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Retain $B$, $S$, $U_t=I+tS$, $\beta$, and $\Phi_t$ from [[additive_combinatorics/adamczewski_2026_erdos1/lemma_5_1|Lemma 5.1]], with $t\in\mathbb Z$. Both $U_t$ and $\beta$ are integer unimodular automorphisms.

Exact kernel

Let $e_0$ be the first standard basis vector and define the integer row vector

$$(a_0(t),\ldots,a_n(t)) =e_0^TU_t^{-1}\beta^{-1}. \tag{1}$$

The inverses are integral: $U_t^{-1}$ is the finite geometric series in the nilpotent matrix $-tS$, and

$$\beta^{-1}(y_0,\ldots,y_n) =(y_0,\ldots,y_{n-1},y_0+\cdots+y_n).$$

For $x\in\mathbb Z^{n+1}$, equation (1) gives

$$\sum_{i=0}^na_i(t)x_i=0 \iff U_t^{-1}\beta^{-1}x\in\{0\}\times\mathbb Z^n \iff x=\Phi_t(z)\text{ for some }z\in\mathbb Z^n. \tag{2}$$

Thus the perturbed rank-$n$ lattice is exactly the integer kernel of this linear form; no saturation assertion is left implicit.

Asymptotic sign and size

For each $i$, the coefficient $a_i(t)$ is the first coordinate of the solution to

$$U_tx=\beta^{-1}e_i.$$

By Cramer's rule and $\det U_t=1$, $a_i(t)$ equals the determinant of the matrix that agrees with $U_t$ except for its first column, which is $\beta^{-1}e_i$. Scaling the last $n$ columns by $t^{-1}$ multiplies it by $t^{-n}$; the first column is unchanged, while each of the other columns becomes

$$t^{-1}e_j+S_j\longrightarrow S_j.$$

The last coordinate of every $\beta^{-1}e_i$ is $1$. The last row of $S$ is zero, and the minor in its first $n$ rows and last $n$ columns is $B$. Expansion along the last row therefore yields

$$\frac{a_i(t)}{t^n}\longrightarrow(-1)^n\det B=D \qquad(0\leq i\leq n). \tag{3}$$

Here the sign normalization of the [[additive_combinatorics/adamczewski_2026_erdos1/lattice_reduction|triangular basis]] makes $D>0$.

For positive integers $s$, put

$$t_s=2^s+E_B.$$

For each of the finitely many indices $i$,

$$\frac{a_i(t_s)}{(2^s)^n} =\frac{a_i(t_s)}{t_s^n} \left(1+\frac{E_B}{2^s}\right)^n \longrightarrow D. \tag{4}$$

Consequently one can choose a single sufficiently large $s$ such that, simultaneously for every $0\leq i\leq n$,

$$0<a_i(t_s)\leq2D(2^s)^n. \tag{5}$$

This direct normalization by $(2^s)^n$ is what gives the exact later bound; an estimate only in terms of $t_s^n$ would not by itself imply (5).

Finally take $R=2^r$ and $q_0=2^{s+r}$. Then

$$(t_s-E_B)R=2^s2^r=q_0,$$

so [[additive_combinatorics/adamczewski_2026_erdos1/proposition_5_2|Proposition 5.2]] applies.

Source and dependencies

An explanation of the proof of Erdős Problem 1, preliminary exposition with no named author (erdosproblems.com, 2026), §6, equations (20)–(24), pp. 7–8. The edition read is named on the source card. The direct row-vector definition in (1), the first-column Cramer determinant, and the uniform finite-index limit in (4) spell out the source's normal-vector argument.

Bears on. #1.