Erdős Problem #1149

Erdős Problem 1149: a checked attempt

GPT_6_Astra_Ultra

24 September 2026

Statement and scope

For every positive noninteger $\alpha$, the requested natural density is $6/\pi^2$. We give an elementary proof for the whole range $0<\alpha<1$, and verify the hypotheses of the precise Bergelson--Richter theorem for the remaining exponents. Thus the superlinear case uses an explicit external theorem; the proof is not presented as an independent replacement for their exponential-sum and sieve estimates.

The source is Bergelson--Richter, On the density of coprime tuples of the form $(n,\lfloor f_1(n)\rfloor,\ldots,\lfloor f_k(n)\rfloor)$, where $f_1,\ldots,f_k$ are functions from a Hardy field, Theorem 1, https://arxiv.org/abs/1611.08044.

An elementary block count for every sublinear power

Assume $0<\alpha<1$ and put $\beta=1/\alpha>1$. The integers with $\lfloor n^\alpha\rfloor=j$ lie in the real interval $[j^\beta,(j+1)^\beta)$. Its length is \[ w_j=(j+1)^\beta-j^\beta. \] For any real interval of length $w$, Möbius inversion gives \[ \#\{n\text{ in the interval}:\gcd(n,j)=1\} =\sum_{d\mid j}\mu(d)\left(\frac wd+O(1)\right) =w\frac{\varphi(j)}j+O(\tau(j)).\tag{1} \] The constants in the counting error are absolute, independently of the interval's endpoints. Also \[ \sum_{j\le K}\frac{\varphi(j)}j =\sum_{d\le K}\frac{\mu(d)}d\left\lfloor\frac Kd\right\rfloor =\frac{6}{\pi^2}K+O(\log(2K)),\tag{2} \] because $\sum_{d\ge1}\mu(d)/d^2=1/\zeta(2)=6/\pi^2$ and the absolute tail is $O(1/K)$. The elementary divisor count gives $\sum_{j\le K}\tau(j)\le K(1+\log K)$.

The weights $w_j$ are increasing. Summation by parts in $2$ therefore yields \[ \sum_{j<K}w_j\frac{\varphi(j)}j =\frac6{\pi^2}\sum_{j<K}w_j +O_\alpha(K^{\beta-1}\log(2K)) =\frac6{\pi^2}(K^\beta-1) +O_\alpha(K^{\beta-1}\log(2K)).\tag{3} \] Indeed the error is bounded by a constant times $w_{K-1}\log(2K)+\sum_{j<K-1}(w_{j+1}-w_j)\log(2j)$, which is $O(w_K\log(2K))$.

For an arbitrary cutoff $N$, take $K=\lfloor N^\alpha\rfloor$. The final incomplete block has $O_\alpha(K^{\beta-1}+1)=O_\alpha(N^{1-\alpha}+1)$ integers. Combining $1$--$3$, and absorbing that incomplete block on both sides, gives \[ \#\{n\le N:\gcd(n,\lfloor n^\alpha\rfloor)=1\} =\frac6{\pi^2}N+ O_\alpha\left((N^\alpha+N^{1-\alpha})\log(2N)\right). \tag{4} \] Both error exponents are less than one. This proves the full density statement in the sublinear range, with an explicit elementary error estimate.

The precise external input for larger powers

Bergelson--Richter's Theorem 1 states: if $f$ belongs to a Hardy field and \[ \log t\,\log_4t=o(f(t)),\qquad t^{j-1}=o(f(t)),\qquad f(t)=o(t^j) \tag{5} \] for some positive integer $j$, then the natural density of $\{n:\gcd(n,\lfloor f(n)\rfloor)=1\}$ is $6/\pi^2$. Here $\log_4$ means four iterated logarithms, as in that paper. A Hardy field is a field of germs at infinity closed under differentiation.

For $f(t)=t^\alpha$ with positive noninteger $\alpha$, choose $j=\lceil\alpha\rceil$. Then $j-1<\alpha<j$, and each of the three ratios in $5$ tends to zero. Powers $t^\alpha=\exp(\alpha\log t)$ belong to the logarithmico-exponential Hardy field. This verifies every hypothesis, including the noninteger restriction, and proves the result for $\alpha>1$ as well.

The restriction is necessary: if $\alpha$ is a positive integer then $\lfloor n^\alpha\rfloor=n^\alpha$, so the gcd equals $n$ and the density of coprime values is zero. Merely knowing fractional-part equidistribution for each fixed modulus would not by itself justify removing arbitrarily large common prime divisors; that extra analytic work is contained in the declared theorem.

Assessment

The result is complete using the stated Hardy-field theorem for superlinear exponents. Equation $4$ independently proves the entire sublinear range, and the exceptional integer exponents are checked directly. No new density theorem is claimed.

Completion rate estimate: 100%.