\documentclass[11pt,reqno]{amsart} \usepackage[letterpaper,margin=1in]{geometry} \usepackage[T1]{fontenc} \usepackage{lmodern} \usepackage{microtype} \usepackage{amsmath,amssymb,amsthm,mathtools} \usepackage{booktabs,array} \usepackage[dvipsnames]{xcolor} \usepackage[unicode,colorlinks=true,linkcolor=MidnightBlue,citecolor=MidnightBlue,urlcolor=MidnightBlue]{hyperref} \hypersetup{ pdftitle={Euler's 1747 conjecture on sums of integers with restricted prime factors}, pdfauthor={Will Howe, Grant Huffman, Jensen Kohlmeyer, Liam Kruer}, pdfsubject={An explicit proof of Euler's restricted additive conjecture}, pdfkeywords={Euler, additive number theory, four squares, Hurwitz quaternions, Hecke operators} } \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{proposition}[theorem]{Proposition} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{corollary}[theorem]{Corollary} \theoremstyle{definition} \newtheorem{definition}[theorem]{Definition} \theoremstyle{remark} \newtheorem{remark}[theorem]{Remark} \newcommand{\Z}{\mathbb Z} \newcommand{\Q}{\mathbb Q} \newcommand{\R}{\mathbb R} \newcommand{\C}{\mathbb C} \newcommand{\F}{\mathbb F} \newcommand{\Af}{\mathbb A_f} \newcommand{\ind}{\mathbf 1} \newcommand{\Pthree}{\mathcal P_3} \newcommand{\HH}{\mathcal H} \newcommand{\nrd}{\operatorname{nrd}} \newcommand{\PGL}{\operatorname{PGL}} \newcommand{\PB}{\operatorname{PB}^{\times}} \newcommand{\norm}[1]{\lVert#1\rVert} \newcommand{\ip}[2]{\langle#1,#2\rangle} \newcommand{\mixed}{\mathrm{mix}} \setlength{\emergencystretch}{1.3em} \allowdisplaybreaks[1] \title[Euler's 1747 conjecture]{Euler's 1747 conjecture on sums of integers\protect\\ with restricted prime factors} \author[W. Howe]{Will Howe} \email{howew@purdue.edu} \author[G. Huffman]{Grant Huffman} \email{huffma97@purdue.edu} \author[J. Kohlmeyer]{Jensen Kohlmeyer} \email{jenwin@purdue.edu} \author[L. Kruer]{Liam Kruer} \email{kruerl@purdue.edu} \date{November 12, 2026} \subjclass[2020]{Primary 11P32; Secondary 11E25, 11F11, 11R52} \keywords{Euler's conjecture, sums of squares, restricted prime factors, Hurwitz quaternions, Hecke operators} \begin{document} \begin{abstract} We prove that every positive integer congruent to $2$ modulo $4$ is a sum of two positive integers whose prime factors are all congruent to $1$ modulo $4$, with $1$ admitted as a summand. Euler proposed this statement in a letter to Goldbach dated May 6, 1747. The proof counts representations by four squares and bounds those containing a common prime divisor congruent to $3$ modulo $4$ in either coordinate pair. An average over a nonsplit torus normalizer converts the quaternionic Hecke bound into an error independent of the sieving prime. Elementary estimates then give the explicit threshold $64^4$. An exact finite arithmetic certificate covers the remaining $4{,}194{,}304$ integers in the required residue class. \end{abstract} \maketitle \section{Introduction} Let \[ \Pthree=\{p\text{ prime}:p\equiv3\pmod4\},\qquad \mathcal A=\{a\in\Z_{>0}:p\mid a\Longrightarrow p\equiv1\pmod4\}. \] The integer $1$ belongs to $\mathcal A$. In his letter to Goldbach of May 6, 1747, Euler proposed that every integer of the form $4m+2$ can be split into two integers of the form $4x+1$, each having no divisor of the form $4p-1$. This is the assertion proved here. The statement appears in Letter 115 of the collected correspondence, together with a discussion in editorial note 10 \cite[pp.