gpt pro 5.2 attempt on Math-900

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 $\le k^{2}$.