% Options for packages loaded elsewhere \PassOptionsToPackage{unicode}{hyperref} \PassOptionsToPackage{hyphens}{url} % \documentclass[ 11pt, ]{article} \usepackage{amsmath,amssymb} \usepackage{iftex} \ifPDFTeX \usepackage[T1]{fontenc} \usepackage[utf8]{inputenc} \usepackage{textcomp} % provide euro and other symbols \else % if luatex or xetex \usepackage{unicode-math} % this also loads fontspec \defaultfontfeatures{Scale=MatchLowercase} \defaultfontfeatures[\rmfamily]{Ligatures=TeX,Scale=1} \fi \usepackage{lmodern} \ifPDFTeX\else % xetex/luatex font selection \fi % Use upquote if available, for straight quotes in verbatim environments \IfFileExists{upquote.sty}{\usepackage{upquote}}{} \IfFileExists{microtype.sty}{% use microtype if available \usepackage[]{microtype} \UseMicrotypeSet[protrusion]{basicmath} % disable protrusion for tt fonts }{} \makeatletter \@ifundefined{KOMAClassName}{% if non-KOMA class \IfFileExists{parskip.sty}{% \usepackage{parskip} }{% else \setlength{\parindent}{0pt} \setlength{\parskip}{6pt plus 2pt minus 1pt}} }{% if KOMA class \KOMAoptions{parskip=half}} \makeatother \usepackage{xcolor} \usepackage[margin=0.9in]{geometry} \usepackage{longtable,booktabs,array} \usepackage{calc} % for calculating minipage widths % Correct order of tables after \paragraph or \subparagraph \usepackage{etoolbox} \makeatletter \patchcmd\longtable{\par}{\if@noskipsec\mbox{}\fi\par}{}{} \makeatother % Allow footnotes in longtable head/foot \IfFileExists{footnotehyper.sty}{\usepackage{footnotehyper}}{\usepackage{footnote}} \makesavenoteenv{longtable} \setlength{\emergencystretch}{3em} % prevent overfull lines \providecommand{\tightlist}{% \setlength{\itemsep}{0pt}\setlength{\parskip}{0pt}} \setcounter{secnumdepth}{-\maxdimen} % remove section numbering \ifLuaTeX \usepackage{selnolig} % disable illegal ligatures \fi \usepackage{bookmark} \IfFileExists{xurl.sty}{\usepackage{xurl}}{} % add URL line breaks if available \urlstyle{same} \hypersetup{ pdftitle={Primitive Integer Solutions of x\^{}4+3y\^{}3+2z\^{}3=0}, hidelinks, pdfcreator={LaTeX via pandoc}} \title{Primitive Integer Solutions of \(x^4+3y^3+2z^3=0\)} \author{} \date{29 September 2026} \begin{document} \maketitle \subsection{Abstract}\label{abstract} We determine the primitive integer solutions of \(x^4+3y^3+2z^3=0\). They are exactly \((1,-1,1)\) and \((-1,-1,1)\), where primitivity means \(\gcd(x,y,z)=1\). A fourth-power descent in \(\mathbb Q(\sqrt[3]{12})\) produces four plane quartics. A congruence modulo \(4\) excludes two of them for primitive input. We determine the rational points on the other two by elliptic descent, finite reduction sieves, and elliptic Chabauty. For the final quartic, an explicit nonzero Cassels--Tate pairing sharpens a rank-three descent bound to rank one. All finite certificates and their verification programs accompany the proof. \subsection{1. The theorem and the cubic-field descent}\label{the-theorem-and-the-cubic-field-descent} \textbf{Theorem.} The complete set of integer solutions satisfying \[ x^4+3y^3+2z^3=0,\qquad \gcd(x,y,z)=1 \] is \[ \boxed{(x,y,z)=(1,-1,1),\quad(-1,-1,1).