GPT 6 Astra Ultra attempt on Math-5603
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Erdős problem #362
1) FORMAL RESTATEMENT. For a set $A$ of $N$ distinct positive integers, uniformly in the integer target $t$, prove that at most $C2^N/N^{3/2}$ subsets sum to $t$. If their cardinality is also prescribed as $l$, prove the uniform stronger bound $C'2^N/N^2$.
2) CONTEXT AND EXPLICIT INPUT. Sárközy--Szemerédi proved the first bound; Halász established the second through a multidimensional theorem. We use only the following scalar Sárközy--Szemerédi input, stated explicitly as Theorem 1.1 in Nguyen's A new approach to an old problem of Erdős and Moser: \[ \sup_z\mathbb P\left(\sum_{i=1}^m\epsilon_i b_i=z\right) \le C m^{-3/2} \quad\text{for distinct integers }b_i \text{ and independent uniform signs }\epsilon_i. \tag{SS} \] Its proof is external. We give the first conversion and a separate pairing argument deriving the two-constraint estimate from this scalar theorem. No novelty is claimed.
3) UNRESTRICTED CARDINALITY. Choose a uniformly random subset of $A=\{a_1<\cdots<a_N\}$, with independent inclusion indicators $\xi_i\in\{0,1\}$. Then \[ 2\sum_i a_i\xi_i-\sum_i a_i =\sum_i a_i(2\xi_i-1). \] Apply (SS) with $b_i=a_i$, and multiply the resulting probability bound by $2^N$. This proves the first assertion, with a constant independent of $t$.
4) PAIRING TO RETAIN THE CARDINALITY GAIN. First suppose $N=2m$. Pair the smallest with the largest, the second smallest with the second largest, and so on: \[ (a_i,a_{2m+1-i}),\qquad1\le i\le m. \] Their positive differences $d_i=a_{2m+1-i}-a_i$ are strictly decreasing, hence distinct. A uniform random subset can be generated independently for each pair as follows. With probability $1/2$ call it active and choose exactly one of its two members, each orientation equally likely. Otherwise call it inactive and choose either both members or neither, each with probability $1/2$. The four pair states are thus equally likely.
Condition on the set $J$ of active pairs, and let $h=|J|$. Inactive both/neither decisions are independent fair bits. The subset cardinality is \[ |S|=h+2Y,\qquad Y\sim\operatorname{Bin}(m-h,1/2), \] independent of all active orientations. If $m/4\le h\le3m/4$, the central binomial bound gives \[ \mathbb P(|S|=l\mid J)\le C_1m^{-1/2}. \tag{2} \] Now additionally fix every inactive decision. The subset sum, as the active orientations vary, has the form \[ c+\sum_{i\in J}d_i\eta_i, \] where the $\eta_i$ are independent fair $\{0,1\}$ bits and $c$ is fixed. By (SS), its largest atom is at most $C_2h^{-3/2}\le C_3m^{-3/2}$. Since the cardinality condition depends only on the inactive decisions and $J$, average this bound over those decisions satisfying it. Combining with $2$ gives \[ \mathbb P\big(\sum_{a\in S}a=t,\ |S|=l\mid J\big) \le C_4m^{-2} \] on the central range of $h$, uniformly in both targets.
The active count is $\operatorname{Bin}(m,1/2)$, so an exponential-moment estimate bounds the probability of $h<m/4$ or $h>3m/4$ by $2e^{-c m}$. Consequently \[ \mathbb P\big(\sum_{a\in S}a=t,\ |S|=l\big) \le C_4m^{-2}+2e^{-cm}\ll N^{-2}. \] For odd $N$, first condition on the inclusion of one element, and apply the even result to the remaining $N-1$ elements, shifting $t$ and $l$ as appropriate. Enlarging the absolute constant handles all small $N$. Multiplication by $2^N$ proves the second assertion.
5) WHY THESE POWERS CANNOT BE IMPROVED. Take $A=[1,N]$. For a uniform random subset, \[ \operatorname{Var}\Big(\sum_{i=1}^Ni\xi_i\Big) =\frac14\sum_{i=1}^Ni^2=O(N^3),\qquad \operatorname{Var}\Big(\sum_i\xi_i\Big)=N/4. \] Chebyshev's inequality places a fixed positive fraction of all subsets in an interval of $O(N^{3/2})$ possible sum values, so one sum occurs at least $c2^N/N^{3/2}$ times. Applying Chebyshev to both coordinates and a union bound similarly puts a fixed positive fraction in a rectangle with $O(N^{3/2})\cdot O(N^{1/2})=O(N^2)$ integer pairs. Hence some pair $(t,l)$ has at least $c'2^N/N^2$ representations. These are sharpness bounds up to constants, not exact extremizer claims.
6) VERIFICATION AND FINAL. Pair differences are distinct even if the original set is highly irregular. The cardinality restriction is imposed before averaging over independent active orientations, preserving the extra $N^{-1/2}$ factor. SOLVED relative to the explicitly stated scalar theorem (SS). The two-constraint reduction and both sharpness arguments are proved here; the deep scalar anti-concentration theorem remains an external dependency.