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Sets with no unique sum mod p

In general

A nonempty finite A⊆GA\subseteq G in an abelian group has a unique sum if some s∈A+As\in A+A has exactly one representation as an unordered pair {a,b}\{a,b\} from AA (repetition allowed). Equivalently: there exist a,b∈Aa,b\in A such that the only solutions of x+y=a+bx+y=a+b in A×AA\times A are the swaps of (a,b)(a,b). The complementary objects — sets in which every sum is multiply represented — are rare. For G=Z/pZG=\mathbb{Z}/p\mathbb{Z} with pp an odd prime, write m(p)m(p) for the least cardinality of such an AA with ∣A∣≥2|A|\ge 2.

Green's Problem 27 asks for the order of m(p)m(p). Unique differences are settled up to a constant (f(p)=Θ(log⁡p)f(p)=\Theta(\log p), Straus / Browkin–Diviš–Schinzel); unique sums are not. Bedert (2023/24) moved the bounds from log⁡p≪m(p)≪p\log p\ll m(p)\ll\sqrt{p} to ω(p)log⁡p≤m(p)≪(log⁡p)2\omega(p)\log p\le m(p)\ll(\log p)^2. Cao–Yuan (arXiv:2608.06728, August 2026) replaced the iterated-log lower bound by a second logarithm,

log⁡p⋅log⁡log⁡p≪m(p)≤(12(log⁡23)2+o(1))(log⁡2p)2,\log p\cdot\log\log p\ll m(p)\le\Bigl(\tfrac{1}{2(\log_2 3)^2}+o(1)\Bigr)(\log_2 p)^2,

and checked the headline inequalities in Lean. The gap between log⁡plog⁡log⁡p\log p\log\log p and (log⁡p)2(\log p)^2 is still the open problem.

The finite campaign is separate from that asymptotic gap. OEIS A398173 now records m(p)m(p) through p=73p=73, including m(59)=15m(59)=15 (checked 2026-09-09). The witnesses in compute/ match its first 15 terms, and a second, progression-driven implementation independently excludes every smaller size through p=53p=53. Named-midpoint exclusions now give the local equalities m(59)=15m(59)=15 and m(61)=15m(61)=15, matching the published table; they are not improvements of the record. The precise claims are in compute/q5/CLAIM.md and compute/q6/CLAIM.md. The first prime absent from the published exact table is 79. Extending the exact table to every prime p≤200p\le 200, plotting it against log⁡p\log p and (log⁡p)2(\log p)^2, and describing the extremal sets remains useful finite work. A new bound here is a checked table extension, not a new asymptotic.

Precise statement

Let pp be an odd prime and A⊆Z/pZA\subseteq\mathbb{Z}/p\mathbb{Z} with ∣A∣≥2|A|\ge 2. A sum s∈A+As\in A+A is unique if there is exactly one unordered pair {a,b}\{a,b\} from AA (repetition allowed) with a+b=sa+b=s. Write rA(s)=#{(a,b)∈A2:a+b=s}r_A(s)=\#\{(a,b)\in A^2:a+b=s\} for the ordered representation function. Because doubling is injective in Z/pZ\mathbb{Z}/p\mathbb{Z} for odd pp, AA has no unique sum if and only if

rA(s)∉{1,2}for every s∈Z/pZ.r_A(s)\notin\{1,2\}\qquad\text{for every }s\in\mathbb{Z}/p\mathbb{Z}.

(The value 00 is allowed: ss need not lie in A+AA+A.) Then

m(p)=min⁡{∣A∣:A⊆Z/pZ,  ∣A∣≥2,  A has no unique sum}.m(p)=\min\bigl\{|A|:A\subseteq\mathbb{Z}/p\mathbb{Z},\;|A|\ge 2,\;A\text{ has no unique sum}\bigr\}.

The minimum is defined: A=Z/pZA=\mathbb{Z}/p\mathbb{Z} works. For p=2p=2 the only two-element set is {0,1}\{0,1\}, and 0+10+1 has ordered multiplicity 22, so m(2)m(2) is undefined; restrict to odd primes.

Finite subquestion. Compute the exact integer m(p)m(p) for every prime 3≤p≤2003\le p\le 200. For each such pp, exhibit at least one extremal set. Record, for the extremal examples, whether they look like intervals, like Nedev/Bedert balanced sets, like Cao–Yuan symmetric squares C+CC+C of a weakly ternary-balanced CC, or like something else. Dump a CSV under compute/ and plot m(p)m(p) against log⁡p\log p and against (log⁡p)2(\log p)^2.

What a solution looks like

Computations

Finite status (2026-09-09): local exact values matching the record. m(59)=15m(59)=15 and m(61)=15m(61)=15. Each 15-set is checked in two languages, and every smaller size is excluded by an AP5 ceiling plus named midpoint classes, with C and Rust in agreement. This matches OEIS A398173 and does not improve it. See compute/q5/CLAIM.md and compute/q6/CLAIM.md.

Figures

Table through p=53p=53: figures/m_p.png.