gpt pro 5.2 attempt on Math-1007

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.