For n∈N, let P(n) mean that for every k∈N, if k^2<n and gcd(n,k)=1, then n-k^2 is prime. Are there infinitely many natural numbers n satisfying P(n)? Equivalently, is the set n∈N:forall k∈N, k^2<n land gcd(n,k)=1 Rightarrow n-k^2 is prime infinite?

Erdős Problem #1141 — Prime Differences from Squares

The problem asks whether infinitely many integers $n$ make $n-k^2$ prime for every admissible $k$ . The April 2026 paper claims that only finitely many such $n$ exist, giving a negative answer.

April 2026 claimed proof

The paper states a stronger result: for every fixed integer $a \ge 1$ , only finitely many $n$ satisfy that $n-ak^2$ is prime whenever $ak^2<n$ and $(k,n)=1$ . It presents this as a deduction from Pollack’s theorem on small prime quadratic residues. The authors say an internal OpenAI model produced the proof, that they checked it, and that ChatGPT-5.4 Pro succeeded on all five attempts; no independent verification is recorded.

Current status (as of September 2026): An arXiv manuscript claims a negative solution, but the proof remains unconfirmed; independent verification is not recorded.

Erdős Problem #1141 — Prime Differences from Squares

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