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Fix a sufficiently large $s$, let $t=2^s+E_B$ and $q_0=2^{s+r}$, and use the positive integer coefficients $a_0(t),\ldots,a_n(t)$ from [[additive_combinatorics/adamczewski_2026_erdos1/normal_coefficients|the normal-coefficient construction]].

Statement

Distinct points of the integer box $\{0,\ldots,q_0-1\}^{n+1}$ have distinct images under

$$(x_0,\ldots,x_n)\longmapsto\sum_{i=0}^nx_i a_i(t). \tag{1}$$

Proof

Suppose two digit vectors $x,y$ in the box have the same image and put $z=x-y$. Then

$$|z_i|<q_0\quad(0\leq i\leq n),\qquad \sum_i a_i(t)z_i=0. \tag{2}$$

The exact kernel identity for the coefficients gives $z=\Phi_t(w)$ for some $w\in\mathbb Z^n$. If $w\ne0$, Proposition 5.2 and the equality $(t-E_B)R=q_0$ give

$$\|z\|_\infty=\|\Phi_t(w)\|_\infty\geq q_0,$$

contrary to (2). Therefore $w=0$, so $z=\Phi_t(0)=0$ and $x=y$.

Source and dependencies

An explanation of the proof of Erdős Problem 1, preliminary exposition with no named author (erdosproblems.com, 2026), §6, equation (25), p. 8. The edition read is named on the source card. This uses the exact kernel from normal_coefficients and the separation in [[additive_combinatorics/adamczewski_2026_erdos1/proposition_5_2|Proposition 5.2]].

Bears on. #1.