GPT 6 Astra Ultra attempt on Math-5463
mathematicsErdős problemscontributionManuscript
Contents
Erdős Problem #154: residue counts in the sumset of a dense Sidon set
Reconstruction with an explicit distribution theorem, 23 September 2026.1) Statement and quantitative objective
Suppose $A\subseteq[1,N]$ is Sidon and $k=|A|\sim\sqrt N$. Must $A+A$ be uniformly distributed over small moduli, in particular parity? We prove that for every sequence of positive moduli $m=o(\sqrt N)$, \[ \max_{t\bmod m}\left| \frac{|\{s\in A+A:s\equiv t\pmod m\}|}{|A+A|/m}-1 \right|\longrightarrow0.\tag{1} \] The external input is Kolountzakis's square-mean residue estimate for $A$. The passage to distinct sums, including doubles, is proved here.2) Precise external estimate
Write $a_x=|\{a\in A:a\equiv x\pmod m\}|$ and $e_x=a_x-k/m$, so $\sum_x e_x=0$. Set $\lambda=\max(0,\sqrt N-k)=o(\sqrt N)$. Theorem 2 of M. N. Kolountzakis, On the uniform distribution in residue classes of dense sets of integers with distinct sums, gives an absolute constant $C$ such that \[ E:=\left(\sum_{x\bmod m}e_x^2\right)^{1/2} \le C\begin{cases} N^{3/8}m^{-1/4},&\lambda\le N^{1/4}m^{1/2},\\ N^{1/4}\lambda^{1/2}m^{-1/2},&\lambda>N^{1/4}m^{1/2}. \end{cases}\tag{2} \] Its assumptions are $k\ge\sqrt N-\lambda$ with $\lambda=o(\sqrt N)$ and $m=o(\sqrt N)$, as satisfied here. Primary source: https://arxiv.org/html/math/9808061. We do not reprove the cosine-polynomial estimate underlying $2$. The modulus one case of $1$ is immediate, so it can also be separated from this input.3) Ordered convolution and the diagonal correction
The number of ordered pairs $(a,b)\in A^2$ with $a+b\equiv t$ is \[ C_t=\sum_xa_xa_{t-x}=\frac{k^2}{m}+\sum_xe_xe_{t-x}. \] The mixed terms vanish because the errors have sum zero. By Cauchy--Schwarz, \[ |C_t-k^2/m|\le E^2\quad\text{uniformly in }t.\tag{3} \] Let $d_t=|\{a\in A:2a\equiv t\pmod m\}|$. Because $A$ is Sidon, each sum with unequal summands has exactly two ordered representations and each double exactly one. Thus the number $S_t$ of distinct sums in residue $t$ is exactly \[ S_t=\frac{C_t+d_t}{2},\qquad |A+A|=\frac{k(k+1)}2. \] In particular, writing $S_t=C_t/2$ would incorrectly drop the doubles, including all their even contributions. Since $0\le d_t\le k$, \[ \left|S_t-\frac{k(k+1)}{2m}\right| \le\frac12E^2+\frac12(k+k/m).\tag{4} \]4) Uniform relative error
Divide $4$ by $k(k+1)/(2m)$. The resulting error is at most \[ \frac{mE^2}{k(k+1)}+\frac{m+1}{k+1}.\tag{5} \] The second term tends to zero since $k\sim\sqrt N$ and $m=o(\sqrt N)$. In the first case of $2$, the first term is \[ O\!\left(\frac{m^{1/2}N^{3/4}}{N}\right) =O\!\left(\sqrt{\frac{m}{\sqrt N}}\right)=o(1). \] In the second case it is $O(\lambda/\sqrt N)=o(1)$. These bounds hold in every residue and also when the two cases alternate along a sequence of $N$. This proves $1$.For parity there is an additional exact check. If $u$ elements of $A$ are even and $v$ are odd, then \[ S_0=\binom{u+1}{2}+\binom{v+1}{2},\qquad S_1=uv, \qquad S_0-S_1=\frac{(u-v)^2+k}{2}. \] At $m=2$, $(u-v)^2=2E^2=o(N)$ by $2$, while $k=o(N)$. Since $|A+A|\sim N/2$, both parities account for asymptotically half of the distinct sums. The exact even bias from doubles is consistent with this conclusion.
5) Outcome and sources
SOLVED, using the stated Kolountzakis estimate. The convolution, diagonal correction and range of moduli are fully justified above. The primary distribution theorem remains an external dependency; no new theorem, independent proof of that input, or formal verification is claimed.https://www.erdosproblems.com/154, checked 23 September 2026; Kolountzakis, primary paper linked above, Theorem 2.
The estimate measures complete coverage using the declared external result, not a correctness probability.