GPT 6 Astra Ultra attempt on Math-6195

Erdős Problem 1141: a small composite witness coprime to $n$

Statement and scope.

For every fixed positive integer $a$, only finitely many positive integers $n$ have $n-ak^2$ prime for every positive integer $k$ satisfying $(k,n)=1$ and $ak^2<n$. Taking $a=1$ answers problem 1141 negatively. This reconstructs the known Pollack-based resolution attributed to an internal OpenAI model in https://arxiv.org/abs/2604.06609; see https://www.erdosproblems.com/1141. It does not determine the largest admissible $n$.

Producing a small prime obstruction.

We use Pollack's Theorem 1.3 from Bounds for the first several prime character nonresidues, https://arxiv.org/abs/1508.05035: for fixed $\varepsilon,A>0$, every quadratic Dirichlet character modulo sufficiently large $M$ has at least $(\log M)^A$ primes $p\leq M^{1/4+\varepsilon}$ with value $1$. Only the nonprincipal case is needed. If $an$ is not a square, use the nonprincipal Kronecker character \[ \chi(t)=(4an/t),\qquad M=4an,\qquad \varepsilon=1/16. \] It yields an odd prime $p\nmid an$ with $p=O_a(n^{5/16})$ and $(an/p)=1$. Therefore $ak^2\equiv n\pmod p$ has a nonzero root $r$ modulo $p$: multiplying this congruence by $a$ reduces it to the residue class $an$.

If $an$ is a square, choose instead an odd prime $p\nmid an$ among the first $\omega(an)+1$ odd primes. Bertrand's postulate gives $p\leq2^{O(\omega(an))}=n^{o(1)}$, and again the congruence has a nonzero root. The estimate $\omega(n)=o(\log n)$ follows from $n\geq(\omega(n)+1)!$.

Making the root coprime to $n$.

Choose $1\leq r<p$ representing the root, and let $s=\omega(n)$. Put $T=2^{2s+1}$. For every squarefree divisor $d$ of $n$, the condition $d\mid r+pt$ specifies one residue class of $t$ modulo $d$, since $p\nmid n$. Inclusion--exclusion therefore counts \[ \#\{0\leq t<T:(r+pt,n)=1\} =T\frac{\varphi(n)}n+E,\qquad |E|\leq2^s. \] Because $\varphi(n)/n=\prod_{q\mid n}(1-1/q)\geq2^{-s}$, this count is at least $2^s>0$. Select a valid $t$, and set $k=r+pt$. Then \[ (k,n)=1,\qquad 1\leq k<pT,\qquad ak^2\equiv n\pmod p. \] Moreover $T=n^{o(1)}$, so in the nonprincipal case \[ ak^2\leq a p^2T^2=O_a(n^{5/8+o(1)})=o(n); \] the bound is even smaller in the principal case. For all sufficiently large $n$, it follows that $ak^2<n$ and $n-ak^2>p$. Thus $n-ak^2$ is a positive multiple of $p$ greater than $p$, and is composite.

This supplies an admissible counterexample $k$ for every sufficiently large $n$, rather than merely for a density-one subsequence. Constants and the final threshold may depend on the fixed $a$, which is precisely the required quantifier order.

Completion estimate.

The general finiteness theorem is proved relative to Pollack's explicitly stated established input, with the coprimality adjustment proved here. COMPLETION: 100%