Erdős Problem #1147

Erdős Problem 1147: an explicit irrational counterexample

Statement and scope.

Must $A=\{n\geq2:\|\alpha n^2\|<1/\log n\}$, for irrational $\alpha>0$, be an asymptotic additive basis of order two? No. We construct an irrational $\alpha$ for which infinitely many integers are absent from $A+A$. The convention at $n=1$ does not matter. The known negative resolution is due to Konieczny; see https://www.erdosproblems.com/1147 and https://arxiv.org/abs/1504.02410. The elementary example below addresses the universal question, without claiming Konieczny's stronger almost-everywhere conclusions.

A parity-preserving approximation.

Let \[ \alpha=\frac12+\sum_{r=1}^{\infty}3^{-r!}, \qquad N_j=3^{j!}. \] Its $j$-th truncation is $a_j/(2N_j)$, where $a_j$ is odd. For $j\geq2$ the positive error $e_j$ satisfies \[ 0<e_j<2\cdot3^{-(j+1)!},\qquad e_jN_j^2<2\cdot3^{(1-j)j!}\longrightarrow0. \] The number $\alpha$ is irrational: if $\alpha=u/v$, its nonzero difference from $a_j/(2N_j)$ is at least $1/(2vN_j)$, contradicting the displayed error bound for large $j$.

An obstruction uniform over all decompositions.

Whenever positive integers $n,m$ satisfy $n+m=N_j$, \[ \alpha n^2-\alpha m^2 =\alpha N_j(2n-N_j) =a_j n-\frac{a_jN_j}{2}+e_jN_j(2n-N_j). \] Both $a_j,N_j$ are odd and the last term has absolute value at most $e_jN_j^2$. Thus, uniformly in such decompositions, \[ \|\alpha n^2-\alpha m^2\|\geq\frac12-e_jN_j^2. \] Choose $K$ so that $1/\log t<1/8$ for $t\geq K$. If both summands are at least $K$ and belong to $A$, the triangle inequality for distance modulo one gives $\|\alpha n^2-\alpha m^2\|<1/4$, contradicting the preceding bound for large $j$.

It remains to exclude a small summand. For each fixed $1\leq s<K$, the identity above gives \[ \alpha(N_j-s)^2\equiv \alpha s^2+\tfrac12+o(1)\pmod1. \] Irrationality implies $\delta_s=\|\alpha s^2+1/2\|>0$. There are only finitely many such $s$, so for all sufficiently large $j$, $\|\alpha(N_j-s)^2\|>\delta_s/2>1/\log(N_j-s)$. Thus the large companion is not in $A$. Every possible decomposition has now been excluded, proving $N_j\notin A+A$ for all sufficiently large $j$.

The same proof works for any prescribed positive threshold $\varepsilon(n)\to0$ in place of $1/\log n$, and tolerates any finite changes to $A$. No external Diophantine approximation theorem is needed.

Completion estimate.

An explicit irrational counterexample and all summand-size cases are proved. COMPLETION: 100%