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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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Statement

Setting (pp. 420, 422). For a finite set AA of non-negative integers, mAm_A is the least size of a set BB with every a∈Aa\in A of the form b+b′b+b', b,b′∈Bb,b'\in B. The paper takes A0={12,22,…,n2}A_0=\{1^2,2^2,\ldots,n^2\}, for which Theorem 1 gives only n1/2≤mA0≤n+1n^{1/2}\le m_{A_0}\le n+1 (p. 422).

Inequality 9 (p. 423, quoted). "$n^{2/3-\epsilon}\le m_{A_0}\le n/\log^Mn$, ϵ\epsilon arbitrarily small, MM arbitrarily large." The print sets the upper bound with MM over log⁡n\log n in the denominator; the proof (p. 423) ends with mA0≤n/log⁡Mnm_{A_0}\le n/\log^Mn "for large nn", and both bounds are read for fixed ϵ\epsilon and MM and all sufficiently large nn.

The upper bound (p. 423). For each odd prime pp the squares fall into exactly (p+1)/2(p+1)/2 residue classes mod pp, so by the Chinese remainder theorem they fall into ∏(p+1)/2\prod(p+1)/2 classes mod P=p⋅q⋅r⋯P=p\cdot q\cdot r\cdots for distinct odd primes p,q,r,…p,q,r,\ldots. Choosing one representative in [0,P)[0,P) of each such class together with all multiples of PP gives a basis, so

mA0≤p+12⋅q+12⋯+n2p⋅q⋅r⋯+1m_{A_0}\le\frac{p+1}{2}\cdot\frac{q+1}{2}\cdots+\frac{n^2}{p\cdot q\cdot r\cdots}+1

for any distinct odd primes p,q,r,…p,q,r,\ldots. Taking the odd primes in increasing order until their product lies between nlog⁡M+1nn\log^{M+1}n and 2nlog⁡M+2n2n\log^{M+2}n, using (pi+1)/2pi≤23(p_i+1)/2p_i\le\frac23 and that there are more than log⁡n/log⁡log⁡n\log n/\log\log n of them, gives the bound for large nn.

The lower bound (p. 423). The paper obtains it as an immediate corollary of Theorem 3, since x2−y2=kx^2-y^2=k has O(kϵ)O(k^\epsilon) solutions for every ϵ\epsilon; for k≤n2k\le n^2 this bounds DA0D_{A_0} by O(n2ϵ)O(n^{2\epsilon}).

Remarks on the same page (p. 423). The paper says that the upper bound shows the squares are not typical, since most sets of type (n,n2)(n,n^2) need more than n/2log⁡nn/2\log n elements by Theorem 2; the full sentence, with its unproved improvement to c nlog⁡log⁡n/log⁡nc\,n\log\log n/\log n, is quoted on the question_p425 page. The same residue-class device applied to the primes below xx gives a basis of size O(x/log⁡log⁡x)1/2O(x/\log\log x)^{1/2}, against the lower bound (x/log⁡x)1/2(x/\log x)^{1/2} from Theorem 1.

Source. P. Erdős and D. J. Newman, Bases for sets of integers, J. Number Theory 9 (1977), no. 4, 420--425: the set A0A_0 and the trivial bounds on p. 422, inequality 9, its proof and the remarks on p. 423. The edition read is identified on the source card.

Read depth. Claims checked: the statement and the residue-class bound were read clause by clause on the page images; the choice of primes and the final estimate were read for structure, not checked step by step. The lower bound rests on Theorem 3 and on the divisor bound the paper cites as known.

Proof pointer

Page 423, as summarized above: the residue-class basis for the upper bound and Theorem 3 with the divisor bound for the lower bound.

Dependencies

Theorem 1 for the comparison bounds; Theorem 3 for the lower bound; the prime number theorem and the bound O(kϵ)O(k^\epsilon) for the number of solutions of x2−y2=kx^2-y^2=k, both cited by the paper as known.

Bears on

  • Problem 333: the paper proves the bound for the first nn squares, a finite set; it says on p. 420 that results for infinite sets generally follow from finite ones by condensation but does not carry this out for the squares. The site's commentary, as the problem's claim page records, credits the paper with a basis of the squares whose counting function is o(N1/2)o(N^{1/2}), the case the problem generalizes. The bound does not bear on the problem's negative answer.
  • Problem 806: the squares are a set of type (n,n2)(n,n^2), and the bound shows that this particular set has a basis of o(n)o(n) elements; the problem asks the same for every set of type (n,n2)(n,n^2) (and every A⊆{1,…,N}A\subseteq\{1,\ldots,N\} with ∣A∣≤N1/2\lvert A\rvert\le N^{1/2}), which the bound does not decide.