} \] This is the \(H=56\) equation in Table 4 of Grechuk's public list {[}G{]}. Put \(X=x\), \(Y=-y\), and \(Z=-z\). Then \[ 2Z^3+3Y^3=X^4. \tag{1} \] A primitive solution has \(X\ne0\) and \(\gcd(Y,Z)=1\). If \(X\) were even, then \(Y\) would be even and \(Z\) odd, giving valuation one at \(2\) on the left. Thus \(X,Y\) are odd. If \(3\mid X\), then \(3\mid Z\), \(3\nmid Y\), and the left side has valuation one at \(3\). Consequently \[ \gcd(X,6)=1. \tag{2} \] Let \[ L=\mathbb Q(s),\quad s^3=12,\quad t=s^2/2, \] \[ \pi=2+s+t,\qquad \rho=3+s+t,\qquad \nu=1+3s-3t. \] The multiplication and norm formulas are \[ s^2=2t,\quad st=6,\quad t^2=3s, \] \[ N(a+bs+ct)=a^3+12b^3+18c^3-18abc. \tag{3} \] The integer ring is \(\mathcal O_L=\mathbb Z[1,s,t]\), of discriminant \(-972\). At \(2\), \(t^3=18\) is Eisenstein; at \(3\), \(s^3=12\) is Eisenstein. These uniformizer bases give local maximality, and the discriminant excludes other index primes. The field has class number one. Indeed, the Minkowski bound is below \(9\), and every prime ideal of norm at most \(8\) is principal: the required generators are \(\pi\), \(\rho\), and \(5+2s+2t\), of norms \(2,3,5\). The polynomial is irreducible modulo \(7\). The full unit group is \(\{\pm\nu^n:n\in\mathbb Z\}\), with \[ \nu^{-1}=55+24s+21t. \] Here is an elementary certificate for the unit assertion. A positive unit \(1<\eta<\nu^{-1}<166\) has its two complex absolute values below \(1\). Fourier inversion, using \(16/70\) in the real embedding. The curves connected by the standard two-isogeny are \[ E:\ V^2=U^3-kU,\qquad E':\ V^2=U^3+4kU. \tag{12} \] Write \(\alpha_E\) and \(\alpha_{E'}\) for the two \(U\)-coordinate descent images, with the usual coefficient class at \((0,0)\). Every class in either image has even valuations outside the primes over \(2,3\). The class-number and unit calculations reduce all possibilities to \[ \mathcal V=\langle-1,\nu,\pi,\rho\rangle \subset L^*/L^{*2},\qquad\dim_{\mathbb F_2}\mathcal V=4. \tag{13} \] \subsubsection{3.1 The dual image has dimension one}\label{the-dual-image-has-dimension-one} At \(3\), use the valuation normalized by \(v_s(s)=1\). The coefficient \(4k\) has valuation \(6\), with leading residue \(1\). For a nonzero \(U\)-coordinate of \(E'(L_3)\), put \(m=v_s(U)\). If \(m<3\), the leading term is \(U^3\); if \(m>3\), it is \(4kU\). In each case the square condition forces \(m\) even and the leading unit of \(U\) to be a square. If \(m=3\), the leading units of \(U^2\) and \(4k\) add to \(2\) modulo \(3\); the right side has odd valuation \(9\), impossible. Thus every local descent class of \(E'\) at \(3\) is trivial. In the basis (13), the local square-class labels at \(3\) are \((1,0,1,2)\), where the first bit denotes the nonsquare unit and the second bit denotes odd valuation. At the real place \(U\) is positive. These two conditions leave only \(1\) and \(\nu\). The torsion point of \(E'\) has class \(4k=(6\pi)^2\nu\). Therefore \[ \alpha_{E'}=\{1,\nu\}. \tag{14} \] On \(E\), let \[ T=(0,0),\qquad P=(12+6s+6t,\ 90+36s+36t). \tag{15} \] Their descent classes are \(-\nu\) and \(\rho\), respectively, because \[ -k=-(3\pi)^2\nu,\qquad U(P)=\rho s^2. \] They are independent in \(\mathcal V\). Thus \(\dim\alpha_E\ge2\). We prove equality with one explicit Cassels--Tate pairing. \subsubsection{3.2 The two covering classes used in the pairing}\label{the-two-covering-classes-used-in-the-pairing} For \(d\in L^*/L^{*2}\), the first covering is \[ C_d:\quad r^2=d u^4-\frac{k}{d}v^4. \tag{16} \] It maps to \(E\) by \(U=d(u/v)^2\), \(V=dru/v^3\), and its Jacobian is \(E'\). Let \(S\) be the corresponding isogeny Selmer group. It is contained in \(\mathcal V\). Both \(d=-1\) and \(d=\pi\) are everywhere locally soluble. Here are explicit Hensel witnesses at the two ramified places; in every row \(v=1\). Triples are coefficients in \(1,s,t\). \begin{longtable}[]{@{} >{\raggedright\arraybackslash}p{(\columnwidth - 10\tabcolsep) * \real{0.1500}} >{\raggedright\arraybackslash}p{(\columnwidth - 10\tabcolsep) * \real{0.1500}} >{\raggedright\arraybackslash}p{(\columnwidth - 10\tabcolsep) * \real{0.1500}} >{\raggedright\arraybackslash}p{(\columnwidth - 10\tabcolsep) * \real{0.1500}} >{\raggedleft\arraybackslash}p{(\columnwidth - 10\tabcolsep) * \real{0.2000}} >{\raggedleft\arraybackslash}p{(\columnwidth - 10\tabcolsep) * \real{0.2000}}@{}} \toprule\noalign{} \begin{minipage}[b]{\linewidth}\raggedright \(d\) \end{minipage} & \begin{minipage}[b]{\linewidth}\raggedright place \end{minipage} & \begin{minipage}[b]{\linewidth}\raggedright \(u\) \end{minipage} & \begin{minipage}[b]{\linewidth}\raggedright approximate root \(r_0\) \end{minipage} & \begin{minipage}[b]{\linewidth}\raggedleft \(v(r_0)\) \end{minipage} & \begin{minipage}[b]{\linewidth}\raggedleft \(v(r_0^2-\mathrm{RHS})\) \end{minipage} \\ \midrule\noalign{} \endhead \bottomrule\noalign{} \endlastfoot \(-1\) & \(2\) & \(1\) & \((25,27,31)\) & 0 & 18 \\ \(-1\) & \(3\) & \(0\) & \((21,15,24)\) & 3 & 12 \\ \(\pi\) & \(2\) & \(1\) & \((28,8,25)\) & 1 & 21 \\ \(\pi\) & \(3\) & \(0\) & \((12,21,0)\) & 3 & 12 \\ \end{longtable} At \(2\) the valuation is normalized by \(v_t(t)=1\); at \(3\) by \(v_s(s)=1\). Each error exceeds twice the derivative valuation, so Hensel's lemma gives an actual root. The respective root-error bounds are \(15,9,17,9\). Outside \(2,3\) the quartics have smooth genus-one reduction; the Hasse bound and Hensel's lemma give local points. Real points are immediate from \(k,\pi>0\), and the complex place poses no restriction. The four classes \(-\nu,\rho,-1,\pi\) are independent and belong to \(S\). Consequently \(S=\mathcal V\) has dimension four. \subsubsection{3.3 An explicit pushout function}\label{an-explicit-pushout-function} Set \[ A=3-3s+3t,\qquad B=9+6s-6t. \] The key identity is \[ A^2+B^2=25k. \tag{17} \] On \(C_{-1}:r^2=-u^4+kv^4\), define \[ g=Au^2+Br-5kv^2,\qquad f=g/v^2. \] Then \[ g(Au^2+Br+5kv^2)=-(Ar-Bu^2)^2. \tag{18} \] The conic \(U^2+R^2=kV^2\) has the point \((A:B:5)\), and \(g=0\) is its tangent pulled back by \((U:R:V)=(u^2:r:v^2)\). Hence \(\operatorname{div}(f)=2\mathfrak d\) for a degree-zero divisor \(\mathfrak d\) of order two. This class is nonzero: if \(f\) were a constant times a square, its square root would have poles bounded by the two points at infinity, whose Riemann--Roch space is spanned by \(1,u/v\). Its square could not have the nonzero \(Br/v^2\) term. The Jacobian \(E'\) has exactly one nonzero rational two-torsion point, since \(U^2+4k\) has no root in the real embedding. Thus \(f\) is a pushout function for that point. The Cassels--Tate pairing on \(S\) is alternating, its radical contains \(\alpha_E\), and its value on \((-1,\pi)\) is the product of Hilbert symbols \((f(P_\mathfrak p),\pi)_\mathfrak p\) {[}F, Theorem 2.1 and Section 3{]}. We now compute that product exactly. \subsubsection{3.4 The local Hilbert-symbol certificate}\label{the-local-hilbert-symbol-certificate} \textbf{Above \(2\).