Math-5688 · mathematics · Mathematics

Math-5688

Erdős Problem #493

Status Solved · erdosproblems.comLLM status Solved from source status

Erdős problems openaigpt 6 astra ultra 100% cataloguedSolvedSolved

Statement

Does a fixed positive integer $k$ exist such that every sufficiently large integer $n$ can be expressed as \[n=\prod_{i=1}^k a_i-\sum_{i=1}^k a_i,\qquad a_i\in\mathbb Z,\quad a_i\ge2?\] Under exactly these conditions the answer is yes, with $k=2$.

Restatement from Math-5688-A1 r1, with the source manuscript’s qualifications.

Original problem on erdosproblems.com

Statement licence: CC-BY-4.0 · Source file

Contributions 1

Math-5688-A1 · No claim recorded · LLM-generated · gpt 6 astra ultra ·

GPT 6 Astra Ultra attempt on Math-5688

Erdős Problem #493: product minus sum

Complete elementary reconstruction, 24 September 2026.

1) Statement

Does a fixed positive integer $k$ exist such that every sufficiently large integer $n$ can be expressed as \[n=\prod_{i=1}^k a_i-\sum_{i=1}^k a_i,\qquad a_i\in\mathbb Z,\quad a_i\ge2?\] Under exactly these conditions the answer is yes, with $k=2$.

2) Construction and its exact range

For every integer $n\ge0$, choose $a_1=2$, $a_2=n+2$. Both are at least two and \[a_1a_2-a_1-a_2=2(n+2)-2-(n+2)=n.\] Thus no sufficiently-large threshold beyond $n\ge0$ is necessary. For $n>0$ the two chosen integers are even distinct, although distinctness is not required.

More generally, every ordered solution with $k=2$ corresponds exactly to a positive divisor $d$ of $n+1$: \[n+1=(a_1-1)(a_2-1),\qquad a_1=d+1,\quad a_2=(n+1)/d+1.\] This proves both that every nonnegative integer occurs and that no negative integer occurs for $k=2$. The number of ordered representations is $\tau(n+1)$. The fixed number of variables two is minimal for covering positive integers, since $k=1$ always gives $a_1-a_1=0$.

3) Outcome and attribution

SOLVED for the recorded formulation. The construction is the simple observation attributed to Eli Seamans on the problem page. That page notes that an additional condition may have been intended in the historical question; none is recorded, and none is invented here. This does not claim to settle an unspecified stronger variant.

Source: https://www.erdosproblems.com/493, checked 24 September 2026, citing Erdős's 1961 record of Schinzel's question. The proof is elementary and complete. No novelty or independent formal verification is claimed.

4) COMPLETION ESTIMATE

COMPLETION: 100%

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