHECK (sample $\alpha$). For two irrationals, I searched primes $p\le 10^6$ and counted those for which $\lfloor \alpha p\rfloor$ is also prime.

• $\alpha=\sqrt2$: there are $6047$ such primes $p\le 10^6$; the first few pairs $(p,\lfloor\sqrt2 p\rfloor)$ are $(2,2),(5,7),(29,41),(31,43),(59,83)$.
• $\alpha=\varphi=(1+\sqrt5)/2$: there are $6036$ such primes $p\le 10^6$; the first few pairs $(p,\lfloor\varphi p\rfloor)$ are $(2,3),(7,11),(11,17),(23,37),(37,59)$.

This does not prove infinitude, but it checks that there is no immediate obstruction for these examples.

Lemma 972.1 (ratio window). Let $\alpha>1$ be real and let $p$ be a positive integer. Put $q=\lfloor \alpha p\rfloor$. Then \[ q\text{ is prime}\ \Longleftrightarrow\ \text{there is a prime }q\text{ with }\alpha-\frac1p<\frac{q}{p}\le \alpha. \]

Proof. By definition, $q=\lfloor\alpha p\rfloor$ is the unique integer satisfying $q\le \alpha p < q+1$. Dividing by $p>0$ gives \[ \frac{q}{p}\le \alpha < \frac{q+1}{p} = \frac{q}{p}+\frac1p, \] which rearranges to $\alpha-\frac1p<\frac{q}{p}\le\alpha$. The equivalence with primality is immediate since $q$ is exactly $\lfloor\alpha p\rfloor$. $\square$

Lemma 972.2 (inversion via fractional parts). Let $\alpha>1$ be irrational. For an integer $q\ge 1$, define $p:=\lfloor q/\alpha\rfloor+1$. Then \[ \lfloor \alpha p\rfloor = q \quad\Longleftrightarrow\quad \Big\{\frac{q}{\alpha}\Big\} > 1-\frac{1}{\alpha}, \] where $\{x\}=x-\lfloor x\rfloor$ is the fractional part. Consequently, the original question is equivalent to asking whether there are infinitely many primes $q$ such that $p=\lfloor q/\alpha\rfloor+1$ is prime and $\{q/\alpha\}>1-1/\alpha$.

Proof. Because $\alpha$ is irrational, $q/\alpha\notin\mathbb{Z}$ for every integer $q$, so $p=\lfloor q/\alpha\rfloor+1$ is the unique integer satisfying \[ \frac{q}{\alpha}<p\le \frac{q}{\alpha}+1. \] Multiplying by $\alpha$ gives \[ q< \alpha p\le q+\alpha. \] Thus $\lfloor\alpha p\rfloor=q$ holds exactly when $\alpha p<q+1$ (since $\alpha p>q$ already), i.e. \[ \alpha\Big(\lfloor q/\alpha\rfloor+1\Big) < q+1. \] Write $q/\alpha=\lfloor q/\alpha\rfloor+\{q/\alpha\}$; then the left-hand side is \[ \alpha\lfloor q/\alpha\rfloor+\alpha = q-\alpha\{q/\alpha\}+\alpha. \] So $\alpha p<q+1$ is equivalent to \[ q-\alpha\{q/\alpha\}+\alpha < q+1 \quad\Longleftrightarrow\quad \{q/\alpha\} > 1-\frac1\alpha. \] This proves the equivalence. The final sentence just substitutes "$q$ prime and $p$ prime" into the equivalence. $\square$

5) VERIFICATION
Lemma lem:972-inversion uses irrationality to guarantee $q/\alpha\notin\mathbb{Z}$, ensuring $p=\lfloor q/\alpha\rfloor+1$ is the unique candidate integer mapped to $q$ by $p\mapsto\lfloor\alpha p\rfloor$. If $\alpha$ were rational, this step could fail when $q/\alpha\in\mathbb{Z}$.

The computation used sieving up to $\lfloor \alpha\cdot 10^6\rfloor$ to test primality of $\lfloor\alpha p\rfloor$.

6) FINAL
UNRESOLVED

(i) Strongest proved partial result. The condition is equivalent to finding infinitely many prime pairs $(p,q)$ with $q/p$ lying in the shrinking interval $(\alpha-1/p,\alpha]$ (Lemma lem:972-ratio), or equivalently prime $q$ with $p=\lfloor q/\alpha\rfloor+1$ prime and $\{q/\alpha\}>1-1/\alpha$ (Lemma lem:972-inversion). Computations for $\alpha=\sqrt2$ and $\alpha=\varphi$ show thousands of solutions up to $p\le 10^6$.

(ii) First gap (crisp). Prove (or disprove) that for every irrational $\alpha>1$, there are infinitely many primes $p$ such that $\lfloor\alpha p\rfloor$ is prime.

(iii) Top 3 next moves.

• Treat $(p,\lfloor\alpha p\rfloor)$ as a two-dimensional prime pattern and attempt a sieve/prime-tuple heuristic in the thin region $q\approx \alpha p$ with error $<1$.
• Use Lemma lem:972-inversion to recast the problem as primes $q$ in a Beatty-type sequence with prime-index constraint, and test whether known distribution results for primes in Beatty sequences can be combined with a second sieve.
• Extend computations to larger $p$ (e.g. $10^7$ or $10^8$) for several $\alpha$ (including badly approximable and very well approximable irrationals) to see if the count matches a stable heuristic.

(iv) Minimal counterexample structure. A counterexample would be an irrational $\alpha>1$ for which all but finitely many primes $p$ yield a composite $\lfloor\alpha p\rfloor$. Any such obstruction would likely have to come from subtle correlations between $p$ and the fractional parts $\{\alpha p\}$ that force $\lfloor\alpha p\rfloor$ into residue classes with small prime factors.

COMPLETION ESTIMATE

The primary source supplies no extractable numerical completion estimate. Its mathematical progress is not numerically assessed here.

Newly verified progress in this collection pass: $0\%$. This measures new verification work, not the amount of mathematics achieved in the reused text.