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Source. Display (9) and the identity before it, p. 3, in §2 "Prime-power residue families" of A Two-Copy Proof of Erdős Problem 126 (2026), a three-page preliminary exposition with no printed author, posted at https://www.erdosproblems.com/static/126-proof.pdf; the edition read is identified on the source card. The step is unlabelled in the print; this page names it.

Statement

For finitely many positive real numbers xix_i, the kernel L(i,j)=log⁡(xi+xj)L(i,j)=\log(x_i+x_j) is conditionally negative semidefinite:

∑i,jzizjL(i,j)≤0whenever∑izi=0.\sum_{i,j}z_iz_jL(i,j)\leq0 \quad\text{whenever}\quad \sum_i z_i=0.

No sign is asserted for vectors whose coordinates do not sum to zero.

Read depth. Claims checked: the identity and display (9) were read on p. 3, and the convergence remark below was added here. Nothing here is independently reviewed.

Proof sketch

P. 3. With ρi=(xi−1)/(xi+1)\rho_i=(x_i-1)/(x_i+1), which lies in (−1,1)(-1,1),

log⁡(xi+xj)=log⁡(xi+1)+log⁡(xj+1)−log⁡2+log⁡(1−ρiρj).\log(x_i+x_j)=\log(x_i+1)+\log(x_j+1)-\log2+\log(1-\rho_i\rho_j).

For a zero-sum zz the first three terms drop out of the quadratic form, and expanding log⁡(1−t)=−∑m≥1tm/m\log(1-t)=-\sum_{m\geq1}t^m/m writes the remainder as −∑m≥11m(∑iziρim)2≤0-\sum_{m\geq1}\frac1m\bigl(\sum_iz_i\rho_i^m\bigr)^2\leq0, which is (9). The print does not justify the rearrangement; it is valid because the mm-th term is at most (∑i∣zi∣)2q2m/m(\sum_i|z_i|)^2q^{2m}/m in absolute value, with q=max⁡i∣ρi∣<1q=\max_i|\rho_i|<1.

Dependencies

The power series of log⁡(1−t)\log(1-t) for ∣t∣<1|t|<1.

Used by. [[arithmetic_functions/adamczewski_2026_erdos126/main_theorem|The main theorem]], with xi=aix_i=a_i.

Bears on

  • Problem 126: the lemma supplies the conditional negativity that the main theorem needs to apply Proposition 1; it bears on the problem only through that theorem.