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Research notes on every problem, and a library of the papers behind them. Built from the open erdos repository.

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Use the parameters and positive coefficients from [[additive_combinatorics/adamczewski_2026_erdos1/normal_coefficients|the normal-coefficient construction]], so q0=2s+rq_0=2^{s+r} and

0<ai(t)≤2D(2s)n(0≤i≤n).(1)0<a_i(t)\leq2D(2^s)^n\qquad(0\leq i\leq n). \tag{1}

For every labeled pair

(i,j),0≤i≤n,0≤j<s+r,(i,j),\qquad 0\leq i\leq n,\quad 0\leq j<s+r,

form the positive integer

bi,j=2jai(t).(2)b_{i,j}=2^ja_i(t). \tag{2}

Distinct elements and subset sums

First the labeled weights in (2) are pairwise distinct. If bi,j=bi′,j′b_{i,j}=b_{i',j'}, use the two digit vectors having respectively the single nonzero digits 2j2^j in coordinate ii and 2j′2^{j'} in coordinate i′i'. Both digits are smaller than q0q_0. Their images under the map of [[additive_combinatorics/adamczewski_2026_erdos1/digit_injectivity|digit injectivity]] agree, so the digit vectors agree. Hence i=i′i=i' and j=j′j=j'.

We may therefore define the set

A={bi,j:0≤i≤n, 0≤j<s+r},A=\{b_{i,j}:0\leq i\leq n,\ 0\leq j<s+r\},

and its cardinality is exactly

∣A∣=(n+1)(s+r).(3)|A|=(n+1)(s+r). \tag{3}

A subset of AA chooses bits εi,j∈{0,1}\varepsilon_{i,j}\in\{0,1\}. For each ii let

xi=∑j=0s+r−1εi,j2j,x_i=\sum_{j=0}^{s+r-1}\varepsilon_{i,j}2^j,

so 0≤xi<q00\leq x_i<q_0, and the subset sum is ∑ixiai(t)\sum_i x_i a_i(t). Different subsets give different bit arrays because the labeled weights are distinct; uniqueness of binary expansion makes their digit vectors different. Digit injectivity then makes their subset sums different. Thus AA is sum-distinct.

Range and ratio

From (1), every element of AA is at most

2s+r−1 2D(2s)n=D2(n+1)s+r.2^{s+r-1}\,2D(2^s)^n =D2^{(n+1)s+r}.

Set

N=D2(n+1)s+r.(4)N=D2^{(n+1)s+r}. \tag{4}

Then A⊆{1,…,N}A\subseteq\{1,\ldots,N\}, and (3)–(4) give the exact cancellation

2∣A∣N=2(n+1)(s+r)D2(n+1)s+r=2nrD.(5)\frac{2^{|A|}}{N} =\frac{2^{(n+1)(s+r)}}{D2^{(n+1)s+r}} =\frac{2^{nr}}{D}. \tag{5}

The lattice reduction supplied kD<2nrkD<2^{nr}, so (5) yields

kN<2∣A∣.kN<2^{|A|}.

Source and dependencies

An explanation of the proof of Erdős Problem 1, preliminary exposition with no named author (erdosproblems.com, 2026), §7, equations (26)–(28), pp. 8–9. The edition read is named on the source card. The proof establishes pairwise distinctness of the labeled weights before treating them as a set; this is needed for the cardinality and subset encoding in (3).

Bears on. #1.