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Statement

Printed p. 34 introduces, for a sequence 1≤a1<a2<⋯1\le a_1<a_2<\cdots of integers, the condition that

∑i=1s1ari=∑i=1s2ali,r1<⋯<rs1,l1<⋯<ls2,s1≠s2,(1’)\sum_{i=1}^{s_1}a_{r_i}=\sum_{i=1}^{s_2}a_{l_i},\qquad r_1<\cdots<r_{s_1},\quad l_1<\cdots<l_{s_2},\quad s_1\ne s_2, \tag{1'}

has no solution for any choice of s1≠s2s_1\ne s_2: two sums of distinct terms with different numbers of summands never coincide, or, as the English summary (p. 38) says of the equation (1'), it "is not solvable for every choice of s1≠s2s_1\ne s_2". This is the property Straus later called admissibility (the definition of the site's Problems 789, 874 and 875).

Theorem IV (printed p. 34). Let AA be a sequence for which (1') has no solution. Then for every xx

A(x)<Cx5/6,A(x)<Cx^{5/6},

where CC is a sufficiently large absolute constant. The text adds that Theorem IV shows condition (1') to give a much sharper bound than the condition (1) of Theorems I–III, and, after the proof (p. 36), that the exponent 5/65/6 can probably be improved and the large sieve probably avoided, but that neither had been achieved.

Construction (printed p. 34, before the theorem). Modifying the recursive construction of pp. 32–33, with Bi+1=2∑r≤kiarB_{i+1}=2\sum_{r\le k_i}a_r as there (the parenthetical on p. 34 says this definition still holds but prints it without the factor 22; display (20) on the same page has ∑r≤kiar=Bi+1/2\sum_{r\le k_i}a_r=B_{i+1}/2), to

aki+l=1+lBi+1+Bi+12,1≤l<Bi+11/2,(16’)a_{k_i+l}=1+lB_{i+1}+B_{i+1}^2,\qquad1\le l<B_{i+1}^{1/2}, \tag{16'}

one obtains, the paper says, a sequence with A(x)>cxαA(x)>cx^\alpha for every xx (the paper says the value of α\alpha would be easy to determine but does not give it) in which (1') has no solution for any s1,s2s_1,s_2; a half-page sketch follows (the minimal-counterexample argument around display (20)). The English summary (p. 38) states it as "There exists such a sequence with A(x)>cxαA(x)>cx^\alpha for every xx if α\alpha is sufficiently small." Erdős, Nicolas and Sárközy cite this passage as the existence of an infinite admissible set with A(x)>xcA(x)>x^c for an unspecified c>0c>0 (their Théorème 2 gives an explicit exponent).

Source. P. Erdős, Számelméleti megjegyzések, III. Néhány additív számelméleti problémáról, Mat. Lapok 13 (1962), 28–38 (Hungarian); printed p. nn is PDF p. n−27n-27 of the eleven-page scan read for this page. Condition (1'), the construction (16') and Theorem IV on printed p. 34 (PDF p. 7), the proof on pp. 35–36 (PDF pp. 8–9), the English summary on p. 38 (PDF p. 11), read on the page images; the Hungarian prose is rendered in the corpus's words, and the displays keep the paper's numbering in modern notation: (1') is printed with the right-hand summand alja_{l_j}, a slip for alia_{l_i}, and (16') ends with a clause aki+1=aki+[Bi+11/2]a_{k_{i+1}}=a_{k_i+[B_{i+1}^{1/2}]} not shown here.

Read depth. Claims checked: condition (1'), the construction (16') with its claimed properties and Theorem IV were read clause by clause on the page images and compared with the English summary. The proof of Theorem IV was read for its structure only (below) and is not checked here; the sketch for the construction was not checked.

