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Source. Bloom, arXiv:2112.03726v2, Proposition 2, printed/PDF pp. 9–12.

Notation and circle-method setup

For a finite set of positive integers $B$, write

$$R(B)=\sum_{n\in B}\frac1n.$$

If $q$ is a prime power, define

$$B_q=\{n\in B:q\mid n\text{ and }(q,n/q)=1\}.$$

Thus $n\in B_{p^r}$ precisely when $p^r\Vert n$. Let

$$Q_B=\{q:q\text{ is a prime power and }B_q\ne\varnothing\}.$$

For a set $\mathcal P$ of prime powers, $[\mathcal P]$ denotes their least common multiple, with $[\varnothing]=1$. In particular, $[Q_B]=\operatorname{lcm}(B)$. Finally put $e(x)=e^{2\pi i x}$.

The paper describes this as a refinement of Croot's Fourier-analytic method: it detects reciprocal sums $1/k$ for arbitrary integer $k$, and its short-interval hypothesis is weighted separately on each exact prime-power class $A_q$. The application regime stated on p. 9 is $\eta=N^{-o(1)}$, $k=N^{o(1)}$, and $M,K=N^{1-o(1)}$.

Proposition 2 (precise statement; printed p. 9)

There is an absolute constant $c>0$ with the following property. Suppose

$$N\ge M\ge N^{3/4},\qquad 1\le k\le cM,$$

where $k$ is an integer, and suppose

$$0<\eta<1,\qquad \frac M2\ge K\ge N^{3/4}.$$

Let $A\subseteq[M,N]$ be a set of integers satisfying all four conditions below.

1. The reciprocal sum lies in the half-open interval

$$R(A)\in\left[\frac2k-\frac1M,\frac2k\right).$$

2. The integer $k$ divides $\operatorname{lcm}(A)$.

3. Every $q\in Q_A$ satisfies

$$q\le c\min\left(\frac Mk, \frac{\eta MK^2}{N^2(\log N)^2}\right).$$

4. For every interval $I$ of length $K$, at least one of the following alternatives holds.

   (a)

$$\#\{n\in A:\text{no element of }I\text{ is divisible by }n\} \ge\frac M{\log N}.$$

(b) Define

$$D_I=\left\{q\in Q_A: \#\{n\in A_q:\text{no element of }I\text{ is divisible by }n\} <\frac{\eta M}{q}\right\}.$$

Then some $x\in I$ is divisible by every $q\in D_I$.

Then there is a subset $S\subseteq A$ for which

$$R(S)=\frac1k.$$

In fact, at least $2^{\Omega(|A|)}$ subsets $S\subseteq A$ have this reciprocal sum.

Rewritten proof

The proof occupies printed pp. 9--12. Decrease the absolute constant $c$ as often as needed, and abbreviate

$$X=c\min\left(\frac Mk, \frac{\eta MK^2}{N^2(\log N)^2}\right), \qquad Q=Q_A,\qquad L=[Q].$$

Thus every member of $Q$ is at most $X$, $L=\operatorname{lcm}(A)$, and $k\mid L$.

1. Fourier detection (printed pp. 9--10)

Let $F(A)$ be the number of subsets $S\subseteq A$ for which $kR(S)$ is an integer. Every such reciprocal sum satisfies

$$0\le R(S)\le R(A)<\frac2k.$$

Moreover, $R(S)=0$ only for $S=\varnothing$. Consequently the nonempty subsets counted by $F(A)$ are exactly the subsets with $R(S)=1/k$, and their number is $F(A)-1$.

