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Claim. The answer to [[problems/arithmetic_functions/E0051/_index|Problem 51]] is yes, with the set $A=\{a_k:k\ge1\}$, $a_k=\prod_{i\le k}(p_i-1)$ over the first $k$ primes $p_1<\dots<p_k$. The manuscript asserts that the least $n$ with $\phi(n)=a_k$ is $n_{a_k}=\prod_{i\le k}p_i$, so that $n_{a_k}/a_k=\prod_{i\le k}p_i/(p_i-1)\to\infty$ by the divergence of the product over the primes. The manuscript, a little over a page long, was posted to the site's discussion thread on 2026-01-11 by the user colossal_noob, who wrote that ChatGPT (free version) had produced it in one attempt; the user could not say whether the argument was already in the literature. The claimant is the user who published it.

Refutation. The minimality step is false, so the ratio $n_{a_k}/a_k$ is not what the manuscript computes. Three replies in the thread on the same day say so: the first (14:27) observes that the proof of minimality is incomplete, since $a_k+1$ may itself be prime, as it is for $k=1$, $2$ and $15$ (for $k\ge2$ that prime is a preimage smaller than $n_k$), and suggests that $n_k$ is in fact the largest preimage; the site's curator, Thomas F. Bloom (14:31), gives the counterexamples $\phi(3)=\phi(2\cdot3)$, so the least preimage of $a_2=2$ is $3$ and not $6$, and, setting the factor $2$ aside, $\phi(3\cdot5\cdot7)=\phi(5\cdot13)=48$, so the least preimage of $a_4=48$ is at most $65$, below both the claimed $2\cdot3\cdot5\cdot7=210$ and its odd half $105$; the third (15:04) confirms that step 1, minimality, is wrong. Terence Tao recorded the attempt in the community wiki page "AI contributions to Erdős problems" (https://github.com/teorth/erdosproblems/wiki/AI-contributions-to-Erd%C5%91s-problems, Section 1(a), AI standalone) as a standalone AI attempt by ChatGPT free version on 11 Jan 2026 whose outcome the row gives as an incorrect proof found. The site's label is OPEN (page last edited 2025-09-30) and the problem's proof-claims tab carries no entry; the claim is recorded as rejected and decides nothing about the question.

Depends on. No page of this wiki.