~935--936]{Euler}. Dietmann and Elsholtz proved that every sufficiently large positive integer congruent to $2$ modulo $4$ has the required decomposition \cite{DE}. Their argument uses a result of Greaves on four squares with two prime variables. We obtain an explicit threshold by counting all four-square representations and estimating the unwanted common divisors of the two coordinate pairs. \begin{theorem}\label{thm:main} Every positive integer $n\equiv2\pmod4$ admits a decomposition \[ n=a+b,\qquad a,b\in\mathcal A. \] \end{theorem} The quantitative part of the proof establishes the result for $n\ge64^4=16{,}777{,}216$ after removing factors from $\mathcal A$. Section~\ref{sec:finite} gives an exact arithmetic certificate for the finite range below this threshold. The spectral ingredient is the weight-two Ramanujan--Petersson bound, applied through the Jacquet--Langlands correspondence. All auxiliary constants are explicit. \section{Four-square representations and prime obstructions} Write $\sigma(u)=\sum_{d\mid u}d$ and $\tau(u)=\sum_{d\mid u}1$. We first remove a multiplicative factor that already belongs to $\mathcal A$. \begin{lemma}\label{lem:reduction} To prove Theorem~\ref{thm:main}, it suffices to consider $n=2m$, where $m$ is odd and every prime divisor of $m$ belongs to $\Pthree$. \end{lemma} \begin{proof} For any odd $m$, write $m=gm_0$, where $g\in\mathcal A$ and every prime divisor of $m_0$ belongs to $\Pthree$. If $2m_0=a+b$ with $a,b\in\mathcal A$, then $2m=ga+gb$ has the same property. \end{proof} For the remainder of the analytic argument, $n=2m$ satisfies Lemma~\ref{lem:reduction}. Put \[ \mathcal R(n)=\{(x_1,x_2,x_3,x_4)\in\Z^4: x_1^2+x_2^2+x_3^2+x_4^2=n\}. \] These are signed, ordered representations. Jacobi's four-square formula gives \begin{equation}\label{eq:four-squares} R(n):=|\mathcal R(n)|=24\sigma(m). \end{equation} Exactly two coordinates are odd, since a square is $0$ or $1$ modulo $4$. Let $\mathcal R_{\mixed}(n)$ consist of the representations in which each of $(x_1,x_2)$ and $(x_3,x_4)$ has opposite parity. Permuting coordinates shows that the six possible placements of the odd coordinates occur equally often. Four of them have the required parity, so \begin{equation}\label{eq:mixed-count} |\mathcal R_{\mixed}(n)|=16\sigma(m)=\frac23R(n). \end{equation} For $p\in\Pthree$, let $b_p(n)$ count the elements of $\mathcal R_{\mixed}(n)$ for which \[ (p\mid x_1\text{ and }p\mid x_2) \quad\text{or}\quad (p\mid x_3\text{ and }p\mid x_4). \] Since $-1$ is a quadratic nonresidue modulo $p$, the implication $p\mid x^2+y^2\Longrightarrow p\mid x,y$ holds. Consequently, any representation outside all these exceptional sets supplies two positive odd summands in $\mathcal A$. It is therefore enough to prove \begin{equation}\label{eq:goal} \sum_{p\in\Pthree}b_p(n)<\frac23R(n). \end{equation} \section{A uniform quaternionic estimate}\label{sec:hecke} \begin{proposition}\label{prop:small-prime} For every $p\in\Pthree$, \begin{equation}\label{eq:small-prime} \frac{b_p(n)}{R(n)} \le \frac{2}{p(p-1)} +\ind_{p>3}\frac{\tau(m)\sqrt m}{3\sigma(m)}. \end{equation} \end{proposition} The estimate follows from a Hecke action on a finite quotient associated with the Hurwitz order. \subsection{The Hurwitz order and its finite quotient} Let $B=\Q+\Q i+\Q j+\Q k$ be the Hamilton quaternion algebra, where $i^2=j^2=-1$ and $k=ij=-ji$. Its reduced norm is \[ \nrd(a+bi+cj+dk)=a^2+b^2+c^2+d^2. \] The Hurwitz order is \[ \HH=\Z^4\ \cup\ (\Z+\tfrac12)^4, \] with coordinates interpreted in the basis $1,i,j,k$. Its 24 units are the eight coordinate units and the sixteen elements $(\pm1\pm i\pm j\pm k)/2$. The order $\HH$ is norm-Euclidean. Indeed, rounding four real coordinates to integers gives a remainder of norm at most $1$. Equality requires all four coordinates to be half-integers, in which case a Hurwitz half-integer gives remainder zero. Euclidean division implies that every right ideal is principal. In particular, the right ideal class number of $\HH$ is one. Let $Q_m=\{\alpha\in\HH:\nrd(\alpha)=m\}$. Right multiplication by $1+i$ is a bijection \begin{equation}\label{eq:hurwitz-bijection} Q_m\longrightarrow\mathcal R(2m). \end{equation} For the inverse, divide an integral quaternion of norm $2m$ by $1+i$. The new coordinates are halves of the sums and differences in two coordinate pairs. The two odd coordinates of the original quaternion ensure that all four new coordinates are integers or all four are strict half-integers. Thus the inverse belongs to $\HH$. Fix $p\in\Pthree$ with $p\nmid m$. The algebra $B$ splits at every odd prime, and reduction identifies $\HH/p\HH$ with $M_2(\F_p)$, with reduced norm corresponding to determinant. Write \[ G_p=\PGL_2(\F_p),\qquad \rho_p:\{\alpha\in\HH:p\nmid\nrd(\alpha)\}\longrightarrow G_p. \] Let $\Gamma_p=\rho_p(\HH^\times)$ and $Y_p=\Gamma_p\backslash G_p$. The only units with scalar reduction are $\pm1$, so \begin{equation}\label{eq:group-sizes} |\Gamma_p|=12,\qquad |Y_p|=\frac{p(p^2-1)}{12}. \end{equation} We equip $\C[Y_p]$ with the counting inner product. \subsection{The Hecke operator and its spectral bound} For $m$ odd and $p\nmid m$, define \begin{equation}\label{eq:hecke-def} (T_m f)(\Gamma_p g) =\frac1{24}\sum_{\alpha\in Q_m} f(\Gamma_p\rho_p(\alpha)g). \end{equation} Right multiplication by a unit permutes the left-unit classes in $Q_m$, so this is well-defined. It commutes with right translation by $G_p$. Quaternion conjugation permutes $Q_m$ and gives projective inversion, which proves self-adjointness. Equations~\eqref{eq:four-squares} and \eqref{eq:hurwitz-bijection} give $T_m\ind=\sigma(m)\ind$. The function \[ \chi_p(\Gamma_p g)=\left(\frac{\det(g)}p\right) \] is well-defined because a scalar changes the determinant by a square and every element of $\Gamma_p$ has square determinant. It satisfies $T_m\chi_p=(\frac mp)\sigma(m)\chi_p$. \begin{lemma}\label{lem:spectrum} On $\{\ind,\chi_p\}^{\perp}$, the operator norm of $T_m$ is at most $\tau(m)\sqrt m$. \end{lemma} \begin{proof} Let $\PB$ denote the algebraic group of projective units of $B$. At a finite prime $\ell\ne p$, take the projective image of $\HH_\ell^\times$ as the local level. At $p$, take the kernel of reduction