# Perfect Cuboid Radical-FDRS R12 ## General Sylow-$2$ Prime Search, the $T_0$ Gate-Swap Symmetry, and Quotient Compression to the $2$-Isogenous Elliptic Curve **Series:** Perfect Cuboid Global Closure **Branch:** Radical-FDRS / Geometry-Coupling Branch R **Phase:** R12 — General Prime Search and Gate-Symmetry Compression **Date:** 2026-10-04 **Status:** Exact computational checkpoint + exact algebraic symmetry theorem — not a proof of Perfect Cuboid existence or nonexistence --- # 0. Executive summary R11 strengthened the historical fixed-slice necessary condition to eight coefficient classes modulo $$ 2^{18} $$ in each of two surviving rational $2$-torsion cosets. R12 follows the proposed general Sylow-$2$ route. A search through $$ 260{,}000 $$ consecutive good prime candidates spanning approximately $$ 4.2\times10^6 $$ to $$ 8.26\times10^6 $$ found one new prime for which the projected known Mordell-Weil direction has order strictly greater than $$ 2^{18}. $$ The prime is $$ \boxed{ p=6{,}291{,}881. } $$ An independent exact character-sum point count gives $$ \boxed{ \#E(\mathbb F_p) = 6{,}291{,}456 = 3\cdot2^{21}. } $$ The odd cofactor is therefore $$ 3. $$ For the known odd-index Mordell-Weil surrogate $$ G_0=(480704,-468686400), $$ the projected point $$ \boxed{ P_2=[3]G_0 } $$ has exact order $$ \boxed{ 2^{19}. } $$ Indeed, $$ \boxed{ [2^{18}]P_2=(0,0), } $$ while $$ \boxed{ [2^{19}]P_2=O. } $$ Thus this prime genuinely sees one coefficient bit beyond the R11 modulus. However, when the R11 coefficient classes are lifted from modulo $$ 2^{18} $$ to modulo $$ 2^{19}, $$ and the square gates are tested correctly **before** the odd-cofactor projection, every lift survives. The reason is structural, not accidental. Let $$ \boxed{ T_0=(0,0)\in E[2]. } $$ Translation by this $2$-torsion point sends $$ X \longmapsto -\frac{AB}{X}. $$ Under this translation the two R8 squareclasses are exchanged: $$ \boxed{ [\mathcal R_1(P+T_0)] = [\mathcal R_2(P)], } $$ and $$ \boxed{ [\mathcal R_2(P+T_0)] = [\mathcal R_1(P)]. } $$ Therefore the simultaneous double-square predicate is invariant under $$ P\mapsto P+T_0. $$ This immediately explains two facts that previously appeared as separate observations: 1. the two R9/R11 surviving torsion cosets $$ T_A=(A,0) $$ and $$ T_{-B}=(-B,0) $$ always have identical double-gate behavior, because $$ T_A+T_0=T_{-B}; $$ 2. the new prime cannot distinguish the coefficient bit $$ c\mapsto c+2^{18}, $$ because $$ [2^{18}]P_2=T_0. $$ Thus projected order alone is not the correct strength metric for future prime selection. The relevant condition is projected order **modulo the gate-swap torsion symmetry**. R12 then quotients by this symmetry. The quotient by translation through $$ T_0 $$ is the standard degree-$2$ isogeny $$ E\longrightarrow\widehat E. $$ Define $$ \boxed{ U=\frac{Y^2}{X^2} = X+M-\frac{AB}{X}. } $$ Then $$ U $$ is invariant under $$ X\mapsto-\frac{AB}{X}. $$ The quotient curve is $$ \boxed{ \widehat E: \qquad V^2 = U^3 - 2MU^2 + L^2U, } $$ where $$ M=B-A=950625, $$ and $$ L=A+B=1352033. $$ Most importantly, the product of the two R8 square gates simplifies to $$ \boxed{ \mathcal R_1(P)\mathcal R_2(P) = X^2 \left( 2MU-L^2 \right). } $$ Therefore its squareclass is $$ \boxed{ [ \mathcal R_1\mathcal R_2 ] = [ 2MU-L^2 ]. } $$ Hence the historical genus-$3$ product-square condition descends to the quotient as the **linear square-value gate** $$ \boxed{ 2MU-L^2 \in \mathbb Q^2. } $$ This is the main R12 structural reduction. --- # 1. Elliptic core Use $$ \boxed{ E: \qquad Y^2 = X(X-A)(X+B), } $$ with $$ A=448^2=200704, $$ and $$ B=1073^2=1151329. $$ Define $$ \boxed{ M=B-A=950625, } $$ and $$ \boxed{ L=A+B=1352033. } $$ Then $$ E: \qquad Y^2 = X^3+MX^2-ABX. $$ --- # 2. R8 square gates The two exact fixed-slice square functions are $$ \boxed{ \mathcal R_1(P) = \frac{2X+M}{L}, } $$ and $$ \boxed{ \mathcal R_2(P) = L\,X(MX-2AB). } $$ A rational point on the complete fixed genus-$5$ fiber must satisfy $$ \mathcal R_1(P)\in\mathbb Q^2, $$ and $$ \mathcal R_2(P)\in\mathbb Q^2. $$ --- # 3. R11 residue state R11 left the coefficient set $$ \boxed{ \mathcal C_{18} = \{ 1, 16385, 114687, 131071, 