} Use \(u=v=1\) and the first Hensel row. Substituting its approximate root gives \[ g_0=462-213s+456t,\qquad v_t(g_0)=2. \] The exact root differs from \(r_0\) by valuation at least \(15\), so this substitution preserves the required square class. Let \[ \begin{aligned} N_1&=(-2-2s-2t)^2-\pi=50+19s+15t,\\ N_2&=(-2-s-t)^2-\pi=14+6s+5t. \end{aligned} \] Both are norms from \(L_2(\sqrt\pi)\). The quotient \(g_0/(N_1N_2)\) is a unit whose coefficient vector modulo \(8\) is \((1,2,6)\). It is the square of \((1,3,2)\) in \(\mathcal O_L/8\mathcal O_L\). Since \(v_t(8)=9>2v_t(2)=6\), this square root lifts. The error in \(g\) preserves the same residue modulo \(8\) after normalization. Thus \(f\) is a local norm and the Hilbert symbol is \(+1\). \textbf{Above \(3\).} Use \(u=0\), \(v=1\) and the second Hensel row. Then \[ g_0=603-216s+270t,\qquad v_s(g_0)=6, \] and \(g_0/s^6\) reduces to \(1\). Its square root lifts. The root-error bound preserves its valuation and leading residue, so \(f\) itself is a square in \(L_3\). The symbol is again \(+1\). \textbf{Above \(5\).} The factorization is \[ s^3-12\equiv(s-3)(s^2+3s+4)\pmod5. \] At the degree-one prime, \(s\equiv8\pmod{25}\) and \(t\equiv7\pmod{25}\). Thus \(A\equiv0\), \(B\equiv15\pmod{25}\), while \(k\equiv4\pmod5\). The local point \(u=0,v=1,r\equiv2\pmod5\) gives \(g/5\equiv2\pmod5\): its valuation is one. Also \(\pi\equiv2\pmod5\) is nonsquare. The Hilbert symbol is \(-1\). At the degree-two prime, \(\pi\equiv2s=(1+2s)^2\) in \(\mathbb F_5[s]/(s^2+3s+4)\), so the symbol is \(+1\). \textbf{Every other place.} At the real place \(\pi>0\), and complex symbols are trivial. At an odd prime outside \(2,3,5\), normalize \(u,v\) integrally with at least one a unit. The element \(k\) is a unit because \(N(k)=2916\). If \(v\) is a unit and \(g\) vanishes modulo the prime, the other factor in (18) is a unit, so \(v(g)\) is even. If \(v\) is not a unit, \(u\) is a unit; \(g\) cannot vanish, by \(A^2+B^2=25k\). Hence \(v(f)\) is even in every case. As \(\pi\) is also a unit, the symbol is \(+1\). The product is therefore \[ \boxed{\langle-1,\pi\rangle=-1.} \tag{19} \] \subsubsection{3.5 Rank and odd index}\label{rank-and-odd-index} A nonzero alternating pairing on a four-dimensional space has rank at least two. Since the two independent actual classes \(-\nu,\rho\) belong to its radical, the radical has dimension exactly two. Hence \(\dim\alpha_E=2\). Together with (14), the two-isogeny rank formula gives \[ \boxed{\operatorname{rank}E(L)=2+1-2=1.