from $\PGL_2(\Z_p)$ to $G_p$. Denote the product of these compact open subgroups by $U(p)$. The class-number-one property gives \begin{equation}\label{eq:adelic-quotient} \PB(\Q)\backslash\PB(\Af)/U(p)\simeq Y_p. \end{equation} To see the stabilizer, choose an integral representative at every finite place. A rational quaternion carrying one such representative to another can, after rational rescaling, be taken to be a Hurwitz unit. Its remaining action at $p$ is precisely $\Gamma_p$. The usual unramified Hecke correspondence of index $m$ on \eqref{eq:adelic-quotient} is \eqref{eq:hecke-def}. In fact, its local right cosets at $\ell^e\parallel m$ specify integral right ideals of reduced norm $\ell^e$. Combining these local ideals gives the integral right ideals of reduced norm $m$. Each is $\alpha\HH$ for an $\alpha\in Q_m$, unique up to right multiplication by a unit. Returning to integral representatives gives $\rho_p(\alpha^{-1})$ at $p$. Conjugation sends these right-unit classes to the left-unit classes in \eqref{eq:hecke-def}, since projectively $\alpha^{-1}=\overline\alpha$. One-dimensional automorphic representations factor through the reduced norm. Their restrictions to $Y_p$ depend only on the determinant in $\F_p^\times/(\F_p^\times)^2$, so lie in the span of $\ind$ and $\chi_p$. The Jacquet--Langlands correspondence transfers the remaining representations to weight-two cuspidal representations, unramified at every prime dividing $m$ \cite[\S16]{JL}. This use of the correspondence and the weight-two spectral bound is also described in \cite[\S8.4]{EMV}. By Ramanujan--Petersson, at an unramified prime $\ell$ the normalized Satake parameters $u_\ell,v_\ell$ have absolute value $1$ \cite{Deligne}. In the normalization of \eqref{eq:hecke-def}, the index-$\ell^e$ eigenvalue is \[ \ell^{e/2}\sum_{r=0}^{e}u_\ell^r v_\ell^{e-r}. \] The triangular cosets with diagonal $(\ell^r,\ell^{e-r})$ have multiplicity $\ell^r$ and spherical factor $\ell^{(e-2r)/2}$. Their product is $\ell^{e/2}$, giving the displayed normalization. The eigenvalue is at most $(e+1)\ell^{e/2}$ in absolute value. Multiplication over $\ell^e\parallel m$ proves the lemma. \end{proof} \subsection{Averaging over a torus normalizer} Since $p\equiv3\pmod4$, the subalgebra $\F_p[i]$ is the field $\F_{p^2}$. Let \[ C_p=\F_p[i]^\times/\F_p^\times\subset G_p,\qquad K_p=C_p\cup C_p\rho_p(j). \] Conjugation by $j$ induces the nontrivial field automorphism, so $K_p$ is a subgroup of order $2(p+1)$. An invertible quaternion has its first or its second coordinate pair zero modulo $p$ exactly when its projective reduction belongs to $K_p$. Since $\rho_p(1+i)\in C_p$, multiplication by $1+i$ preserves this condition. The intersection $\Gamma_p\cap K_p$ consists of the four projective classes of $1,i,j,k$. Let \[ \mathcal O_p=\{\Gamma_p k:k\in K_p\},\quad f_p=\ind_{\mathcal O_p}, \quad \mu_p=\frac{|\mathcal O_p|}{|Y_p|}=\frac6{p(p-1)}. \] There are $(p+1)/2$ points in $\mathcal O_p$. Each left-unit class whose reduction lies in $\mathcal O_p$ has exactly eight elements reducing into $K_p$. Counting all representations therefore gives \begin{equation}\label{eq:bad-operator} b_p(n)\le 8(T_m f_p)(\Gamma_p). \end{equation} The