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Let $C$ be one of the matrices supplied by [[additive_combinatorics/adamczewski_2026_erdos1/proposition_3_2|Proposition 3.2]], of order $n+1$: it is admissible, and its columns all sum to $q$. Let $R=2^r$ be a power of $2$ that clears its denominators, put $G=RC$, an integer matrix, and define the integer matrix $B$ by

$$B_{ij}=G_{ij}-G_{i0}\qquad(1\leq i,j\leq n).$$

Statement

The only $z\in\mathbb Z^n$ for which both

$$|(Bz)_i|<R\quad(1\leq i\leq n),\qquad \left|\sum_{i=1}^n(Bz)_i\right|<R$$

hold is $z=0$.

Equivalently, with

$$L_B(z)=\left(Bz,-\sum_i(Bz)_i\right),$$

every nonzero $z$ has $\|L_B(z)\|_\infty\geq R$.

The proof below uses only the admissibility of $C$, its common column sum and the integrality of $G$.

Proof

Set

$$w=(-z_1-\cdots-z_n,z_1,\ldots,z_n)\in\mathbb Z^{n+1}.$$

In each row $i\geq1$, the definition of $B$ gives

$$(Gw)_i=\sum_{j=1}^n(G_{ij}-G_{i0})z_j=(Bz)_i. \tag{1}$$

Summing the coordinates of $Gw$ multiplies the coordinate sum of $w$, which is zero, by the common column sum $Rq$:

$$\sum_{i=0}^n(Gw)_i =Rq\sum_{j=0}^nw_j =0.$$

Together with (1), this yields

$$(Gw)_0=-\sum_{i=1}^n(Bz)_i.$$

So the two hypotheses make every coordinate of $Gw=RCw$ smaller than $R$ in absolute value, that is, $\|Cw\|_\infty<1$, and admissibility of $C$ makes $w$, and with it $z$, zero.

Source and dependencies

An explanation of the proof of Erdős Problem 1, preliminary exposition with no named author (erdosproblems.com, 2026), §4, Lemma 4.1, p. 6, with the notation $L_B$ set on p. 5. The edition read is named on the source card. Admissibility is supplied by [[additive_combinatorics/adamczewski_2026_erdos1/proposition_3_1|Proposition 3.1]].

Bears on. #1.