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Choose from [[additive_combinatorics/adamczewski_2026_erdos1/proposition_3_2|Proposition 3.2]] an admissible matrix $C$ of order $n+1$, common column sum $q>0$, positive determinant, and dyadic denominator. Choose $r\in\mathbb N_0$ so that $R=2^r$ clears its denominators, put $G=RC\in M_{n+1}(\mathbb Z)$, and suppose

$$k\det C<q. \tag{1}$$

The columns of $G$ all have sum $Rq$.

Determinant reduction

For $1\leq i,j\leq n$, define

$$(B_0)_{ij}=G_{ij}-G_{i0}. \tag{2}$$

To compute its determinant, subtract column $0$ of $G$ from every other column. Each new nonfirst column has total coordinate sum zero, while the first column has sum $Rq$. Now replace row $0$ by the sum of all rows. This row operation preserves the determinant and makes the first row

$$(Rq,0,\ldots,0).$$

Expansion along it gives

$$\det G=(Rq)\det B_0. \tag{3}$$

Since $\det G=R^{n+1}\det C$,

$$q\det B_0=R^n\det C. \tag{4}$$

Thus $\det B_0>0$, and (1) implies

$$k\det B_0<R^n. \tag{5}$$

The addition of the lower rows to row $0$ is required before the expansion in (3); it is implicit in the source's compressed determinant sentence.

For $z\in\mathbb Z^n$, define the balanced lift

$$L_{B_0}(z)=\left(B_0z,-\sum_{i=1}^n(B_0z)_i\right)\in\mathbb Z^{n+1}. \tag{6}$$

[[additive_combinatorics/adamczewski_2026_erdos1/lemma_4_1|Lemma 4.1]] proves that every nonzero $z$ satisfies $\|L_{B_0}(z)\|_\infty\geq R$.

Upper-triangular basis

We next change the domain basis using integer column operations. In the first row of $B_0$, Euclid's algorithm, implemented by column swaps, column signs, and integer column transvections, replaces the row by $(g,0,\ldots,0)$, where $g$ is the nonzero gcd of its entries. The lower-right minor remains nonsingular because the whole determinant is nonzero. Repeating there gives a lower-triangular matrix. Altogether this right-multiplies by a unimodular integer matrix, so it preserves the absolute determinant and bijects $\mathbb Z^n$ with itself in the cube-exclusion statement.

Reverse both the row and column orders to obtain an upper-triangular matrix. The column reversal is another integer basis change. The row reversal merely permutes the coordinates of $Bz$: it preserves their maximum absolute value and their sum, so it preserves the balanced norm in (6). Finally, if necessary, multiply one column by $-1$. This also preserves the lattice and orients the determinant so that, for the resulting upper-triangular matrix $B$,

$$(-1)^n\det B=D>0,\qquad D=|\det B_0|. \tag{7}$$

The final matrix therefore satisfies

$$z\ne0\Longrightarrow \left\|\left(Bz,-\sum_i(Bz)_i\right)\right\|_\infty\geq R, \qquad kD<R^n. \tag{8}$$

Source and dependencies

An explanation of the proof of Erdős Problem 1, preliminary exposition with no named author (erdosproblems.com, 2026), §4, equations (11)–(16), pp. 5–6. The edition read is named on the source card. The displayed row operation and the effect of row reversal on the balanced sum make explicit details suppressed by the exposition. The Euclidean reduction is proved here at the level needed: every operation is unimodular, termination follows from the integer Euclidean algorithm, and nonsingularity permits induction on the lower-right minor.

Bears on. #1.