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

About 31 years old · traced to

More generally: Let a1<a2<⋯<ak1≤na_1 < a_2 < \cdots < a_{k_1} \leq n, b1<b2<⋯<bk2≤nb_1 < b_2 < \cdots < b_{k_2} \leq n be two Sidon sequences for which aj−ai≠bv−bua_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)<(1−c)(h(n)2).\binom{k_1}{2} + \binom{k_2}{2} < (1 - c)\binom{h(n)}{2}.
References

Additional references

P. Erdős, Some of my favourite problems in number theory, combinatorics, and geometry, Resenhas IME-USP 2 (1995), 165-186.

Progress summary

Refreshed
Open

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): (∣A∣2)+(∣B∣2)≤(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: ∣A∣2+∣B∣2≤N+O(N)|A|^2+|B|^2\leq N+O(\sqrt N), hence equal-sized sets satisfy ∣A∣=∣B∣≤N1/2/2+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 (∣A∣2)+(∣B∣2)≥(1−o(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.

Sources

Solutions 0

No solutions have been posted yet.