} \tag{20} \] The reductions at degree-one primes over \(5,11\) have orders \(4,12\). They bound global torsion by four. The only nonzero rational two-torsion point is \(T\), since \(k=(3\pi)^2\nu\) is nonsquare. A half of \(T\) would have \(U^2=-k\) by the doubling formula, impossible in the real embedding. Thus \(E(L)_{\rm tors}=\{O,T\}\). The natural map \(E(L)/2E(L)\to\alpha_E\) is a surjection between two-dimensional spaces. The classes of \(P,T\) span the target, so they span the source. Therefore \(\langle P,T\rangle\) has finite odd index in \(E(L)\). \subsection{\texorpdfstring{4. The complete \(331\) sieve and local uniqueness on \(D_1\)}{4. The complete 331 sieve and local uniqueness on D\_1}}\label{the-complete-331-sieve-and-local-uniqueness-on-d_1} The prime \(331\) splits completely in \(L\), with roots \[ s\equiv7,107,217\pmod{331}. \tag{21} \] Each elliptic reduction has \(332=4\cdot83\) points. The reduction of \(P\) in the product has order \(166\); adding \(T\) gives a subgroup \(H\) of order \(332\). Its quotient in the product has order \[ 332^3/332=110224=16\cdot83^2. \] The point \([4]P\) reduces in the three factors to \[ (276,40),\quad(118,216),\quad(94,179). \tag{22} \] These are nonzero. If the odd index of \(\langle P,T\rangle\) were divisible by \(83\), then \(P=83R+eT\) for a global \(R\), and \([4]P\) would reduce to zero in all factors. Thus the index is prime to \(83\). It is consequently prime to \(|E(\mathbb F_{331})^3/H|\), and \[ \operatorname{red}_{331}E(L)=H. \tag{23} \] Complete enumeration of all \(331^2+331+1=109893\) projective residue points finds \(332\) on \(D_1\). Under (10), only \[ (1:0:0) \tag{24} \] maps into \(H\). Polynomial evaluation and independent multiplication in \(\mathbb F_{331}[s]/(s^3-12)\) give the same projective list. All \(3\cdot332^2=330672\) elliptic additions agree in affine and Jacobian representations. The rational formula, homogeneous formula and interpolation of the three conjugates give identical images, including the apparent poles. It remains to prove uniqueness in the disk of (24). The quartic is smooth modulo \(331\): the equation and its partial derivatives generate the unit ideal in each of its three affine charts. In the chart \(a=1\), use \(c\) as parameter. At the center, the partial derivative with respect to \(b\) is \(8\), and the tangent vector is \((0,1,1)\). For \(\omega=dU/(2V)\), compute the three elliptic logarithms by \(\log P=\log([166]P)/166\). The roots in (21) lift modulo \(331^3\) to \[ 13040745,\quad28244337,\quad31244300. \] The normalized logarithm is \[ \frac{\log P}{331}\equiv (99490,6698,96118)\pmod{331^2}, \] and reduces to \((190,78,128)\). An exact integral annihilator is \((0,L_3,-L_2)\), where \(L_i\) are the exact entries of \(\log P/331\). It reduces to \[ (0,128,253). \tag{25} \] Direct differentiation of (9) gives the exact coefficient \(-s\) at the center in the parameter \(c\). Its three residues are \((324,224,114)\), and their dot product with (25) is \[ \boxed{251\not\equiv0\pmod{331}.} \tag{26} \] All map denominators and elliptic \(V\)-coordinates are units in this disk. The rank-one calculation implies that the annihilator vanishes on every global elliptic logarithm. Section 6's local zero lemma therefore gives exactly one rational point in the disk. Equations (23)--(26) prove \[ D_1(\mathbb Q)=\{(1:0:0)\}. \tag{27} \] \subsection{\texorpdfstring{5. The \(D_2\) certificate, included in the proof}{5. The D\_2 certificate, included in the proof}}\label{the-d_2-certificate-included-in-the-proof} For completeness this section gives the other rational-point classification in (7), with its executable certificates in \texttt{prior\_D2/}. Its elliptic curve is \[ E_2:\ V^2=U(U-r)(U+r),\qquad r=3\pi\nu. \] \subsubsection{5.1 Full two-descent}\label{full-two-descent} The Kummer map is \((U,U-r)\) modulo squares, with values \((-r^2,-r)\) and \((r,2r^2)\) at the two generators of full two-torsion. Each factor belongs to \(\mathcal V\) of (13), so there are \(256\) global candidate pairs. At \(2\), the local Kummer image has dimension \([L_2:\mathbb Q_2]+2=5\). At \(3\) it has dimension two. Here are three further actual local points at \(2\), besides torsion. Their right sides have valuation zero, and the last column supplies a unit square root modulo \(8\) in the basis \(1,s,t\). \begin{longtable}[]{@{}lll@{}} \toprule\noalign{} \(U\) & \(U(U^2-r^2)\) modulo \(8\) & square root modulo \(8\) \\ \midrule\noalign{} \endhead \bottomrule\noalign{} \endlastfoot \((-7,-7,-4)\) & \((1,4,0)\) & \((1,2,0)\) \\ \((-7,-6,-6)\) & \((1,7,6)\) & \((1,2,3)\) \\ \((-7,-4,-4)\) & \((5,1,0)\) & \((1,3,3)\) \\ \end{longtable} The five Kummer vectors are independent and exhaust the local image. The two torsion vectors exhaust the image at \(3\). At infinity, the two signs must agree. With masks in the order \((-1,\nu,\pi,\rho)\), the complete global survivor list is \begin{verbatim} (0,0) (0,2) (1,13) (1,15) (6,8) (6,10) (7,5) (7,7) (10,12) (10,14) (11,1) (11,3) (12,4) (12,6) (13,9) (13,11) \end{verbatim} They are spanned by torsion and the two actual points \[ \begin{aligned} P_2&=(-2+2s-t,\ 4+12s-12t),\\ R_2&=(9-6s+2t,\ -33+18s-3t). \end{aligned} \] The four global Kummer pairs are \((1,13),(12,6),(7,7),(0,2)\). Their span has size \(16\), proving rank two and odd index for these points and full two-torsion. Finite orders \(8,12\) at degree-one primes over \(5,11\) show that the global torsion is exactly the full two-torsion. \subsubsection{5.2 Two compatible reduction sieves}\label{two-compatible-reduction-sieves} The primes \(211\) and \(7207\) split completely. At both, the elliptic factors have order \(p+1\) and structure \(\mathbb Z/((p+1)/2)\times\mathbb Z/2\). In the product, the subgroup generated by \(P_2,R_2\) and torsion has the following index: \begin{longtable}[]{@{} >{\raggedleft\arraybackslash}p{(\columnwidth - 6\tabcolsep) * \real{0.2500}} >{\raggedleft\arraybackslash}p{(\columnwidth - 6\tabcolsep) * \real{0.2500}} >{\raggedleft\arraybackslash}p{(\columnwidth - 6\tabcolsep) * \real{0.2500}} >{\raggedleft\arraybackslash}p{(\columnwidth - 6\tabcolsep) * \real{0.2500}}@{}} \toprule\noalign{} \begin{minipage}[b]{\linewidth}\raggedleft \(p\) \end{minipage} & \begin{minipage}[b]{\linewidth}\raggedleft \(\#D_2(\mathbb F_p)\) \end{minipage} & \begin{minipage}[b]{\linewidth}\raggedleft product subgroup index \end{minipage} & \begin{minipage}[b]{\linewidth}\raggedleft surviving points \end{minipage} \\ \midrule\noalign{} \endhead \bottomrule\noalign{} \endlastfoot \(211\) & 212 & \(424=8\cdot53\) & 3 \\ \(7207\) & 7208 & \(14416=16\cdot17\cdot53\) & 7 \\ \end{longtable} The minors of the free-generator coordinate matrix are \((15,27,2)\) modulo \(53\) at \(211\), and \((10,10,3)\) modulo \(17\) and \((3,8,43)\) modulo \(53\) at \(7207\). Their nonvanishing proves saturation at \(17,53\). Odd index then proves that the computed product subgroups are the full global reduction images. At \(211\), the surviving projective