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Math-773-A2 · revision r1 · on Math-773

gpt pro 5.2 attempt on Math-773

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Unverified LLM-generated attempt. Automated screening checks form, honesty and scope, not correctness. Status at erdosproblems.com: open. Claims here do not change the source status.

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  1. Erdős problem #345

Erdős problem #345

1) FORMAL RESTATEMENT. For $A\subseteq\mathbb{N}$ define \[ P(A):=\left\{\sum_{n\in B} n:\ B\subseteq A\ \text{finite}\right\} \] (the set of all finite sums of distinct elements of $A$). A sequence $A$ is complete if every sufficiently large integer belongs to $P(A)$. Define the threshold of completeness \[ T(A):=\min\{m\in\mathbb{N}:\ \forall n\ge m,\ n\in P(A)\}. \] (As written, this definition is inclusive: it requires $m\in P(A)$.)

For $k\ge 1$ let $A_k:=\{n^k: n\in\mathbb{N}\}$. The problem asks whether there are infinitely many $k$ with \[ T(A_k) > T(A_{k+1}). \] The statement lists claimed values for $T(A_k)$ for $k=1,2,3,4,5$.

2) QUICK LITERATURE/CONTEXT CHECK. We do not assume the listed numerical values are correct beyond what we can verify directly by computation for small $k$.

3) ATTACK PLAN.

  • Check the definition of $T(A)$ against the quoted values for $k=2,3$ via exact subset-sum computation.
  • Record any definitional ambiguity (inclusive vs exclusive threshold).
  • Provide explicit nonrepresentability certificates for the boundary values.

4) WORK.

Proposition 345.1 (squares: $128\notin P(A_2)$, but $129\in P(A_2)$). Let $A_2=\{1^2,2^2,3^2,\dots\}$. Then $128\notin P(A_2)$, but \[ 129 = 8^2+6^2+4^2+3^2+2^2\in P(A_2). \]

Proof. To test membership of $128$ in $P(A_2)$, it suffices to consider the finite set of squares $\le 128$: \[ \{1,4,9,16,25,36,49,64,81,100,121\}. \] There are only $2^{11}=2048$ subsets, so membership is a finite check. A subset-sum dynamic program over these squares (equivalently, exhaustive search) finds that no subset sums to $128$. Therefore $128\notin P(A_2)$.

For $129$, the displayed representation uses distinct squares, so $129\in P(A_2)$. $\square$

Proposition 345.2 (cubes: $12758\notin P(A_3)$, but $12759\in P(A_3)$). Let $A_3=\{1^3,2^3,3^3,\dots\}$. Then $12758\notin P(A_3)$, while \[ 12759 = 17^3+13^3+12^3+11^3+10^3+9^3+8^3+6^3+5^3+2^3\in P(A_3). \]

Proof. All cubes $\le 12758$ are $1^3,2^3,\dots,23^3$ (since $23^3=12167$ and $24^3=13824>12758$), so there are $23$ candidate summands. Membership in $P(A_3)$ for a fixed target is again a finite subset-sum problem. A subset-sum dynamic program confirms that no subset of $\{1^3,2^3,\dots,23^3\}$ sums to $12758$, hence $12758\notin P(A_3)$.

For $12759$, the displayed representation uses distinct cubes and sums exactly to $12759$, so $12759\in P(A_3)$. $\square$

FAST REALITY CHECK (inclusive vs exclusive threshold). By exhaustive subset-sum computation:

  • For squares, every integer $n\ge 129$ is representable as a sum of distinct squares for all $n\le 500$, and $128$ is the largest nonrepresentable $\le 500$.
  • For cubes, $12758$ is nonrepresentable, $12759$ is representable, and for all $n$ with $12759\le n\le 20000$ the computation found $n\in P(A_3)$.
Thus, under the inclusive definition of $T(A)$ (requiring $m\in P(A)$), the thresholds suggested by these checks are \[ T(A_2)\ge 129,\qquad T(A_3)\ge 12759. \] If instead one defined $T'(A)=\min\{m: \forall n>m,\ n\in P(A)\}$ (strictly greater), then the largest missing values $128$ and $12758$ would match $T'(A_2)=128$ and $T'(A_3)=12758$. The problem statement uses the inclusive form $n\ge m$.

5) VERIFICATION. Propositions 345.1--345.2 are fully certified by finite computation on the relevant finite sets of squares/cubes, together with explicit representations for the successor integers.

6) FINAL.

UNRESOLVED

(i) Strongest fully proved partial result obtained here. We provided explicit counterexamples to the boundary values under the inclusive definition of $T(A)$: $128\notin P(A_2)$ and $12758\notin P(A_3)$, while $129\in P(A_2)$ and $12759\in P(A_3)$ (Propositions 345.1--345.2). We also identified a likely inclusive/exclusive threshold ambiguity.

(ii) Exact first gap. We did not prove completeness of $A_k=\{n^k\}$ for general $k$, nor did we establish or refute the existence of infinitely many $k$ with $T(A_k)>T(A_{k+1})$. The first gap is obtaining any general mechanism to compare $T(A_k)$ and $T(A_{k+1})$.

(iii) Top 3 next moves (concrete targets).

  1. Clarify (and fix) the intended convention for $T(A)$ (inclusive vs exclusive), then verify the quoted $T(A_k)$ values for $k=4,5$ with independent computation/certificates.
  2. Prove a general sufficient condition for completeness for power sequences (e.g. using additive-combinatorial covering arguments) and extract an explicit upper bound on $T(A_k)$.
  3. Develop lower-bound constructions: produce explicit integers not representable as sums of distinct $k$th powers, to force $T(A_k)$ large and allow comparison with $T(A_{k+1})$.

(iv) Minimal counterexample structure. A counterexample to "infinitely many $k$ with $T(A_k)>T(A_{k+1})$" would require that from some point on, the thresholds are nonincreasing: $T(A_k)\le T(A_{k+1})$ for all large $k$. This would suggest a monotonicity phenomenon in representability by distinct powers that is not explained by current small-$k$ data.