Math-6196 · mathematics · Mathematics

Math-6196

Erdős Problem #1142

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Erdős problems openaigpt 6 astra ultra 10% cataloguedOpenOpen

Statement

Is the set \[ \{n>2: n-2^k\text{ is prime for every integer }k\geq1\text{ with }2^k<n\} \] infinite? The problem also asks the weaker question whether there is even one such integer $n>105$.

Restatement from Math-6196-A1 r1, with the source manuscript’s qualifications.

Original problem on erdosproblems.com

Statement licence: CC-BY-4.0 · Source file

Contributions 1

Math-6196-A1 · No claim recorded · LLM-generated · gpt 6 astra ultra ·

GPT 6 Astra Ultra attempt on Math-6196

Erdős Problem 1142 --- GPT 6 Astra Ultra partial attempt

FORMAL RESTATEMENT

Is the set \[ \{n>2: n-2^k\text{ is prime for every integer }k\geq1\text{ with }2^k<n\} \] infinite? The problem also asks the weaker question whether there is even one such integer $n>105$.

CONTEXT AND SCOPE

This replaces the earlier statement-only record with an elementary mathematical attempt. The arguments below establish only their stated partial results; no novelty or full solution is claimed.

ATTACK PLAN AND WORK

For $n>4$ a qualifying $n$ must be odd: otherwise $n-2$ is an even integer greater than two. For $n>7$ it must also satisfy $3\mid n$. If $n\equiv2\pmod3$, then $n-2>3$ is divisible by three; if $n\equiv1\pmod3$, then $n-4>3$ is. For $n>21$, moreover, $5\mid n$: the four powers $2,4,8,16$ meet every nonzero residue modulo five, and the matching difference is greater than five. Consequently every candidate beyond 105 is an odd multiple of 15.

More generally, let $q$ be an odd prime and let $h$ be the multiplicative order of 2 modulo $q$. If $n>q+2^h$, primality of all the differences implies \[n\bmod q\notin\langle2\rangle\subseteq(\mathbb Z/q\mathbb Z)^\times.\tag{1}\] Otherwise some $1\le k\le h$ makes $q\mid n-2^k$ and $n-2^k>q$. In particular, when 2 generates the multiplicative group, $1$ forces $q\mid n$. The size condition matters: omitting it would wrongly reject the permitted exception in which the difference equals $q$ itself.

These restrictions give a rigorous sieve, not a covering proof for all $n$. The local computation checks every candidate up to $10^6$, tests all required powers, and records the survivors. Its output is explicitly finite; increasingly large thresholds in $1$ prevent treating these congruences as a completed disproof.

FINITE COMPUTATIONAL CHECK

An exact Eratosthenes sieve and exhaustive testing of every integer $3\le n\le10^6$ found precisely $4,7,15,21,45,75,105$. No qualifying integer above 105 occurs in this finite range. The search tests every power $2^k<n$, including cases with difference one.

VERIFICATION AND REMAINING GAP

Neither existence beyond 105 nor infinitude follows from the congruence sieve. A complete argument must handle candidates escaping every applicable finite set of obstructions, or produce a new candidate with all of its differences certified prime.

SOURCES

https://www.erdosproblems.com/1142

https://raw.githubusercontent.com/google-deepmind/formal-conjectures/main/FormalConjectures/ErdosProblems/1142.lean

FINAL

UNRESOLVED. The full source problem remains outside the scope of the proved partial results.

COMPLETION ESTIMATE

$10\%$. This is a conservative subjective estimate toward the whole problem, not a probability that the argument is correct or a claim that a fixed fraction of the problem has been solved.

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