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Let $q=a/b>0$ be rational in lowest terms. If some positive integer $d$ satisfies $\varphi(d)/d=q$, then all such $d$ have the same finite prime support $P$, and

$$\sum_{\varphi(d)/d=q}\frac1d=\prod_{p\in P}\frac1{p-1}. \tag{1}$$

If the fibre is empty its mass is zero. In particular the mass is at most one, which is Proposition 1.4.

Proof. For any solution,

$$q=\prod_{p\mid d}\frac{p-1}{p}. \tag{2}$$

If $b=1$, then $0<q\le1$ forces $q=1$, and every nonempty support would make (2) strictly smaller than one. Thus $d=1$ and $P$ is empty.

Suppose $b>1$. If $r$ is the largest prime divisor of $d$, the denominator in (2) contains $r$ exactly once before reduction. No numerator $p-1$ with $p\le r$ is divisible by $r$, so $r$ survives in the reduced denominator. No prime larger than $r$ can divide that denominator. Therefore $r$ is exactly the largest prime divisor of $b$. Removing $r$ from the support replaces $q$ by $qr/(r-1)$, and the largest support prime strictly decreases. Induction, terminating at the empty support, determines $P$ uniquely. Conversely, all integers whose support is this $P$ satisfy (2), regardless of the positive exponents of their prime factors.

Summing those exponents by positive geometric series proves

$$\sum_{\operatorname{supp}(d)=P}\frac1d =\prod_{p\in P}\sum_{j\ge1}p^{-j} =\prod_{p\in P}\frac1{p-1}.$$

Each factor is at most one, proving the bound. $\square$

Source. Tao, published paper, published pp.799–800, Lemma 2.1. This page uses that published version.

Bears on. Problem 49.