Reused writeup on Math-900 (gpt pro 5.2)

Erdős Problem 681 --- collection record

PROVENANCE AND STATUS

Official problem: https://www.erdosproblems.com/681

Saved collaborative-database status: open. Snapshot checked: 2026-09-23T17:22:24Z. Latest status and formalization links: https://teorth.github.io/erdosproblems/

This is an attributed collection of existing writeups. The model-folder names below identify their original attribution; placement in GPT_6_Astra_Ultra does not attribute their mathematical work to that model. All locally available source versions selected for this problem are reproduced below, without editing their text.

Proof claims, computations, literature statements, and completion estimates in the reused text have not been independently verified in this collection pass. A source claiming a solution does not change the saved upstream status.

Primary source (latest numbered version in the preferred model folder): attacks/erdos/gpt_pro_5.2/681.tex.

REUSED SOURCE 1

Original attribution: gpt_pro_5.2. Original file: attacks/erdos/gpt_pro_5.2/681.tex.

FORMAL RESTATEMENT. Is it true that for all sufficiently large integers $n$ there exists an integer $k\ge1$ such that \[ \text{$n+k$ is composite}\qquad\text{and}\qquad p(n+k)>k^{2}\ ? \]

QUICK LITERATURE/CONTEXT CHECK. No additional results are stated in the problem text beyond the definition of $p(m)$. I do not use external literature.

ATTACK PLAN. The requirement "$n+k$ composite" is restrictive because for composite $m$ we always have $p(m)\le\sqrt m$. Thus $p(n+k)>k^{2}$ forces $k^{2}<\sqrt{n+k}$, i.e. $k\lesssim n^{1/4}$. So for each $n$ there are only very few $k$ worth checking. There is also a trivial family of $n$ for which $k=1$ works.

WORK. Lemma 1 (Many $n$ work with $k=1$). If $n$ is odd and $n\ge3$, then $k=1$ satisfies the condition. More generally, if $n+1$ is composite, then $k=1$ satisfies the condition.

Proof. If $n+1$ is composite, then $p(n+1)\ge 2$ and $k^{2}=1$, hence $p(n+1)>1$. Also $n+1$ is composite by assumption, so both requirements hold. In particular, if $n$ is odd and $n\ge3$, then $n+1$ is even and greater than $2$, hence composite, so the above applies. $\square$

Lemma 2 (If $n+1$ is an odd prime, any solution $k$ is small and odd). Assume $n+1$ is an odd prime (so $n$ is even). If $k\ge2$ is even, then $n+k$ is even and composite, hence $p(n+k)=2\le k^{2}$; therefore no even $k\ge2$ can work. Moreover, if $n+k$ is composite and $p(n+k)>k^{2}$, then necessarily $k^{4}<n+k$.

Proof. If $n$ is even and $k$ is even, then $n+k$ is even. For $k\ge2$ we have $n+k\ge n+2>2$, so $n+k$ is composite and its least prime divisor is $2$. Since $k^{2}\ge4$, we get $p(n+k)=2\le k^{2}$, contradicting $p(n+k)>k^{2}$.

For the second statement: if $m=n+k$ is composite, then $p(m)\le \sqrt m$ as in Lemma 3 of Problem 680 (write $m=ab$ with $a\le\sqrt m$ and note $p(m)\le a$). If also $p(m)>k^{2}$, then $k^{2}<\sqrt m$, i.e. $k^{4}<m=n+k$. $\square$

VERIFICATION (FAST REALITY CHECK). Because any solution must satisfy $k^{4}<n+k$ (Lemma 2), it is enough to search $k\le \lfloor (n+k)^{1/4}\rfloor$, i.e. $k=O(n^{1/4})$. I exhaustively searched all $n\le 200\,000$ and found:

Total n tested: 200000
Number of n with no k found: 9523
Largest n with no k found: 199998
Every failure satisfies n+1 is prime: True
Among n with n+1 prime (17983 values), successes: 8460, failures: 9523
Maximum minimal k among the successes with n+1 prime: 19 (at n=145680)

So, unlike Problem 680, there are many large $n$ (up to $200\,000$) for which no such $k$ exists. This suggests the statement might be false, but this is only numerical evidence.

FINAL. UNRESOLVED.

(i) Strongest proved partial result. If $n+1$ is composite then $k=1$ works (Lemma 1). Hence any counterexample must satisfy $n+1$ prime (and then $n$ is even). For such $n$, any potential witness $k$ must be odd and satisfy $k^{4}<n+k$ (Lemma 2), so there are only $O(n^{1/4})$ possible $k$.

(ii) First gap (crisp). For $n=p-1$ with $p$ an odd prime, one needs to decide whether there always exists an odd $k\ll n^{1/4}$ such that $p-1+k$ is composite but has all prime factors $>k^{2}$. I do not have a method to guarantee such a "$k^{2}$-rough composite" in the extremely short interval $[p,\,p+O(p^{1/4})]$.

(iii) Top 3 next moves. $1$ Develop a sieve argument for odd $k\le n^{1/4}$ to show that not all of the numbers $n+k$ can have a small prime factor $\le k^{2}$. $2$ Search computationally for structural patterns in the failing $n$ (e.g. congruence classes of $n$ or of $n+1$) that might suggest an infinite family. $3$ Try to produce either (a) an unconditional lower bound on the density of $k^{2}$-rough composites in short intervals, or (b) a conditional construction (e.g. via primes in arithmetic progressions) that yields infinitely many failing $n$.

(iv) Minimal counterexample structure. A minimal counterexample $n$ must be even with $n+1$ prime, and for every odd $k$ with $k^{4}<n+k$ the number $n+k$ must either be prime (disallowed) or be composite with least prime factor $\le k^{2}$.

COMPLETION ESTIMATE

The primary source supplies no extractable numerical completion estimate. Its mathematical progress is not numerically assessed here.

Newly verified progress in this collection pass: $0\%$. This measures new verification work, not the amount of mathematics achieved in the reused text.