gpt pro 5.2 attempt on Math-824

1) FORMAL RESTATEMENT.

Let $p(m)$ denote the least prime factor of $m$ (for $m\ge 2$), and interpret the quantifier "$m$ is composite" as part of the condition (otherwise the auxiliary quantity $F(n)=\min_{m>n}(m-p(m))$ would be identically $0$ because primes have $p(m)=m$).

Question: Does there exist a function $f:\mathbb{N}\to\mathbb{N}$ with $f(n)\to\infty$ such that for all sufficiently large integers $n$ there exists a composite integer $m$ satisfying \[ n+f(n)<m<n+p(m)? \] Equivalently, writing $m=n+k$, we ask whether for all large $n$ there exists an integer $k>f(n)$ such that $n+k$ is composite and $p(n+k)>k$.

In terms of $F(n):=\min\{m-p(m): m>n\ \text{composite}\}$, the existence of such an $f$ is equivalent to the statement that $n-F(n)\to\infty$ (since $m<n+p(m)\iff m-p(m)<n$).

2) QUICK LITERATURE/CONTEXT CHECK.

The problem statement cites Erdős' question in [Er92e] about whether $n-F(n)\sim c\,n^{1/2}$ for some $c>0$. I will not use any results beyond what is explicitly stated in the problem text.

3) ATTACK PLAN.

• Rephrase the inequality in terms of a shift $k=m-n$ and the roughness condition $p(n+k)>k$.
• Prove unconditional structural bounds on any such $k$ (in particular, show $k\ll \sqrt{n}$ always).
• Exhibit an explicit infinite family of $n$ for which a solution exists with $k\asymp \sqrt{n}$ (prime squares).
• Run an exact brute-force check for small $n$ using the structural bound on $k$.

4) WORK.

Lemma 463.1 (shift/roughness equivalence). For an integer $n\ge 1$ and an integer $k\ge 1$, set $m=n+k$. Then \[ n<m<n+p(m)\quad\Longleftrightarrow\quad k<p(n+k). \] Consequently, the original problem is equivalent to: find $f(n)\to\infty$ such that for all large $n$ there exists an integer $k>f(n)$ with $n+k$ composite and $p(n+k)>k$.

Proof. The inequalities $n<m$ and $m<n+p(m)$ are respectively $k>0$ and $n+k<n+p(n+k)$, i.e. $k<p(n+k)$. Conversely, $k<p(n+k)$ implies $n<n+k<n+p(n+k)$. $\square$

Lemma 463.2 (prime-square intervals give $k\asymp\sqrt{n}$ infinitely often). Let $p$ be prime and let $n$ be any integer with \[ p^2-p<n<p^2. \] Then taking $m=p^2$ (composite) yields \[ n<m<n+p(m). \] In particular, for $n=p^2-p+1$ we obtain a valid shift $k=m-n=p-1\asymp \sqrt{n}$.

Proof. Let $m=p^2$. Then $p(m)=p$ and $m-n=p^2-n=:k$ satisfies $1\le k\le p-1$ because $p^2-p<n<p^2$. Hence $k<p=p(m)$, i.e. $m<n+p(m)$, and clearly $m>n$. $\square$

Lemma 463.3 (universal upper bound $k\ll\sqrt{n}$). Suppose $m=n+k$ is composite and satisfies $k<p(m)$. Then \[ k^2+k\le n. \] Equivalently, \[ k\le \left\lfloor\frac{\sqrt{4n+1}-1}{2}\right\rfloor<\sqrt{n}. \]

Proof. Since $k<p(m)$ and both are integers, $p(m)\ge k+1$. As $m$ is composite and $p(m)$ is its least prime factor, we can write $m=p(m)\cdot r$ with $r\ge p(m)$, hence $m\ge p(m)^2\ge (k+1)^2$. Using $m=n+k$ gives $n+k\ge (k+1)^2=k^2+2k+1$, i.e. $n\ge k^2+k+1$, hence $k^2+k\le n-1<n$ and in particular $k^2+k\le n$. $\square$

FAST REALITY CHECK (exact for small $n$). Define \[ h(n):=\max\{k\ge 1: n+k\ \text{composite and }\ p(n+k)>k\} \] (with $h(n)=\text{None}$ if no such $k$ exists). By Lemma 463.3 it suffices to test $1\le k\le \lfloor(\sqrt{4n+1}-1)/2\rfloor$.

