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Statement

$S_x$ is the Schinzel–Szekeres set defined on the page of Lemma 2.1, and $\tau(n)$ is the number of divisors of $n$.

Lemma 2.2 (printed p. 262). If $1<n\le x$ and $n$ is divisible by no element of $S_x$, then

$$\tau(n)\ge\frac{\log n}{\log(x/n)}$$

(display (2.3)).

At $n=x$ the right side has a zero denominator; the proof's inequality $n\le x^{1-1/\tau(n)}$ (p. 263), of which (2.3) is a rearrangement, shows that $n=x$ cannot occur under the hypothesis.

Source. I. Z. Ruzsa, On the small sieve. II. Sifting by composite numbers, J. Number Theory 14 (1982), 260–268; Lemma 2.2 on printed p. 262, proof pp. 262–263. The edition is identified in the source digest.

Read depth. Claims checked: the statement and the final inequality of the proof were read on the page images. The proof was not checked.

Proof pointer

Write $n=p_1^{\alpha_1}\cdots p_k^{\alpha_k}$ with $p_1<\cdots<p_k$. Since no divisor of $n$ lies in $T_x$, each $p_j^{\alpha_j+1}p_{j+1}^{\alpha_{j+1}}\cdots p_k^{\alpha_k}$ is at most $x$ (2.4). Raising these $k$ inequalities to suitable powers and multiplying gives $n\le x^{1-b}$ with $b=1/\tau(n)$, which is (2.3).

Dependencies

• Lemma 2.1 (for the definition of $T_x$ and $S_x$ only).

Bears on

No problem directly. It is the input to Lemma 2.5.