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Let $F_{m,h}$ be the matrix produced from the $1\times1$ identity matrix by $h$ successive applications of the lift of [[additive_combinatorics/adamczewski_2026_erdos1/proposition_3_1|Proposition 3.1]], all with the same odd block size $d=2m+1$.

The lift formulas and [[additive_combinatorics/adamczewski_2026_erdos1/lemma_2_4|Lemma 2.4]] give:

$$\begin{aligned} \operatorname{ord}(F_{m,h})&=d^h,\\ q_h&=\left(\frac32\right)^h,\\ 2^hF_{m,h}&\in M_{d^h}(\mathbb Z),\\ z\in\mathbb Z^{d^h},\ \|F_{m,h}z\|_\infty<1&\Longrightarrow z=0,\\ \det F_{m,h}&=(1+2^{-d})^{1+d+\cdots+d^{h-1}}. \end{aligned} \tag{1}$$

With $h$ held fixed, the determinant converges to $1$ as $m\to\infty$, since

$$(1+2^{-d})^{1+d+\cdots+d^{h-1}} \leq \exp\!\left((1+d+\cdots+d^{h-1})2^{-d}\right),$$

and for fixed $h$ the exponent is at most $hd^{h-1}2^{-d}$, which tends to $0$.

Statement

For every $k\in\mathbb N$ there are integers $m,h\geq1$ with

$$k\det F_{m,h}<\left(\frac32\right)^h.$$

The proof below covers $k=0$ as well, so the statement holds whether or not $\mathbb N$ is read to contain $0$.

Proof

Put $K=\max(1,k)$. Choose $h\geq1$ with $(3/2)^h>2K$. Keeping this $h$, take $m$ for which the determinant in (1) is below $2$, as the limit above allows. Then

$$k\det F_{m,h}\leq K\det F_{m,h}<2K<\left(\frac32\right)^h.$$

The use of $K$ handles $k=0$; the source's displayed chain $k\det F_{m,h}<2k$ is strict only for $k>0$.

Source and dependencies

An explanation of the proof of Erdős Problem 1, preliminary exposition with no named author (erdosproblems.com, 2026), §3, equations (8)–(10) and Proposition 3.2, pp. 4–5. The edition read is named on the source card. The endpoint repair is elementary and leaves the construction unchanged for positive $k$.

Bears on. #1.