For integers $a$ and positive $b$, additive-character orthogonality gives

$$\mathbf 1_{a/b\in\mathbb Z} =\frac1b\sum_{-b/2<h\le b/2}e\left(\frac{ha}{b}\right),$$

where the summation interval contains exactly $b$ integers. Since every $n\in A$ divides $L$, each $kR(S)$ has the form $km/L$ with $m\in\mathbb Z$. Apply the orthogonality formula and then sum independently over the choice of each element of $S$:

$$F(A)=\frac1L\sum_{-L/2<h\le L/2} \prod_{n\in A}\left(1+e\left(\frac{kh}{n}\right)\right). \tag{1}$$

The term $h=0$ is $2^{|A|}/L$. If $L$ is even, the endpoint $h=L/2$ contributes

$$\frac1L\prod_{n\in A}\left(1+e\left(\frac{kL}{2n}\right)\right)\ge0,$$

because $L/n$ is an integer and hence every exponential in the product is $1$ or $-1$. Set

$$J=(-L/2,L/2)\cap\mathbb Z\setminus\{0\}.$$

After discarding only the nonnegative endpoint term from (1), it follows that

$$F(A)\ge \frac{2^{|A|}}L+ \frac1L\sum_{h\in J}\prod_{n\in A} \left(1+e\left(\frac{kh}{n}\right)\right). \tag{2}$$

2. Nonnegative major arcs (printed pp. 10--11)

For each $t\in\mathbb Z$, define

$$\mathcal M(t)=\left\{h\in J: \left|h-\frac{tL}{k}\right|\le\frac K{2k}\right\}.$$

The centers are integers because $k\mid L$. They are separated by $L/k$, whereas each arc has radius $K/(2k)$. Since $L\ge\min A\ge M\ge2K$, the arcs are disjoint. Let

$$\mathfrak m=J\setminus\bigcup_{t\in\mathbb Z}\mathcal M(t)$$

be the minor arcs.

If $h=tL/k+r\in\mathcal M(t)$, then $r$ is an integer and $e(tL/n)=1$ for every $n\in A$. The identity

$$1+e(\theta)=2e(\theta/2)\cos(\pi\theta)$$

therefore turns the contribution of $\mathcal M(t)$ to the second term in (2) into

$$\frac{2^{|A|}}L \sum_{\substack{|r|\le K/(2k)\\r\in J-tL/k}} \left(\prod_{n\in A}\cos\left(\frac{\pi kr}{n}\right)\right) e\left(\frac{kr}{2}R(A)\right). \tag{3}$$

Put

$$\nu(r)=\sum_{t\in\mathbb Z}\mathbf 1_{r\in J-tL/k}.$$

Both $\nu(r)$ and the product of cosines in (3) are even functions of $r$. Pairing $r$ with $-r$, the total contribution of all major arcs is

$$\frac{2^{|A|}}L \sum_{\substack{0\le r\le K/(2k)\\r\in\mathbb Z}} (2-\mathbf 1_{r=0})\nu(r) \left(\prod_{n\in A}\cos\left(\frac{\pi kr}{n}\right)\right) \cos\bigl(\pi krR(A)\bigr). \tag{4}$$

Every summand in (4) is nonnegative. Indeed,

$$0\le\frac{kr}{n}\le\frac K{2n}\le\frac12$$

for $n\in A$, so each cosine in the product is nonnegative. Also condition 1 lets us write

$$kR(A)=2-\varepsilon, \qquad 0<\varepsilon\le\frac kM.$$

Because $r$ is an integer,

$$\cos\bigl(\pi krR(A)\bigr) =\cos(2\pi r-\pi r\varepsilon) =\cos(\pi r\varepsilon)\ge0,$$

the final inequality following from $0\le r\varepsilon\le K/(2M)\le1/4$. Hence all the major arcs make a nonnegative contribution to (2).

For later use, set

$$C(B;h)=\prod_{n\in B}\left|\cos\left(\frac{\pi kh}{n}\right)\right|.$$

It is enough to prove

$$\sum_{h\in\mathfrak m}C(A;h)\le\frac12. \tag{5}$$

Indeed, the absolute value of the total minor-arc contribution in (2) is at most $2^{|A|}/L$ times the left side of (5), so (5) gives

$$F(A)\ge\frac{2^{|A|-1}}L. \tag{6}$$

The paper displays the stronger target $1/4$ at its equation (5), but the bound $1/2$ is the one needed for its stated conclusion (6); see the source note below about signed frequencies.

