Erdős Problem #43 — Maximizing total pair counts of two difference-disjoint Sidon sets

Erdős

More generally: Let a1<a2<<ak1na_1 < a_2 < \cdots < a_{k_1} \leq n, b1<b2<<bk2nb_1 < b_2 < \cdots < b_{k_2} \leq n be two Sidon sequences for which ajaibvbua_j - a_i \neq b_v - b_u for all i<ji < j and u<vu < v. How large can

max((k12)+(k22))\max\left(\binom{k_1}{2} + \binom{k_2}{2}\right)

be? I guessed [31] that it is less than (h(n)2)+O(1)\binom{h(n)}{2} + O(1). I offer 100 dollars for a proof or disproof. Assume next that k1=k2k_1 = k_2. I am sure that then

(k12)+(k22)<(1c)(h(n)2).\binom{k_1}{2} + \binom{k_2}{2} < (1 - c)\binom{h(n)}{2}.
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Progress summary

Refreshed
Claimed solved

Reported constructions now refute both proposed statements, but no retrieved paper confirms the claims.

Erdős offered a prize in 1995 for proving or disproving the first assertion. The problem also asks whether equal-sized sets permit a fixed positive improvement.

Known results

  • Erdős (1982): (A2)+(B2)(1+o(1))N/2(f(N)2)\binom{|A|}{2}+\binom{|B|}{2}\leq(1+o(1))N/2\sim\binom{f(N)}{2}, though the O(1)O(1) error term may be too optimistic.
  • Tao: A2+B2N+O(N)|A|^2+|B|^2\leq N+O(\sqrt N), hence equal-sized sets satisfy A=BN1/2/2+O(1)|A|=|B|\leq N^{1/2}/\sqrt{2}+O(1), without a fixed positive improvement.

Barreto's negative answer (recorded May 2026)

The construction solving Problem #42 gives A=f(N)|A|=f(N) with B|B|\to\infty, refuting the first assertion. Barreto is credited with infinitely many equal-sized examples satisfying (A2)+(B2)(1o(1))(f(N)2)\binom{|A|}{2}+\binom{|B|}{2}\geq(1-o(1))\binom{f(N)}{2}, refuting any fixed proportional improvement; a finite N=24N=24 example even exceeds (f(N)2)\binom{f(N)}{2}.

Current status (as of May 2026): negative answers to both questions are recorded, but formal independent corroboration remains absent.

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