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Source. The unnumbered question following the Theorem, pp. 1--2, of Stefan Steinerberger, On an iterated arithmetic function problem of Erdős and Graham, arXiv:2504.08023v1 (2025), as identified on the source card.

Statement

Question (p. 1). Dropping the primality condition, the paper asks whether the equation

$$\prod_{p\mid n}\left(1-\frac{1}{p}\right)=\frac{\phi(n)}{n}=\frac{2}{3}+\frac{2}{3n}$$

has infinitely many solutions $n$.

The primality constraint dropped is the requirement in part (2) of the paper's Theorem that $8m+7$ be prime. The equation is $\phi(n)=\tfrac23(n+1)$, the equation $\phi(q)=\tfrac23(q+1)$ to which §§2.5 and 2.8 of the proof reduce, and the paper notes that every solution of $\phi(6m+5)=4m+4$ gives the solution $n=6m+5$ (p. 1).

What the paper records (pp. 1--2). The solutions $n=5,35,1295,1679615$, with $35=5\cdot7$, $1295=5\cdot7\cdot37$ and $1679615=5\cdot7\cdot37\cdot1297$. It says that the equation does not seem to have any other 'small' solutions, and maybe none at all, without stating a search range for this equation. It relates the question to whether the product $\prod_{p\mid n}(1-1/p)$ over a finite set of primes can be unusually close to $2/3$ for the number of primes involved. The paper proves nothing about the question.

A check made here, not in the paper. Conversely, every solution $n$ of $\phi(n)=\tfrac23(n+1)$ has the form $6m+5$: an even $n$ has $\phi(n)\le n/2$, too small, so $n$ is odd, and $3\mid n+1$. So the question is the same as whether $\phi(6m+5)=4m+4$ has infinitely many solutions $m\ge0$. The four known solutions are $m=0,5,215,279935$, for which $8m+7$ is $7$, $47$, $1727=11\cdot157$ and $2239487=23\cdot97369$; only the first two are prime.

Read depth. Claims checked: the question and the recorded solutions were read on pp. 1--2; the check above was computed for this page.

Bears on

• Problem 411: a member of branch (2) of the Theorem for the shift $r=2$ needs a solution $n=6m+5$ of this equation with $8m+7$ prime and at least $10^{10}$. If the four known solutions were the only ones, branch (2) would be empty, and the Theorem would confine every solution of $\phi(m)+\phi(m+\phi(m))=m$ to odd parts in $\{1,3,5,7,35,47\}$. Infinitely many solutions would not by themselves give a member of branch (2), which also needs the primality of $8m+7$. The question is open in the paper, and neither answer would say which $n$ reach a solution or anything about other shifts.