function $f_p$ is invariant under right translation by $K_p$. The same is true of $T_mf_p$, hence \begin{equation}\label{eq:orbit-average} (T_mf_p)(\Gamma_p) =\frac{\ip{f_p}{T_mf_p}}{|\mathcal O_p|}. \end{equation} The norm map $\F_{p^2}^\times\to\F_p^\times$ is surjective. The orbit $\mathcal O_p$ therefore has equally many points in the two determinant classes. Thus $f_p-\mu_p\ind$ is orthogonal to both $\ind$ and $\chi_p$, and \[ \norm{f_p-\mu_p\ind}_2^2=|\mathcal O_p|(1-\mu_p). \] Lemma~\ref{lem:spectrum} and \eqref{eq:orbit-average} imply \begin{equation}\label{eq:torus-bound} (T_mf_p)(\Gamma_p) \le \sigma(m)\mu_p+\tau(m)\sqrt m(1-\mu_p). \end{equation} For $p=3$ one has $\mu_p=1$, so the second term vanishes. Combining \eqref{eq:bad-operator} and \eqref{eq:torus-bound} with $R(n)=24\sigma(m)$ proves Proposition~\ref{prop:small-prime} when $p\nmid m$. \begin{proof}[Completion of the proof of Proposition~\ref{prop:small-prime}] Suppose $p\mid m$. If a coordinate pair is divisible by $p$, the other pair has sum of squares zero modulo $p$, so all four coordinates are divisible by $p$. Hence $b_p(n)=0$ when $p^2\nmid m$ and $b_p(n)=16\sigma(m/p^2)$ otherwise. Writing $m=p^e u$ gives \[ \frac{\sigma(m/p^2)}{\sigma(m)} =\frac{p^{e-1}-1}{p^{e+1}-1}<\frac1{p^2}. \] It follows that $b_p(n)/R(n)\le2/(3p^2)\le2/[p(p-1)]$. \end{proof} \section{Large primes and elementary lattice counting} For $t\ge0$, let $N_{\mixed}(t)$ be the number of signed, ordered pairs $(x,y)\in\Z^2$ with opposite parity and $x^2+y^2\le t$. \begin{lemma}\label{lem:lattice} For every $t\ge0$, one has $N_{\mixed}(t)\le2(t+1)$. \end{lemma} \begin{proof} Each orbit under quarter-turns has four elements and a unique representative with $x>0$ and $y\ge0$. Write \[ x+y=2u+1,\qquad x-y=2v+1, \qquad u\ge0,\quad -u\le v\le u. \] Assign to this pair the positive integer $L=u^2+u+v+1$. For fixed $u$, these labels run from $u^2+1$ to $(u+1)^2$. The labels are therefore distinct. Moreover, \[ \frac{x^2+y^2+1}{2}=L+v^2\ge L. \] At most $\lfloor(t+1)/2\rfloor$ labels can occur. Multiplication by four gives the claimed upper bound. \end{proof} Define \[ M(n)=\max_{1\le u\le n} \prod_{\substack{q^e\parallel u\\q\equiv1\ (4)}}(e+1). \] The two-square formula implies $r_2(u)\le4M(n)$ for $1\le u\le n$, where $r_2$ counts signed, ordered representations. \begin{lemma}\label{lem:large-prime} For every $P\ge3$, \begin{equation}\label{eq:large-prime} \frac1{R(n)}\sum_{\substack{p>P\\p\in\Pthree}}b_p(n) \le\frac43M(n) \left( \sum_{\substack{p>P\\p\in\Pthree}}\frac1{p^2} +\frac{\#\{p\in\Pthree:PP\\p\in\Pthree}}\frac1{p^2} &\le\frac2{15P}+\frac2{P^2}, \label{eq:prime-tail}\\ 2\sum_{p\in\Pthree}\frac1{p(p-1)}&<\frac{21}{50}. \label{eq:prime-density} \end{align} \end{lemma} \begin{proof} Every prime in $\Pthree$ exceeding $3$ belongs to one of the eight residue classes \[ \mathcal S=\{7,11,19,23,31,43,47,59\}\pmod{60}. \] Let $A(t)$ count the positive integers at most $t$ in these classes. Checking the eight jumps in one period gives \[ -\frac{14}{15}\le A(t)-\frac{2t}{15}\le\frac{14}{15}. \] Periodicity extends the bound to every $t\ge0$, and proves \eqref{eq:prime-count}. For $t\ge P$ it also gives \[ A(t)-A(P)\le\frac2{15}(t-P)+2. \] Partial