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Use B,R,LB,EBB,R,L_B,E_B, and Φt\Phi_t from [[additive_combinatorics/adamczewski_2026_erdos1/lattice_reduction|the lattice reduction]] and [[additive_combinatorics/adamczewski_2026_erdos1/lemma_5_1|Lemma 5.1]]. Thus every nonzero z∈Znz\in\mathbb Z^n satisfies ∥LB(z)∥∞≥R\|L_B(z)\|_\infty\geq R.

Statement

Let t≥EBt\geq E_B be an integer and q0q_0 a real number with

q0≤(t−EB)R.q_0\leq(t-E_B)R.

Then every nonzero z∈Znz\in\mathbb Z^n satisfies

∥Φt(z)∥∞≥q0.\|\Phi_t(z)\|_\infty\geq q_0.

Proof

If q0≤0q_0\leq0, the conclusion follows from nonnegativity of the norm. Suppose therefore that q0>0q_0>0. Let z≠0z\ne0 and put M=∥LB(z)∥∞≥RM=\|L_B(z)\|_\infty\geq R. Fix an index jj with ∣LB(z)j∣=M|L_B(z)_j|=M. From Φt=tLB+E\Phi_t=tL_B+E and Lemma 5.1,

tM=∣tLB(z)j∣≤∣Φt(z)j∣+∣E(z)j∣≤∣Φt(z)j∣+EBM.tM =|tL_B(z)_j| \leq|\Phi_t(z)_j|+|E(z)_j| \leq|\Phi_t(z)_j|+E_BM.

Were ∥Φt(z)∥∞<q0\|\Phi_t(z)\|_\infty<q_0, it would follow that

(t−EB)M<q0.(t-E_B)M<q_0.

But M≥RM\geq R and t−EB≥0t-E_B\geq0, so the left side is at least (t−EB)R≥q0(t-E_B)R\geq q_0, a contradiction.

Source and dependencies

An explanation of the proof of Erdős Problem 1, preliminary exposition with no named author (erdosproblems.com, 2026), §5, Proposition 5.2, p. 7. The edition read is named on the source card. Dependencies are Lemmas 4.1 and 5.1 and the triangle inequality.

Bears on. #1.