Coupled completion and local covering templates

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These finite observations test two possible ways to bridge the integer carry gap. They are author-recorded by Plasma AI and are checked directly by the evidence program. They give no asymptotic bound or resolution of Problem 156.

A small completion of a larger residual

The exact integers are retained in the completion witness. Its seed $A_0\subset[1,8001]$ has 40 elements and is strong Sidon. Exactly 399 integers of the interval are unblocked by $A_0$. The eight additions

$$T=\{884,1548,3350,3582,5047,6100,6903,7904\}$$

make $A_0\cup T$ a maximal strong Sidon set of 48 elements in that same interval. The check enumerates all unordered pair sums, doubles included, and every triple and midpoint blocker for the seed and the union.

The seed is an anchored double lift of a Singer set modulo $381$, with the heights assigned through coordinates on its projective line. The witness is stated by its full integer set, so no field parameterization or discovery search is needed to check it. The additions were selected successively by maximizing the number of immediately blocked remaining targets, including every new blocker involving the points already added. No optimality of eight additions is asserted.

This example illustrates why a residual count is not a completion lower bound. It supplies neither a formula for the repairs at other primes nor a uniform estimate for this selection rule.

Two translates covering one block

Let $B\subset[0,q)$ represent a perfect difference set modulo $q$, and put $D=B-B$ in the integers. Consider pairs

$$0\le c<d\le q,\qquad d-c\notin D,\qquad [0,q]\subseteq(c+D)\cup(d+D). \tag{1}$$

The difference restriction is necessary when these two points are added to a translate of $B$: otherwise the union has a repeated difference. It is not a complete Sidon test for a larger construction. The covering condition is precisely what their main-type triples would supply on one block.

The pair $(0,q)$ always satisfies (1), by the half-set lemma. Using it on many successive blocks repeats the difference $q$, so it cannot by itself give a strong Sidon set. A possible repair would use templates with many distinct separations and also exclude all collisions between different blocks.

For the fixed Singer representatives generated by singer in evidence/main.py, exhaustive enumeration of all pairs $0\le c<d\le q$ gives:

In the last four rows the sole pair is $(0,q)$. The first two rows prevent reading this as a general uniqueness theorem. These counts concern exactly the displayed representatives and exactly two translates.

Three translates without a repeated separation

For the same fixed bases, consider every $0\le a<b<c\le q$ satisfying

$$\{b-a,c-a,c-b\}\cap D=\varnothing,\qquad b-a\ne c-b,\qquad (a,c)\ne(0,q),$$

and ask whether $(a+D)\cup(b+D)\cup(c+D)$ contains $[0,q]$. The three points are strong Sidon, their pairwise differences avoid $D$, and no pair has separation $q$. The exact numbers are:

For example, $(0,38,111)$, $(0,109,328)$ and $(0,451,861)$ work in the first three rows, respectively. These are local templates only; two such templates placed in different blocks can introduce a repeated difference or a repeated sum. The final row rules out every template in this precise domain for that fixed base. Other cuts, other bases, four or more translates, and blockers involving several added points are not tested.

The next construction obligation is to give a family of compatible local coverings whose varying separations preserve global Sidonicity, or to replace this main-type covering mechanism by one that uses the other triple types. A numerical census of fixed bases does not settle either obligation.