summation yields \[ \sum_{\substack{k>P\\k\bmod60\in\mathcal S}}\frac1{k^2} =2\int_P^\infty\frac{A(t)-A(P)}{t^3}\,dt \le\frac2{15P}+\frac2{P^2}. \] For $P\ge3$, this set contains all the primes in \eqref{eq:prime-tail}. Finally, direct summation through $60$ gives \begin{multline*} 2\sum_{p\in\Pthree}\frac1{p(p-1)}\\ \le\sum_{p\in\{3,7,11,19,23,31,43,47,59\}}\frac2{p(p-1)} +\frac{120}{59}\left(\frac2{15\cdot60}+\frac2{60^2}\right) <\frac{21}{50}. \end{multline*} Here $p/(p-1)\le60/59$ for $p>60$, so \eqref{eq:prime-tail} bounds the remainder. \end{proof} \section{The analytic range} \begin{proposition}\label{prop:analytic} Suppose $n=2m\ge64^4$, where $m$ is odd and all its prime divisors belong to $\Pthree$. Then $n\in\mathcal A+\mathcal A$. \end{proposition} \begin{proof} Set $X=n^{1/4}\ge64$ and $P=2X$. Proposition~\ref{prop:small-prime}, Lemma~\ref{lem:divisors}, and \eqref{eq:prime-count} give \begin{align} \frac1{R(n)}\sum_{\substack{p\le P\\p\in\Pthree}}b_p(n) &\le\frac{21}{50} +\frac{\tau(m)\sqrt m}{3\sigma(m)}\left(\frac{4X}{15}+1\right) \notag\\ &\le\frac{21}{50} +\frac{23}{30}\,2^{3/10}X^{-1/5} \left(\frac4{15}+\frac1X\right). \label{eq:small-total} \end{align} For the large primes, Lemmas~\ref{lem:large-prime}--\ref{lem:prime-sums} give \begin{align} \frac1{R(n)}\sum_{\substack{p>P\\p\in\Pthree}}b_p(n) &\le\frac{44}{15}X^{4/5} \left(\frac1{15X}+\frac1{2X^2} +\frac2{15X^2}+\frac1{X^4}\right) \notag\\ &=\frac{44}{15}X^{-1/5} \left(\frac1{15}+\frac{19}{30X}+\frac1{X^3}\right). \label{eq:large-total} \end{align} The prime-count term in \eqref{eq:large-prime} was bounded by $2/(15\sqrt n)+1/n$. For $X\ge64$, \[ X^{-1/5}\le\frac49,\qquad 2^{3/10}<\frac54,\qquad X^{-1}\le\frac1{64}. \] The first two inequalities follow by raising to the fifth and tenth powers. Adding \eqref{eq:small-total} and \eqref{eq:large-total} now gives \begin{align*} \frac1{R(n)}\sum_{p\in\Pthree}b_p(n) &\le\frac{21}{50} +\frac{23}{30}\cdot\frac54\cdot\frac49 \left(\frac4{15}+\frac1{64}\right)\\ &\hspace{1.7em} +\frac{44}{15}\cdot\frac49 \left(\frac1{15}+\frac{19}{30\cdot64} +\frac1{64^3}\right)\\ &=\frac{21235493}{33177600}<\frac23. \end{align*} Equations~\eqref{eq:mixed-count} and \eqref{eq:goal} therefore supply the required representation. \end{proof} \section{The finite range}\label{sec:finite} We give the finite arithmetic certificate in terms of binary indicators and finite sets. Put $H=2^{22}-1=4{,}194{,}303$ and define \begin{equation}\label{eq:epsilon} \varepsilon_j= \prod_{\substack{p\le4096\\p\in\Pthree}} \left(1-\left\lfloor\frac{4j+1}{p}\right\rfloor +\left\lfloor\frac{4j}{p}\right\rfloor\right) \quad(0\le j\le H), \qquad \varepsilon_j=0\quad(j<0). \end{equation} Each factor is $0$ if $p\mid4j+1$ and is $1$ otherwise. Thus $\varepsilon_j\in\{0,1\}$. \begin{lemma}\label{lem:indicator} For $0\le j\le H$, one has $\varepsilon_j=1$ exactly when $4j+1\in\mathcal A$. \end{lemma} \begin{proof} The integer $4j+1\le64^4-3$ is odd and congruent to $1$ modulo $4$. Its prime factors belonging to $\Pthree$ occur an even number of times, counting multiplicity. If any occur, at least two occur. At least one of these factors is then no larger than $\sqrt{4j+1}<4096$. Formula~\eqref{eq:epsilon} detects