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Source. Theorem 1.1, p. 1, of W. D. Banks, J. B. Friedlander, C. Pomerance and I. E. Shparlinski, Counting integers with a smooth totient, arXiv:1809.01214v1 (2018), the version named on the source card; later published in Q. J. Math. 70 (2019), no. 4, 1371--1386, which was not compared.

Read depth. Claims checked: the statement, its notation and the two propositions it merges were read clause by clause on the page images; the proofs (Sections 2 and 3, pp. 3--14) were read for structure only. Nothing here is independently reviewed.

Statement

Setting (pp. 1--3). An integer is $y$-smooth when none of its prime factors exceeds $y$; $P(n)$ is the largest prime factor of $n>1$ and $P(1)=1$. $\Phi(x,y)$ is the number of positive integers $n\le x$ with $\varphi(n)$ $y$-smooth, that is $P(\varphi(n))\le y$. The paper writes $\log_k$ for the $k$-fold iterated natural logarithm (p. 3).

Theorem 1.1 (p. 1). Fix $\varepsilon>0$. For $x,y$ with $y\ge(\log\log x)^{1+\varepsilon}$ and $u=\log x/\log y\to\infty$,

$$\Phi(x,y)\le x\exp\bigl(-u(\log\log u+\log\log\log u+o(1))\bigr).$$

The range has no upper limit on $y$ beyond $u\to\infty$. The theorem strengthens the bound $\Phi(x,y)\le x/\exp((1+o(1))u\log\log u)$ asserted in the same range as Theorem 3.1 of the authors' 2004 paper (Multiplicative structure of values of the Euler function, Fields Inst. Commun. 41), whose proof had a gap pointed out by Paul Kinlaw (p. 1).

The paper gives no matching lower bound. It notes (p. 2) that under a weak form of a conjecture on primes $p$ with $p-1$ smooth, Lamzouri has shown that $\Phi(x,x^{1/u})\sim\sigma(u)x$ for bounded $u$ with $\sigma(u)=\exp(-u(\log\log u+\log\log\log u+o(1)))$ as $u\to\infty$, with $\sigma$ continuous and monotonic, and says equality in Theorem 1.1 may hold. That remark is conditional and is not part of the theorem.

Proof pointer

The theorem is the union of two propositions with overlapping ranges (p. 2).

• Proposition 2.3 (p. 4): for fixed $\varepsilon>0$, $(\log_2x)^{1+\varepsilon}\le y\le x^{1/\log_2x}$ and $u\to\infty$, the same bound. The proof is a Rankin-type argument: bound $\Phi(x,y)$ by $x^c$ times an Euler product over primes $p\le x$ with $P(p-1)\le y$, choose $c=1-(\log_2u+\log_3u-\delta)/\log y$, and control the prime sum with Lemma 2.1 (p. 3, a bound for $\sum_n A^n\rho(n)$ with $\rho$ the Dickman-de Bruijn function) and Lemma 2.2 (p. 4, an upper bound for the number of primes $p\le t$ with $p-1$ $y$-smooth, cited from Pomerance and Shparlinski).
• Proposition 3.2 (p. 8): for $y\ge\exp(\sqrt{\log x\log_2x})$ and $u\to\infty$, the same bound. The proof uses Lemma 3.1 (p. 8), an inequality analogue of Hildebrand's identity for $\Phi(x,y)$, and an induction on $u$ in steps of one (pp. 8--14).

Section 4 (pp. 14--15) notes that Proposition 2.3 at $y=(\log_2x)^{1+\varepsilon}$ gives $\Phi(x,(\log_2x)^{1+\varepsilon})\le x^{\varepsilon/(1+\varepsilon)+o(1)}$.

Dependencies

Lemmas 2.1, 2.2 and 3.1 of the paper; de Bruijn's estimates for $\rho$ and for $\psi(x,y)$, cited from the literature.

Bears on

• Problem 1003: the theorem says nothing about $\varphi(n)=\varphi(n+1)$ and the paper does not mention the problem. In a comment of 23 February 2026 on the erdosproblems.com forum thread for the problem, Steve Fan sketched how using this theorem for one step, item (iv), of the Erdős-Pomerance-Sárközy argument bounds the number of $n\le x$ with $\varphi(n)=\varphi(n+1)$ by $x\exp(-c_0((\log x)(\log\log x)(\log\log\log x))^{1/3})$ for some $c_0>0$. That sketch is not checked here, and an upper bound on the count does not decide whether there are infinitely many solutions.