points are \[ (1:22:49),\quad(1:106:106),\quad(1:141:141). \] Their coordinates in the global free generators modulo \(53\) are \[ (30,32),\quad(1,0),\quad(52,0). \] The seven points at \(7207\) have coordinates \[ (2,4),(27,51),(1,0),(52,12),(52,0),(9,30),(4,3). \] Compatibility leaves only \((1,0),(52,0)\), eliminating the first \(211\) disk. The remaining disks are the reductions of \((2:1:1),(3:1:1)\). Every projective point is checked; an independently organized C++ field-multiplication enumeration agrees with the polynomial enumeration. \subsubsection{\texorpdfstring{5.3 Local uniqueness at \(211\)}{5.3 Local uniqueness at 211}}\label{local-uniqueness-at-211} The three roots are \(38,63,110\), with lifts modulo \(211^3\) \[ 614892,\quad953783,\quad7825256. \] The normalized logarithms, using the finite exponent \(106\), are \[ \begin{aligned} \log(P_2)/211&\equiv(19201,14981,36081)\pmod{211^2},\\ \log(R_2)/211&\equiv(9872,44497,23895)\pmod{211^2}. \end{aligned} \] The first row is divisible by \(211\). Taking the cross product of the exact rows \(\log(P_2)/211^2\) and \(\log(R_2)/211\) gives an integral annihilator reducing to \((200,22,167)\). In the chart \(b=1\), the two pullback coefficients from (11) are \begin{longtable}[]{@{}lllr@{}} \toprule\noalign{} point & parameter & exact coefficient & combined residue \\ \midrule\noalign{} \endhead \bottomrule\noalign{} \endlastfoot \((2:1:1)\) & \(a-2\) & \(3+s+t\) & 178 \\ \((3:1:1)\) & \(a-3\) & \(2+s+2t/3\) & 94 \\ \end{longtable} The dependent-coordinate derivatives are \(-16,-24\), both units. The maps are integral and the elliptic \(V\)-coordinates are units on both disks. Section 6 proves uniqueness in each disk, giving \[ D_2(\mathbb Q)=\{(2:1:1),(3:1:1)\}. \tag{28} \] \subsection{6. The common local zero lemma and the logarithm precision}\label{the-common-local-zero-lemma-and-the-logarithm-precision} Let \(p\) be either \(211\) or \(331\). Compose the three elliptic logarithms with the exact integral annihilator. On a surviving disk, center a regular parameter \(T\) at the known rational point. The pullback of the invariant differential is integral with unit constant coefficient. Its integral is \[ F(T)=uT+\sum_{n\ge2}\frac{a_{n-1}}nT^n, \qquad u\in\mathbb Z_p^*,\quad a_{n-1}\in\mathbb Z_p, \quad T\in p\mathbb Z_p. \] It vanishes at every rational point in the disk. For \(T\ne0\) and \(n\ge2\), \[ v_p\left(\frac{a_{n-1}}nT^n\right) \ge n v_p(T)-v_p(n)>v_p(T). \] These valuations tend to infinity. The ultrametric inequality implies that the convergent tail cannot cancel the linear term. Hence \(F(T)=0\) forces \(T=0\). This is the all-heights uniqueness argument used in both cases. For the models \(V^2=U^3+bU\) and parameter \(\tau=-U/V\), the exact formal differential is \[ \frac{dU}{2V}=\frac{d\tau}{\sqrt{1-4b\tau^4}},\qquad \log(\tau)=\tau+O(\tau^5). \] The scalar entering the formal group is prime to \(p\). Thus computations modulo \(p^3\) give the normalized logarithms to the stated precision. The programs compute each value using both Jacobian projective arithmetic and independent division-polynomial recurrences. \subsection{7. Reconstruction}\label{reconstruction} By (6), only \(j=1,2\) can occur for primitive input. The complete point lists (27)--(28) give