A brute-force script checked $h(n)$ for $2\le n\le 50$ and produced the following (format: $n$, $h(n)$, witness $(m,p(m))$, and the tested bound $k_{\max}$):

 n h(n) witness(m,p) kmax
 2 None         None    1
 3    1       (4, 2)    1
 4 None         None    1
 5    1       (6, 2)    1
 6 None         None    2
 7    2       (9, 3)    2
 8    1       (9, 3)    2
 9    1      (10, 2)    2
10 None         None    2
11    1      (12, 2)    2
12 None         None    3
13    2      (15, 3)    3
14    1      (15, 3)    3
15    1      (16, 2)    3
16 None         None    3
17    1      (18, 2)    3
18 None         None    3
19    2      (21, 3)    3
20    1      (21, 3)    4
21    4      (25, 5)    4
22    3      (25, 5)    4
23    2      (25, 5)    4
24    1      (25, 5)    4
25    2      (27, 3)    4
26    1      (27, 3)    4
27    1      (28, 2)    4
28 None         None    4
29    1      (30, 2)    4
30 None         None    5
31    4      (35, 5)    5
32    3      (35, 5)    5
33    2      (35, 5)    5
34    1      (35, 5)    5
35    1      (36, 2)    5
36 None         None    5
37    2      (39, 3)    5
38    1      (39, 3)    5
39    1      (40, 2)    5
40 None         None    5
41    1      (42, 2)    5
42 None         None    6
43    6      (49, 7)    6
44    5      (49, 7)    6
45    4      (49, 7)    6
46    3      (49, 7)    6
47    2      (49, 7)    6
48    1      (49, 7)    6
49    2      (51, 3)    6
50    1      (51, 3)    6

In particular, one sees the prime-square phenomenon: for $n\in\{43,44,45,46,47,48\}$ the witness is $m=49=7^2$ and the gap $k=49-n$ ranges from $6$ down to $1$.

For $n\le 500$, the ten largest values found were:

(bestk, n, (m,p(m)))
(18, 419, (437, 19))
(18, 343, (361, 19))
(17, 420, (437, 19))
(17, 344, (361, 19))
(16, 477, (493, 17))
(16, 421, (437, 19))
(16, 375, (391, 17))
(16, 345, (361, 19))
(16, 307, (323, 17))
(16, 273, (289, 17))

All of these correspond to witnesses $m=p^2$ with $p\in\{17,19\}$.

5) VERIFICATION.

• Lemma 463.3 uses only the inequality $m\ge p(m)^2$ (true for composite $m$) and the integer step $p(m)\ge k+1$ when $k<p(m)$.
• The brute-force check is exact for each tested $n$ because Lemma 463.3 reduces the search to finitely many $k$.

6) FINAL. UNRESOLVED

(i) Strongest proved partial result. Unconditionally, any admissible gap $k=m-n$ must satisfy $k^2+k\le n$ (Lemma 463.3), so $k=O(\sqrt{n})$. Conversely, for infinitely many $n$ (namely $p^2-p<n<p^2$ with $p$ prime) the choice $m=p^2$ gives a solution with $k$ as large as $p-1\asymp\sqrt{n}$ (Lemma 463.2). Computations up to $n\le 500$ match this prime-square behavior.

(ii) First gap (crisp). Prove (or disprove) that $h(n):=\max\{k: n+k\text{ composite and }p(n+k)>k\}$ satisfies $h(n)\to\infty$ as $n\to\infty$. Equivalently: produce an explicit $f(n)\to\infty$ such that for every sufficiently large $n$ there exists a composite $m=n+k$ with $k>f(n)$ and $p(m)>k$.

(iii) Top 3 next moves.

1. Develop a sieve-theoretic lower bound for the existence of a $k$-rough composite in the short interval $(n+g(n),\,n+\lfloor(\sqrt{4n+1}-1)/2\rfloor]$ for some $g(n)\to\infty$.
2. Try a structured construction beyond prime squares: e.g. semiprimes $m=pq$ with $p\approx q\approx \sqrt{n}$, aiming to force $p(m)=p>m-n$.
3. Computation: for each $n$ search all $k$ up to the exact bound in Lemma 463.3 and study the distribution of $h(n)$ (especially the frequency of $h(n)=\text{None}$) to guess the true order of $\min_{n\le x} h(n)$.

(iv) Minimal counterexample structure. A counterexample to the existence of such an $f$ would require infinitely many large $n$ for which every $k$ in a long initial range (say $1\le k\le K(n)$ with $K(n)\to\infty$) fails: for each such $k$, either $n+k$ is prime or else $p(n+k)\le k$. Any "minimal" obstruction would thus be an $n$ for which all composites in $[n+1,\,n+K(n)]$ have unusually small least prime factor relative to their distance from $n$.