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Source. Wouter van Doorn and GPT-6 Astra Pro (the author line as printed), Practical numbers and Egyptian fractions, Lemma 3.2 with its proof and the definition of $M_d(X)$, physical p. 3 of the seven-page PDF held by van Doorn (2026). The displays were read on the page image, since the text extraction garbles them. Consumed by the Lemma 3.3 reconstruction. The note describes itself as a simplified and explicit version of the bound of the Price claim (abstract and Section 1, physical p. 1; Section 2, p. 2) and does not say which step of that argument this lemma replaces; the description of the Price claim's analytic input as an exponential-sum theorem comes from the site comment recorded on the Price card, not from the note.

Standing. Author-recorded reconstruction of a claimed result (see the Lemma 3.1 page for the note's standing); not an independent review; changes no status and assigns no tier. The proof uses only Cauchy–Schwarz, Plancherel's identity and character orthogonality on $\mathbb Z/d\mathbb Z$, all written out below.

Definitions

$D(n)$ is the set of positive divisors of $n$, and $e_q(z)=\exp(2\pi iz/q)$. For a finite nonempty set $X$ of integers and an integer $d\ge1$,

$$M_d(X)=\frac1{|X|^2}\,\bigl|\{(x,y)\in X^2:x\equiv y\pmod d\}\bigr| =\sum_{a\bmod d}\Bigl(\frac{|\{x\in X:x\equiv a\}|}{|X|}\Bigr)^2 ,$$

the sum of the squares of the residue probabilities of $X$ modulo $d$. For $d\ge1$ and $\xi\in\mathbb Z$ put

$$f_d(\xi)=\frac1{|X|}\sum_{z\in X}e_d(\xi z),$$

so that $|f_d(\xi)|\le1$ and $f_d(0)=1$.

Statement

Let $V_1,V_2$ be coprime positive odd integers, $V=V_1V_2$, $X_i=D(V_i)$ and $X=D(V)$. Let $A>1$ be odd and suppose that

$$S:=\sum_{\substack{d\mid A\\ d>1}}d^{2/3} \bigl(M_d(X_1)\,M_d(X_2)\,M_d(X)\bigr)^{1/3}<1 . \tag{3.1}$$

Then every residue $c$ modulo $A$ has a representation

$$c\equiv z_0+2z_1+4z_2+8z_3\pmod A,\qquad z_0,z_1,z_2,z_3\in D(V). \tag{3.2}$$

Proof

Fix a divisor $d$ of $A$ with $d>1$; $d$ is odd. Three estimates are established for $f_d$, then combined.

Step 1: the pointwise bound at primitive frequencies. Since $V_1$ and $V_2$ are coprime, every divisor of $V$ factors uniquely as a divisor of $V_1$ times a divisor of $V_2$, so multiplication is a bijection $X_1\times X_2\to X$ and $|X|=|X_1|\,|X_2|$. Grouping the elements of $X_1$ by residue class $a$ modulo $d$, with $N_1(a)=|\{x\in X_1:x\equiv a\}|$,

$$f_d(\xi)=\frac1{|X_1|\,|X_2|}\sum_{a\bmod d}N_1(a)\sum_{y\in X_2}e_d(\xi ay).$$

By Cauchy–Schwarz over $a$,

$$|f_d(\xi)|^2\le\frac1{|X_1|^2|X_2|^2}\Bigl(\sum_aN_1(a)^2\Bigr) \sum_{a\bmod d}\Bigl|\sum_{y\in X_2}e_d(\xi ay)\Bigr|^2 =\frac{M_d(X_1)}{|X_2|^2} \sum_{a\bmod d}\Bigl|\sum_{y\in X_2}e_d(\xi ay)\Bigr|^2 ,$$

since $\sum_aN_1(a)^2=|X_1|^2M_d(X_1)$. Expanding the square and summing over $a$ first, character orthogonality gives $\sum_{a\bmod d}e_d(\xi a(y-y'))=d$ if $d\mid\xi(y-y')$ and $0$ otherwise. When $(\xi,d)=1$ the condition is $y\equiv y'\pmod d$, so the inner sum is $d\,|X_2|^2M_d(X_2)$ and

$$|f_d(\xi)|^2\le d\,M_d(X_1)\,M_d(X_2)\qquad\bigl((\xi,d)=1\bigr).$$

Step 2: Plancherel. Expanding $|f_d(\xi)|^2$ and summing over all $\xi$ modulo $d$, orthogonality gives

$$\sum_{\xi\bmod d}|f_d(\xi)|^2 =\frac1{|X|^2}\sum_{z,z'\in X}\sum_{\xi\bmod d}e_d(\xi(z-z')) =d\,M_d(X).$$

Step 3: the four-fold product. Since $d$ is odd, multiplication by $2^\ell$ permutes the residues modulo $d$ and permutes the residues coprime to $d$. Hence for $(\xi,d)=1$ both $\xi$ and $2\xi$ are coprime to $d$, and Step 1 bounds $|f_d(\xi)|\,|f_d(2\xi)|\le d\,M_d(X_1)M_d(X_2)$; and for $\ell\in\{2,3\}$, $\sum_{\xi\bmod d}|f_d(2^\ell\xi)|^2=d\,M_d(X)$ by Step 2. Bounding the first two factors pointwise, extending the sum to all $\xi$, and applying Cauchy–Schwarz to the last two,

$$\sum_{\substack{\xi\bmod d\\(\xi,d)=1}}\prod_{\ell=0}^{3}|f_d(2^\ell\xi)| \le d\,M_d(X_1)M_d(X_2) \Bigl(\sum_{\xi\bmod d}|f_d(4\xi)|^2\Bigr)^{1/2} \Bigl(\sum_{\xi\bmod d}|f_d(8\xi)|^2\Bigr)^{1/2} =d^2M_d(X_1)M_d(X_2)M_d(X).$$

Step 4: orthogonality modulo $A$. Suppose that some residue $c$ has no representation (3.2). The number of quadruples $(z_0,z_1,z_2,z_3)\in X^4$ with $z_0+2z_1+4z_2+8z_3\equiv c\pmod A$ equals

$$\frac1A\sum_{h\bmod A}e_A(-hc)\sum_{z_0,\dots,z_3\in X} e_A\bigl(h(z_0+2z_1+4z_2+8z_3)\bigr) =\frac{|X|^4}A\sum_{h\bmod A}e_A(-hc)\prod_{\ell=0}^{3}f_A(2^\ell h),$$

and by assumption it is $0$. The term $h=0$ equals $1$. Every nonzero $h$ modulo $A$ is uniquely $h=(A/d)\xi$ with $d=A/\gcd(h,A)>1$ a divisor of $A$ and $1\le\xi<d$, $(\xi,d)=1$; then $e_A(hz)=e_d(\xi z)$ for every $z$, so $f_A(2^\ell h)=f_d(2^\ell\xi)$. Moving the $h=0$ term to the left, taking absolute values, and grouping by $d$,

$$1\le\sum_{\substack{d\mid A\\d>1}}\ \sum_{\substack{\xi\bmod d\\(\xi,d)=1}} \prod_{\ell=0}^{3}|f_d(2^\ell\xi)| \le\sum_{\substack{d\mid A\\d>1}}d^2M_d(X_1)M_d(X_2)M_d(X)$$

by Step 3. Each term on the right is the cube of the corresponding nonnegative term $d^{2/3}(M_d(X_1)M_d(X_2)M_d(X))^{1/3}$ of $S$, and a sum of cubes of nonnegative reals is at most the cube of their sum, so the right side is at most $S^3<1$. This contradiction proves the lemma.