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Source. Theorem 2, p. 5, with the constructions of Section 2.2 (pp. 6--7), of Greg Martin and Kevin O'Bryant, Constructions of Generalized Sidon Sets, J. Combin. Theory Ser. A 113 (2006), no. 4, 591-607, read in the arXiv edition arXiv:math/0408081v2 (21 Feb 2005) named on the source card.

Read depth. Claims checked: the statement, the constructions it rests on and the definitions it uses were read clause by clause on the page images; the proof (Section 3.2, pp. 9--13) was read for structure only. Nothing here is independently reviewed.

Statement

Setting (pp. 1--3). S∗S(k)S*S(k) counts ordered pairs (s1,s2)∈S×S(s_1,s_2)\in S\times S with s1+s2=ks_1+s_2=k (sums modulo nn for subsets of Zn\mathbb Z_n), and ∥S∗∥∞=max⁡kS∗S(k)\lVert S^*\rVert_\infty=\max_k S*S(k). With [n]={1,…,n}[n]=\{1,\ldots,n\},

R(g,n)=max⁡{∣S∣:S⊆[n], ∥S∗∥∞≤g}(equation (1), p. 2),R(g,n)=\max\{\lvert S\rvert : S\subseteq[n],\ \lVert S^*\rVert_\infty\le g\} \quad\text{(equation (1), p. 2)},

and C(g,n)C(g,n) is the same maximum over S⊆ZnS\subseteq\mathbb Z_n (equation (2), p. 3).

Theorem 2 (p. 5). Let qq be a prime power, and let k,g,f,x,yk,g,f,x,y be positive integers with k<qk<q.

  • (i) if pp is a prime, then C(2k2,p2−p)≥k(p−1)C(2k^2,p^2-p)\ge k(p-1);
  • (ii) C(2k2,q2−1)≥kqC(2k^2,q^2-1)\ge kq;
  • (iii) C(2k2,q2+q+1)≥kq+1C(2k^2,q^2+q+1)\ge kq+1;
  • (iv) if gcd⁡(x,y)=1\gcd(x,y)=1, then C(gf,xy)≥C(g,x) C(f,y)C(gf,xy)\ge C(g,x)\,C(f,y);
  • (v) R(gf,xy)≥R(gf, xy+1−⌈y/C(f,y)⌉)≥R(g,x) C(f,y)R(gf,xy)\ge R\bigl(gf,\,xy+1-\lceil y/C(f,y)\rceil\bigr)\ge R(g,x)\,C(f,y);
  • (vi) R(g, 3g−⌊g/3⌋+1)≥g+2⌊g/3⌋+⌊g/6⌋R\bigl(g,\,3g-\lfloor g/3\rfloor+1\bigr)\ge g+2\lfloor g/3\rfloor+\lfloor g/6\rfloor.

The hypothesis k<qk<q is printed for the whole theorem; part (i) involves no qq, and its construction (Section 2.2.1, p. 6) takes kk of the p−1p-1 sets Ruzsa(p,θ,i)\mathtt{Ruzsa}(p,\theta,i), 1≤i<p1\le i<p.