There is at most one maximal member of $Q$ above each underlying prime, so

$$L\le X^{\pi(X)}\le e^{O(X)}.$$

Here the second inequality is Chebyshev's estimate $\pi(X)\ll X/\log X$. On the other hand,

$$|A|\ge MR(A)\ge\frac Mk$$

after decreasing $c\le1$, while $X\le cM/k$. Choosing $c$ sufficiently small therefore makes

$$L\le2^{|A|/2}.$$

Together with (6), this gives

$$F(A)\ge2^{|A|/2-1}>1. \tag{7}$$

The choice $k\le cM$, with $c$ small, also gives an absolute lower bound on $|A|$, so the final strict inequality causes no small-cardinality issue. Once (5) is proved, (7) yields the desired nonempty subset. It also gives $F(A)-1\ge2^{|A|/2-2}$ after another harmless decrease of $c$, proving the claimed $2^{\Omega(|A|)}$ count.

3. Minor arcs of type (a) (printed p. 11)

It remains to prove (5). Since $C(A;-h)=C(A;h)$ and $0\notin\mathfrak m$, it is enough to bound positive minor-arc frequencies and double the result. For $h>0$, take the closed interval $I_h=[kh-K/2,kh+K/2]$, which has length $K$. For each $n\in A$, choose a residue $h_n$ satisfying

$$kh\equiv h_n\pmod n, \qquad |h_n|\le\frac n2.$$

The distance from $kh$ to the nearest multiple of $n$ is $|h_n|$. Consequently no element of $I_h$ is divisible by $n$ exactly when $|h_n|>K/2$. Define

$$D_h=D_{I_h} =\left\{q\in Q: \#\{n\in A_q:|h_n|>K/2\}<\frac{\eta M}{q}\right\}.$$

Partition $\mathfrak m^+=\mathfrak m\cap\mathbb Z_{>0}$ into $\mathfrak m_1$, where alternative 4(a) holds for $I_h$, and $\mathfrak m_2=\mathfrak m^+\setminus\mathfrak m_1$.

For $0\le x\le1/2$,

$$\cos(\pi x)\le1-x^2\le e^{-x^2}.$$

Using the chosen residue of $kh$ modulo $n$, this implies

$$\left|\cos\left(\frac{\pi kh}{n}\right)\right| \le \exp\left(-\frac{h_n^2}{n^2}\right). \tag{8}$$

If $h\in\mathfrak m_1$, then $|h_n|>K/2$ for at least $M/\log N$ members of $A$. Since every such $n\le N$, (8) gives

$$C(A;h) \le\exp\left(-\sum_{n\in A}\frac{h_n^2}{n^2}\right) \le\exp\left(-\frac{K^2M}{4N^2\log N}\right).$$

There are fewer than $L$ possible frequencies in $J$, and $L\le e^{O(X)}$, hence

$$\sum_{h\in\mathfrak m_1}C(A;h) \le e^{O(X)} \exp\left(-\frac{K^2M}{4N^2\log N}\right). \tag{9}$$

Because $0<\eta<1$, the definition of $X$ gives

$$X\le \frac{cK^2M}{N^2(\log N)^2} \le \frac{cK^2M}{N^2\log N}.$$

Thus, after taking $c$ small enough, (9) is at most $1/8$. The lower bounds $M,K\ge N^{3/4}$ make the remaining negative exponent grow with $N$; decreasing $c$ also disposes of the bounded admissible cases.

4. Minor arcs of type (b) (printed pp. 11--12)

We first prove the pointwise estimate

$$C(A;h)\le N^{-4|Q\setminus D_h|} \qquad(h\in\mathfrak m_2). \tag{10}$$

Assume (10) for the moment. Since alternative 4(a) fails for $h\in\mathfrak m_2$, alternative 4(b) gives a multiple of $[D_h]$ within distance $K/2$ of $kh$. Fix $D\subseteq Q$. Each multiple of $[D]$ in $[1,kL]$ can lie within $K/2$ of $kh$ for at most $K/k+1\le M$ positive integers $h$. Therefore

$$\#\{h\in\mathfrak m_2:D_h=D\} \le \frac{MkL}{[D]} \le Mk\prod_{q\in Q\setminus D}q \le kN^{|Q\setminus D|+1}. \tag{11}$$

The loose interval $[1,kL]$ contains every relevant multiple: for positive $h\in J$, one has $kh<kL/2$, and the minor-arc condition keeps the interval $I_h$ away from zero.