Proof pointer

Pages 35–36. The Lemma on p. 35 is Rényi's sharpening of Linnik's large sieve (A. Rényi, Compositio Math. 8 (1950), 68–75): for integers a1<⋯<aZ≤Na_1<\cdots<a_Z\le N and functions 0<f(p)<10<f(p)<1, Q(p)<1Q(p)<1 (so printed; the application below takes f(p)=p/3f(p)=p/3 and Q(p)=2Q(p)=2, outside these ranges) on the primes p<12N1/3p<\tfrac12N^{1/3}, the number Z(p,h)Z(p,h) of ai≡h(modp)a_i\equiv h\pmod p satisfies ∣Z(p,h)−Z/p∣<Z/(pQ(p))|Z(p,h)-Z/p|<Z/(pQ(p)) for all but at most 9NQ2/(Zτ)9NQ^2/(Z\tau) primes pp (with τ=min⁡f(p)/p\tau=\min f(p)/p, Q=max⁡Q(p)Q=\max Q(p)) and, for the other primes, all but possibly f(p)f(p) of the classes hh. With f(p)=p/3f(p)=p/3, Q(p)=2Q(p)=2 and Z=CN5/6Z=CN^{5/6} there is a prime pp in (14N1/3,12N1/3)(\tfrac14N^{1/3},\tfrac12N^{1/3}) with (21) Z(p,h)>Z/(2p)Z(p,h)>Z/(2p) for at least 23p\tfrac23p residue classes; the congruence ai+aj≡0(modp)a_i+a_j\equiv0\pmod p then has at least (Z/2p)2p/6=Z2/(24p)>C2N4/3/12(Z/2p)^2p/6=Z^2/(24p)>C^2N^{4/3}/12 solutions (22), while the sums ai+aj<2Na_i+a_j<2N fall into fewer than 8N2/38N^{2/3} multiples of pp (23); so two distinct multiples m1=pn1m_1=pn_1, m2=pn2m_2=pn_2 each have at least C2N2/3/100C^2N^{2/3}/100 representations as ai+aja_i+a_j (24), and for CC large this exceeds max⁡(n1,n2)\max(n_1,n_2), so that the number n1pn2=n2pn1n_1pn_2=n_2pn_1 can be written as a sum of n1n_1 representations of m2m_2 and as a sum of n2n_2 representations of m1m_1, two sums of distinct terms with 2n1≠2n22n_1\ne2n_2 summands, contradicting (1'). The finite form: the Lemma and the argument concern the terms a1<⋯<aZ≤Na_1<\cdots<a_Z\le N only, so the proof applies as written to a finite set B⊆{1,…,N}B\subseteq\{1,\ldots,N\} in which (1') has no solution and gives ∣B∣<CN5/6|B|<CN^{5/6}; this reading is made here from the proof's structure and is the one Erdős's 1965 survey gives the theorem ("It is known that h(n)<c8n5/6h(n)<c_8n^{5/6}", printed p. 188 of Proc. Sympos. Pure Math. VIII, citing this paper).

Dependencies

Rényi's form of the large sieve (the Lemma, p. 35), quoted from Compositio Math. 8 (1950), 68–75; not held here.

Bears on

  • Problem 789: the site's "Erdős [Er62c] proved h(n)≪n5/6h(n)\ll n^{5/6}". Theorem IV bounds the counting function of an admissible sequence; applied to the admissible subsets of A={1,…,n}A=\{1,\ldots,n\} (the finite form above) it gives h(n)≤Cn5/6h(n)\le Cn^{5/6} for the problem's h(n)h(n), a one-line deduction made here. The paper contains no lower bound for h(n)h(n); the site's attribution to [Er62c] of the improvement h(n)≫(nlog⁡n)1/3h(n)\gg(n\log n)^{1/3} is not supported by its eleven pages, which carry no statement of that form.
  • Problem 874: the origin of the admissibility condition (Deshouillers and Freiman, p. 141: "introduced by P. Erdős in 1962 ... and called admissibility by E.G. Straus in 1966"), and the bound k(N)<CN5/6k(N)<CN^{5/6} for the problem's k(N)k(N), superseded by Straus's (4/3+o(1))N1/2(4/\sqrt3+o(1))N^{1/2}.
  • Problem 875: the construction (16') is the earliest infinite admissible sequence of polynomial growth on record here; its exponent is unspecified, and the paper says nothing about the gaps an+1−ana_{n+1}-a_n.