these primitive integral representatives: \begin{longtable}[]{@{}rlr@{}} \toprule\noalign{} \(j\) & \((a,b,c)\) & \(N\beta\) \\ \midrule\noalign{} \endhead \bottomrule\noalign{} \endlastfoot 1 & \((1,0,0)\) & 1 \\ 2 & \((2,1,1)\) & 2 \\ 2 & \((3,1,1)\) & 3 \\ \end{longtable} A primitive integral representative of a rational projective point is unique up to sign, and the fourth power removes that sign. Conditions (2) and (5) eliminate the last two rows. The first gives \[ 2Z+sY=s-2,\qquad(X,Y,Z)=(\pm1,1,-1). \] Undoing the sign changes gives precisely \((x,y,z)=(\pm1,-1,1)\). Direct substitution verifies both points. This proves the theorem. \subsection{8. Reproduction}\label{reproduction} Run from the directory containing this note: \begin{verbatim} python run_all.py \end{verbatim} The requirements are Python 3.10 or later, SymPy, NumPy, and a C++17 compiler. Run with Python assertions enabled. The driver regenerates the certificates from the displayed constants and runs these stages: \begin{longtable}[]{@{} >{\raggedright\arraybackslash}p{(\columnwidth - 2\tabcolsep) * \real{0.5000}} >{\raggedright\arraybackslash}p{(\columnwidth - 2\tabcolsep) * \real{0.5000}}@{}} \toprule\noalign{} \begin{minipage}[b]{\linewidth}\raggedright program \end{minipage} & \begin{minipage}[b]{\linewidth}\raggedright finite certificate \end{minipage} \\ \midrule\noalign{} \endhead \bottomrule\noalign{} \endlastfoot \texttt{prior\_D2/verify\_mixed\_descent.py} & integer ring, class number, units, \(E_2\) descent \\ \texttt{prior\_D2/verify\_mixed\_sieve.py} & complete \(211,7207\) sieves and compatibility \\ \texttt{prior\_D2/verify\_mixed\_chabauty.py} & both \(D_2\) local uniqueness certificates \\ \texttt{verify\_reduction.py} & all four quartics, mod-\(4\) exclusions, reconstruction \\ \texttt{verify\_rank.py} & exact pairing witnesses, isogeny images, rank-one data \\ \texttt{verify\_sieve.py} & complete \(331\) sieve in independent representations \\ \texttt{verify\_chabauty.py} & smoothness, logarithms, and the unit derivative \(251\) \\ \end{longtable} The mathematical implications use unique ideal factorization, Minkowski's bound, Mordell--Weil, Kummer descent and the two-isogeny rank formula, Hensel's lemma, the Hasse bound, Hilbert symbols, the Cassels--Tate pairing, and elliptic logarithms. Sections 1--7 specify their applications and connect the finite certificates to the all-heights conclusion. \subsection{References}\label{references} {[}G{]} B. Grechuk, \emph{A systematic approach to Diophantine equations: open problems}, arXiv:2404.08518v9, 30 August 2026, Table 4, p.~4. https://arxiv.org/abs/2404.08518v9 {[}F{]} T. Fisher, \emph{Higher descents on an elliptic curve with a rational 2-torsion point}, Mathematics of Computation \textbf{86} (2017), no. 307, 2493--2518. Theorem 2.1, Sections 3 and 5. https://arxiv.org/abs/1509.03234 {[}S{]} M. Stoll, \emph{Implementing 2-descent for Jacobians of hyperelliptic curves}, Acta Arithmetica \textbf{98} (2001), 245--277, especially Lemmas 4.1--4.4. DOI: 10.4064/aa98-3-4. {[}B{]} N. Bruin, \emph{Chabauty methods using elliptic curves}, Journal fuer die reine und angewandte Mathematik \textbf{562} (2003), 27--49. https://www.cecm.sfu.ca/\textasciitilde nbruin/bruin\_crelle.pdf \end{document}