If $D_h=Q$, alternative 4(b) would put a multiple of $[Q]=L$ within $K/2$ of $kh$, contradicting $h\in\mathfrak m$. Hence $D_h\ne Q$. Combining (10) and (11), and using $|Q|\le N$, gives

$$\begin{aligned} \sum_{h\in\mathfrak m_2}C(A;h) &\le kN\sum_{D\subsetneq Q}N^{-3|Q\setminus D|}\\ &=kN\left((1+N^{-3})^{|Q|}-1\right)\\ &\ll\frac{k}{N}\le c. \end{aligned}$$

Take $c$ small enough that this is at most $1/8$. Together with the type (a) estimate,

$$\sum_{h\in\mathfrak m^+}C(A;h)\le\frac14.$$

Doubling by symmetry proves (5).

It remains to establish (10). For each $n\in A$, the number of $q\in Q$ for which $n\in A_q$ is exactly $\omega(n)$, because there is one exact prime power $p^{v_p(n)}$ for each prime divisor $p$ of $n$. In particular it is at most $(\log N)/(\log2)$. Choose an absolute $a>0$ so small that

$$\frac{a}{\log N}\#\{q\in Q:n\in A_q\}\le1$$

for every $n\in A$. Since every cosine factor lies in $[0,1]$,

$$C(A;h) \le\prod_{q\in Q}C(A_q;h)^{a/\log N}. \tag{12}$$

Now let $q\in Q\setminus D_h$. The definition of $D_h$ says that at least $\eta M/q$ members $n\in A_q$ have $|h_n|>K/2$. Applying (8) on this class and using $n\le N$,

$$\begin{aligned} C(A_q;h) &\le\exp\left(-\sum_{n\in A_q}\frac{h_n^2}{n^2}\right)\\ &\le\exp\left(-\frac{\eta MK^2}{4N^2q}\right). \end{aligned} \tag{13}$$

The smoothness hypothesis $q\le X$ gives

$$\frac{\eta MK^2}{4N^2q}\ge\frac{(\log N)^2}{4c}.$$

For any prescribed large absolute $B$, a sufficiently small choice of $c$ therefore makes (13) at most

$$e^{-B(\log N)^2}=N^{-B\log N}.$$

In (12), discard the factors indexed by $D_h$, since they are at most one, and use this estimate for every $q\notin D_h$. Choosing $B$ so that $aB\ge4$ yields

$$C(A;h) \le\prod_{q\in Q\setminus D_h} \left(N^{-B\log N}\right)^{a/\log N} \le N^{-4|Q\setminus D_h|},$$

which is (10). This completes the minor-arc estimate, hence the proof of the proposition and its exponential multiplicity assertion.

External dependencies

The proof does not invoke an earlier numbered result from the paper. Its only external analytic input is the standard Chebyshev bound $\pi(x)\ll x/\log x$, used to obtain $[Q]\le e^{O(X)}$. Additive-character orthogonality is written explicitly above. The cosine estimates and $\omega(n)\ll\log n$ are elementary. Croot [2] is methodological background, not a logical dependency of Proposition 2.

Source details

The source defines its fiber sets for nonnegative frequencies but later sums over signed frequencies. The rewrite explicitly estimates positive frequencies and doubles, obtaining a signed bound $1/2$. This suffices for $F(A)\ge2^{|A|-1}/L$, although the paper displays the stronger target $1/4$. The parameter mismatch in the downstream application is recorded on Proposition 1; it does not alter the present proposition.

Bears on

• Problem 298
• Problem 299