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Call a square real matrix $C$ admissible if

$$z\in\mathbb Z^I,\quad \|Cz\|_\infty<1\quad\Longrightarrow\quad z=0.$$

Fix an integer $m\geq1$ and the odd cyclic matrix $C_m$ of order $d=2m+1$. Given $C$ with rows and columns indexed by a finite set $I$, split each input coordinate $i$ into $d$ coordinates $x_{i,0},\ldots,x_{i,d-1}$, with sum

$$t_i=\sum_{j=0}^{d-1}x_{i,j}.$$

Define a change of variables by

$$y_{i,0}=(Ct)_i-\sum_{j=1}^{d-1}x_{i,j},\qquad y_{i,j}=x_{i,j}\quad(1\leq j<d). \tag{1}$$

Then

$$\sum_{j=0}^{d-1}y_{i,j}=(Ct)_i. \tag{2}$$

The matrix $\operatorname{Lift}_m(C)$ first makes this change and then applies $C_m$ separately to every $y$-block.

Structural formulas

Ordering the block-zero coordinates before all other coordinates makes the change in (1) block triangular with diagonal blocks $C$ and the identity. The block-diagonal second map has one copy of $C_m$ for every $i\in I$. Therefore

$$\det(\operatorname{Lift}_m(C))=(\det C_m)^{|I|}\det C. \tag{3}$$

When all columns of $C$ have sum $q$, summing (2) over $i$ shows that the change map multiplies the total coordinate sum by $q$, so its columns have sum $q$ as well. The block map has column sums $3/2$, and column sums multiply under composition, so the lifted matrix has column sums

$$\frac32q. \tag{4}$$

Finally, if $r\in\mathbb N_0$ and $2^rC$ has integer entries, then the change map has denominator dividing $2^r$, and the block map has denominator $2$. Hence

$$2^{r+1}\operatorname{Lift}_m(C)$$

has integer entries.

Statement

If $C$ is admissible, then $\operatorname{Lift}_m(C)$ is admissible.

Proof

Take an integer vector $x$ with $\|\operatorname{Lift}_m(C)x\|_\infty<1$. By (1), in each block $i$ every coordinate $y_{i,j}$ with $j\geq1$ is an integer. Applying [[additive_combinatorics/adamczewski_2026_erdos1/lemma_2_3|Lemma 2.3]] to that block gives

$$\left|\sum_jy_{i,j}\right|<1.$$

By (2), $|(Ct)_i|<1$ for every $i$. The vector $t$ is integral, so admissibility of $C$ gives $t=0$.

Now (1) gives

$$y_{i,0}=-\sum_{j=1}^{d-1}x_{i,j}=x_{i,0},$$

where the last equality uses $t_i=0$. Thus $y=x$ is integral. Each block $x_i$ has $\|C_mx_i\|_\infty<1$, and [[additive_combinatorics/adamczewski_2026_erdos1/corollary_2_2|Corollary 2.2]] forces every block, and hence $x$, to vanish.

Source and dependencies

An explanation of the proof of Erdős Problem 1, preliminary exposition with no named author (erdosproblems.com, 2026), §3, equations (3)–(7) and Proposition 3.1, pp. 3–4. The edition read is named on the source card. The exact lift formulas, determinant, column-sum, and denominator calculations are included because all four are used later.

Bears on. #1.