GPT 6 Astra Ultra attempt on Math-5596
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Erdős problem #350
1) FORMAL RESTATEMENT. For a finite dissociated set $A$ of positive integers, prove $\sum_{a\in A}a^{-1}<2$. Here dissociated means that all its subset sums are distinct.
2) CONTEXT AND PLAN. We reconstruct Ryavec's generating-function argument, published by Benkoski and Erdős in On weird and pseudoperfect numbers, Theorem 1, and prove its sharp finite refinement. No novelty is claimed. The equality case is the set of powers of two, not a consecutive interval of integers.
3) PROOF WITH THE SHARP FINITE BOUND. Let $|A|=r\ge1$, and arrange the $2^r$ subset sums as $0=s_0<s_1<\cdots<s_{2^r-1}$. Since these are distinct nonnegative integers, $s_j\ge j$. For $0<x<1$, \[ \prod_{a\in A}(1+x^a)=\sum_{j=0}^{2^r-1}x^{s_j} \le\sum_{j=0}^{2^r-1}x^j=\frac{1-x^{2^r}}{1-x}. \] Take logarithms, integrate with respect to $dx/x$, and substitute $y=x^a$ in each term on the left. The integrals converge at both endpoints. If \[ I=\int_0^1\frac{-\log(1-x)}x\,dx, \] then $0<I<\infty$ and \[ \int_0^1\frac{\log(1+x)}x\,dx =I-\int_0^1\frac{-\log(1-x^2)}x\,dx=I/2. \] Thus no evaluation of the Basel sum is needed. On the right, the substitution $y=x^{2^r}$ gives \[ \frac I2\sum_{a\in A}\frac1a\le(1-2^{-r})I, \qquad \sum_{a\in A}\frac1a\le2-2^{1-r}<2. \] The empty set also satisfies the requested strict inequality.
4) EQUALITY CHARACTERIZATION. If equality holds in the finite bound, then $s_j=j$ for every $j$: otherwise the generating-function inequality is strict on the entire interval $(0,1)$ and its integral is strict. Write $A=\{a_1<\cdots<a_r\}$. The smallest positive subset sum forces $a_1=1$. Inductively, if $a_i=2^{i-1}$ for $i\le h$, those terms fill exactly $[0,2^h-1]$. Dissociation forces $a_{h+1}\ge2^h$, while representing $2^h$ forces $a_{h+1}\le2^h$. Hence $a_{h+1}=2^h$. Conversely, the powers of two have distinct subset sums and reciprocal sum $2-2^{1-r}$.
5) VERIFICATION AND FINAL. The proof uses the number of subset sums, not the sum of the elements, so it applies to arbitrary dissociated positive integer sets. Every analytic step is justified by finite products and convergent positive integrals. SOLVED. The sharp bound and its equality case are fully proved here, with literature attribution.