precisely this possibility. \end{proof} Let $J=\{0\le i\le78:\varepsilon_i=1\}$. The corresponding possible summands are \begin{align*} \{4i+1:i\in J\}=\{&1,5,13,17,25,29,37,41,53,61,65,73,85,89,97,\\ &101,109,113,125,137,145,149,157,169,173,181,185,\\ &193,197,205,221,229,233,241,257,265,269,277,281,\\ &289,293,305,313\}. \end{align*} For $1\le a\le313$, $a\equiv1\pmod4$, define the exact integer \begin{equation}\label{eq:uncovered} U(a)=\sum_{k=0}^{H} \prod_{i=0}^{(a-1)/4} (1-\varepsilon_i\varepsilon_{k-i}). \end{equation} It counts the indices still unrepresented when the first summand is at most $a$. Table~\ref{tab:finite} records successive values of \eqref{eq:uncovered}; all quantities are integers. \begin{table}[htbp] \centering \caption{The finite arithmetic certificate. Each column pair gives an upper bound $a$ for the first summand and the number $U(a)$ of indices remaining.}\label{tab:finite} \begin{tabular}{r r@{\qquad}r r@{\qquad}r r} \toprule $a$ & $U(a)$ & $a$ & $U(a)$ & $a$ & $U(a)$\\ \midrule 1 & 2\,849\,643 & 101 & 6\,561 & 229 & 8\\ 13 & 1\,261\,334 & 149 & 756 & 277 & 2\\ 29 & 364\,708 & 193 & 57 & 289 & 1\\ 61 & 71\,867 & 205 & 23 & 313 & 0\\ \bottomrule \end{tabular} \end{table} The evaluation in the table can equivalently be performed by the following finite-set recurrence, which specifies every removal. Start with $C_0=\{0,1,\ldots,H\}$. In increasing order, for each $i\in J$ replace the current set $C$ by \begin{equation}\label{eq:finite-recurrence} C\setminus\{k:i\le k\le H,\ \varepsilon_{k-i}=1\}. \end{equation} The indicators themselves are obtained by starting with all indices retained and, for each of the $287$ primes $p\le4096$ in $\Pthree$, removing the progression \[ \frac{3p-1}{4},\quad\frac{3p-1}{4}+p,\quad \frac{3p-1}{4}+2p,\quad\ldots\quad\text{through }H. \] There are $1{,}344{,}661$ retained indices. The recurrence \eqref{eq:finite-recurrence} ends with the empty set, as recorded by $U(313)=0$. \begin{proposition}\label{prop:finite} For every $n\equiv2\pmod4$ with $2\le n<64^4$, there are $a,b\in\mathcal A$ such that $n=a+b$. The smaller summand can be chosen at most $313$. \end{proposition} \begin{proof} Write $n=4k+2$, where $0\le k\le H$. Since $U(313)=0$, one of the factors in the $k$th product in \eqref{eq:uncovered} vanishes. Thus $\varepsilon_i=\varepsilon_{k-i}=1$ for some $0\le i\le78$ with $i\le k$. Lemma~\ref{lem:indicator} gives \[ n=(4i+1)+(4(k-i)+1)\in\mathcal A+\mathcal A. \] One of these summands is at most $313$. \end{proof} \begin{proof}[Proof of Theorem~\ref{thm:main}] Use Lemma~\ref{lem:reduction} to write $n=g(2m_0)$ with $g\in\mathcal A$ and every prime divisor of $m_0$ in $\Pthree$. Proposition~\ref{prop:analytic} applies if $2m_0\ge64^4$. Proposition~\ref{prop:finite} applies otherwise. In both cases, multiply the resulting two summands by $g$. \end{proof} \section*{Acknowledgment} OpenAI Codex generated the mathematical argument, performed the exact calculations, and drafted the manuscript under human direction. \begin{thebibliography}{9} \raggedright \bibitem{Deligne} P.~Deligne, \emph{La conjecture de Weil.~I}, Publ. Math. Inst. 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