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Fix m≥1m\geq1, put d=2m+1d=2m+1, and read subscripts modulo dd.

Statement

Integers z0,…,zd−1z_0,\ldots,z_{d-1} satisfying

∣2zi+zi−1∣<2(0≤i<d)|2z_i+z_{i-1}|<2\qquad(0\leq i<d)

are all zero.

Proof

Let zi≠0z_i\ne0. If zi−1z_{i-1} were zero or of the same sign as ziz_i, then ∣2zi+zi−1∣=2∣zi∣+∣zi−1∣≥2|2z_i+z_{i-1}|=2|z_i|+|z_{i-1}|\geq2, against the hypothesis; so zi−1z_{i-1} is nonzero, of the sign opposite to that of ziz_i. Walking back from index ii through all dd indices to ii again reverses the sign dd times, and as dd is odd, ziz_i would have the sign opposite to its own. Hence no coordinate is nonzero.

Source and dependencies

An explanation of the proof of Erdős Problem 1, preliminary exposition with no named author (erdosproblems.com, 2026), §2, Lemma 2.1, p. 2. The edition read is named on the source card. Only integrality, the triangle order on Z\mathbb Z, and oddness of dd are used.

Bears on. #1.