GPT 6 Astra Ultra attempt on Math-5464

Erdős Problem #157: density constraints and the greedy-construction gap

Independent partial reconstruction, 23 September 2026.

1) Statement and known resolution

Does an infinite Sidon set exist which is an asymptotic basis of order three? A basis of order three represents every sufficiently large integer as a sum of three elements, with repetition permitted. Pilatte proved existence. This attempt proves necessary density estimates and an elementary greedy Sidon construction at the cubic-root scale. It does not establish the required three-sum coverage.

2) A necessary cubic-root lower bound

Write $A(N)=|A\cap[1,N]|$. If all integers from $N_0$ to $N$ are sums of three elements of $A$, positivity puts all their summands in $A\cap[1,N]$. There are only $\binom{A(N)+2}{3}$ unordered triples with repetition. Therefore \[ N-N_0+1\le\binom{A(N)+2}{3}\le\frac{(A(N)+2)^3}{6}, \] and hence $A(N)\ge(6(N-N_0+1))^{1/3}-2$. This is a necessary counting condition, not a covering argument.

3) A sharp leading upper density bound for Sidon sets

Let $B\subseteq[1,N]$ be Sidon with $k$ elements. Its positive differences are distinct, so $k(k-1)/2\le N-1$ and initially $k=O(\sqrt N)$. Fix $h=\lceil N^{3/4}\rceil$ and let $r(z)$ count pairs $(b,t)\in B\times[1,h]$ with $b+t=z$. We have $\sum_zr(z)=kh$, over at most $N+h$ positions. Also \[ \sum_zr(z)^2 =kh+2\sum_{d=1}^{h-1}(h-d)|\{b\in B:b+d\in B\}| \le kh+h(h-1), \] since every positive difference has at most one ordered representation. Cauchy--Schwarz gives \[ \frac{k^2h^2}{N+h}\le kh+h^2, \qquad k^2\le(N+h)(1+k/h)=N+O(N^{3/4}). \] Thus $k\le\sqrt N+O(N^{1/4})$.

This also rules out an asymptotic Sidon basis of order two. The number of sums at most $N$ is at most $\binom{A(N)+1}{2}\le N/2+O(N^{3/4})$, which cannot cover $N-O(1)$ integers. Order three avoids this obstruction because its number of potential representations is cubic in $A(N)$.

4) What the elementary greedy construction supplies

Start with $a_1=1$ and repeatedly choose the least positive integer which can be adjoined while preserving the Sidon property. For a current $k$-element set $B$, every forbidden new value $x\notin B$ must satisfy either \[ x+a=b+c\quad(a,b,c\in B),\qquad\text{or}\qquad 2x=b+c\quad(b,c\in B). \] Indeed, a collision between two different new sums $x+a$ and $x+b$ would require $a=b$, and $2x=x+a$ would require $x=a$. All other collisions involve an old sum and fall into one of the displayed forms. There are at most $k^3+k^2$ values in these forms, and $k$ old values. Among the first $k^3+k^2+k+1$ positive integers there is therefore an admissible new value.

Admissibility can only be lost when the set grows. Every value smaller than the last chosen element was either previously chosen or already forbidden and stays forbidden. Thus this least-admissible procedure is increasing and yields an infinite Sidon sequence with \[ a_{k+1}\le k^3+k^2+k+1. \] If $A(N)=k$, then $N<a_{k+1}\le(k+1)^3$, so $A(N)>N^{1/3}-1$. This proves the cubic-root lower order for the greedy sequence.

The exclusion $x=b+c-a$ contains a negative summand. It does not imply $x\in A+A+A$. Consequently saturation of this greedy Sidon set provides no proof that its three-fold sumset contains all sufficiently large integers. Nor does a lower counting bound guarantee that all the potential triple sums are distinct or cover every target.

5) A concrete failure of mere infinitude and the remaining gap

The powers $\{3^j:j\ge0\}$ form an infinite Sidon set: two-term sums have unique base-three digits, with a double contributing one digit two. They are not a basis of order three. The base-three digit sum of a sum of three powers is at most three, since carrying only decreases digit sum, whereas $1+3+\cdots+3^r$ has digit sum $r+1$. These give unbounded omitted integers for $r\ge3$.

ATTEMPT UNRESOLVED for construction of the required basis. The density bounds and the greedy construction are proved, but neither establishes the compatibility of Sidon uniqueness with eventual three-sum coverage. That is the essential missing step in this reconstruction.

6) Sources

Problem and current status: https://www.erdosproblems.com/157, checked 23 September 2026. C. Pilatte, A solution to the Erdős--Sárközy--Sós problem on asymptotic Sidon bases of order 3, https://arxiv.org/abs/2303.09659. Pilatte's function-field argument is cited for the known affirmative resolution and is not assumed in the partial proofs above. These are elementary known constructions and estimates, with no novelty claim.

The estimate credits the density constraints, the order-two obstruction and the precise greedy-construction gap; it is subjective, not a correctness probability.

7) COMPLETION ESTIMATE

COMPLETION: 20%