The constructions (pp. 6--7). Parts (i)--(iii) come from unions of kk disjoint Sidon sets drawn from one classical family: Ruzsa's sets modulo p2−pp^2-p, Bose's sets modulo q2−1q^2-1 indexed by nonzero k∈Fqk\in\mathbb F_q, and Singer's sets modulo q2+q+1q^2+q+1 indexed by pairs in Fq×Fq\mathbb F_q\times\mathbb F_q of which none is an Fq\mathbb F_q-multiple of another. In each case the paper shows the union of ∣K∣\lvert\mathcal K\rvert such sets has ∥⋅∗∥∞≤2∣K∣2\lVert\cdot^*\rVert_\infty\le2\lvert\mathcal K\rvert^2 and the stated size; each case has a worked example, a union of two sets with p=q=11p=q=11, and for the Ruzsa example the paper notes ∥⋅∗∥∞=8\lVert\cdot^*\rVert_\infty=8. Parts (iv) and (v) come from the Cilleruelo-Ruzsa-Trujillo construction (Section 2.2.4, p. 7): for S⊆ZxS\subseteq\mathbb Z_x with ∥S∗∥∞≤g\lVert S^*\rVert_\infty\le g and M⊆ZyM\subseteq\mathbb Z_y with ∥M∗∥∞≤f\lVert M^*\rVert_\infty\le f, the set M+yS⊆ZxyM+yS\subseteq\mathbb Z_{xy} has ∥(M+yS)∗∥∞≤gf\lVert(M+yS)^*\rVert_\infty\le gf. The paper places this in the line of Kolountzakis's observation that ∥(S∪(S+1))∗∥∞≤4\lVert(S\cup(S+1))^*\rVert_\infty\le4 for a Sidon set SS (p. 7). Part (vi) is witnessed by an explicit set in [0,3g−⌊g/3⌋][0,3g-\lfloor g/3\rfloor], a union of three integer intervals and one arithmetic progression of step 22 (p. 13). The print states that this set has ∥S∗∥∞\lVert S^*\rVert_\infty equal to g+2⌊g/3⌋+⌊g/6⌋g+2\lfloor g/3\rfloor+\lfloor g/6\rfloor, which is its cardinality; part (vi) needs ∥S∗∥∞≤g\lVert S^*\rVert_\infty\le g, and a direct computation here for 1≤g≤391\le g\le39 gives ∥S∗∥∞=g\lVert S^*\rVert_\infty=g and the printed cardinality, so the displayed value reads as a misprint for gg.

Proof pointer

Section 3.2 (pp. 9--13). For a disjoint union S=⋃SiS=\bigcup S_i of kk sets, ∥S∗S∥∞≤k2max⁡i,j∥Si∗Sj∥∞\lVert S*S\rVert_\infty\le k^2\max_{i,j}\lVert S_i*S_j\rVert_\infty, so parts (i)--(iii) reduce to showing the sets in each family are disjoint and that ∥Si∗Sj∥∞≤2\lVert S_i*S_j\rVert_\infty\le2 for every i,ji,j, including i=ji=j; this is done by unique factorization in Fp[x]\mathbb F_p[x], Fq[x]\mathbb F_q[x] and Fq2[x]\mathbb F_{q^2}[x] respectively (pp. 9--12). Part (iv) reduces a coincidence of gf+1gf+1 sums modulo yy and then modulo xx, using gcd⁡(x,y)=1\gcd(x,y)=1 (p. 12). Part (v) lifts the construction of part (iv) to the integers and shifts MM so that its largest gap, at least ⌈y/C(f,y)⌉\lceil y/C(f,y)\rceil, sits at the end of [y][y] (p. 12). Part (vi) states the size and ∥S∗∥∞\lVert S^*\rVert_\infty of the explicit set without further argument (p. 13).

Dependencies

The classical Sidon constructions of Ruzsa, Bose and Singer, which the paper reproves in the generality it needs, and the Cilleruelo-Ruzsa-Trujillo product construction.

Bears on

  • Problem 158: the problem's sets are those with ∥S∗∥∞≤4\lVert S^*\rVert_\infty\le4, and part (v) with g=f=2g=f=2 gives R(4,xy)≥R(2,x) C(2,y)R(4,xy)\ge R(2,x)\,C(2,y), the finite interleaved Sidon constructions behind the paper's σ(4)\sigma(4) bound (Theorem 3). These are finite sets, one for each nn; the paper does not combine them into one infinite set and says nothing about the lower limit the problem asks about.
  • Problem 30: just after the proof of part (v) the paper recalls Erdős's question, from Guy's problem C9, whether R(2,n)=n+O(1)R(2,n)=\sqrt n+O(1), and remarks that a gap not O(p)O(p) in Bose's Sidon set Bose(p,θ,1)\mathtt{Bose}(p,\theta,1) would answer it negatively (p. 12). That is a remark, not a result; and a negative answer to the O(1)O(1) question would not decide Problem 30, which asks for an error Oϵ(Nϵ)O_\epsilon(N^\epsilon).