131073, 147457, 245759, 262143 \} } $$ modulo $$ 2^{18}. $$ Equivalently, $$ \boxed{ \mathcal C_{18} = \pm \{ 1, 16385, 114687, 131071 \}. } $$ Only the torsion cosets $$ \boxed{ T_A=(A,0), } $$ and $$ \boxed{ T_{-B}=(-B,0) } $$ survived. --- # 4. General-prime search objective R11 suggested searching for arbitrary good primes $$ p $$ such that the Sylow-$2$ projection of the known Mordell-Weil direction has order exceeding $$ 2^{18}. $$ Write $$ \#E(\mathbb F_p) = 2^s m, $$ where $$ m $$ is odd. Define $$ \boxed{ P_2=[m]G_0. } $$ The search target is $$ \boxed{ \operatorname{ord}(P_2)\ge2^{19}. } $$ --- # 5. Exact point-count method For each prime candidate, the group order was first determined by an exact Hasse-interval baby-step/giant-step trace computation. The search does not estimate the trace numerically. It determines the unique integer $$ t $$ inside the Hasse interval $$ |t|\le2\sqrt p $$ satisfying $$ [p+1-t]P=O $$ for enough independent test points. For the final hit, the resulting group order was then independently verified by a complete character sum. Thus the reported hit does not rely on one point-order computation alone. --- # 6. Search range The exact scan processed $$ \boxed{ 260{,}000 } $$ consecutive prime candidates beginning above approximately $$ 4.2\times10^6 $$ and continuing through approximately $$ 8.26\times10^6. $$ Inside this range, exactly one prime was found for which the projected known Mordell-Weil direction had order at least $$ 2^{19}. $$ That prime is $$ \boxed{ 6{,}291{,}881. } $$ This is a computational result for the stated finite search interval only. No density claim is made. --- # 7. Independent group-order certificate At $$ p=6{,}291{,}881, $$ compute the exact character sum $$ S = \sum_{x\in\mathbb F_p} \chi \left( x(x-A)(x+B) \right). $$ The result is $$ \boxed{ S=-426. } $$ Therefore $$ \#E(\mathbb F_p) = p+1+S. $$ Hence $$ \boxed{ \#E(\mathbb F_p) = 6{,}291{,}456. } $$ Factorization gives $$ \boxed{ 6{,}291{,}456 = 3\cdot2^{21}. } $$ Thus the odd cofactor is exactly $$ \boxed{ m=3. } $$ --- # 8. Projected Mordell-Weil point Reduce $$ G_0 = (480704,-468686400) $$ modulo $$ p. $$ Project by the odd cofactor: $$ \boxed{ P_2=[3]G_0. } $$ Its coordinates are $$ \boxed{ P_2 = (3750844,1435474) \in E(\mathbb F_p). } $$ Repeated doubling gives $$ \boxed{ [2^{18}]P_2=(0,0), } $$ and $$ \boxed{ [2^{19}]P_2=O. } $$ Therefore $$ \boxed{ \operatorname{ord}(P_2)=2^{19}. } $$ --- # 9. The odd-primary kernel Since $$ \#E(\mathbb F_p) = 3\cdot2^{21}, $$ the kernel of multiplication by $$ 3 $$ contains one nontrivial cyclic subgroup of order $$ 3. $$ One explicit order-$3$ point is $$ \boxed{ R_3 = (2460118,6009634). } $$ It satisfies $$ \boxed{ [3]R_3=O. } $$ Thus every projected point $$ Q=[3]P $$ has exactly three preimages: $$ \boxed{ P_0, \quad P_0+R_3, \quad P_0+2R_3, } $$ where $$ P_0 $$ is the unique preimage lying in the $2$-primary subgroup. --- # 10. Correct gate-before-projection test For each R11 residue $$ c_0\in\mathcal C_{18}, $$ there are two lifts modulo $$ 2^{19}: $$ $$ c_0, $$ and $$ c_0+2^{18}. $$ This gives $$ 16 $$ candidate coefficient classes. For each coefficient and each surviving torsion coset, define the projected state $$ \boxed{ Q = [c]P_2+T. } $$ Then compute its three preimages under multiplication by $$ 3. $$ The square gates are tested on the preimages, not on $$ Q. $$ A projected coefficient state survives if at least one of its three original preimages satisfies both gates. This is the proof-safe arbitrary-prime procedure. --- # 11. Result: no coefficient elimination All $$ 16 $$ coefficient lifts survive for both torsion cosets. Thus the new prime produces: $$ \boxed{ \text{no additional coefficient exclusion}. } $$ This remains true despite $$ \operatorname{ord}(P_2)=2^{19}. $$ Therefore: $$ \boxed{ \text{larger projected order alone is insufficient}. } $$ The odd-primary preimage structure matters. --- # 12. Exact odd-preimage signatures Choose the order-$3$ generator $$ R_3 $$ above and label the three preimages by $$ j=0,1,2. $$ For the eight base R11 residues modulo $$ 2^{18}, $$ the gate-passing branches are: | $c\bmod2^{18}$ | passing $j$ | |---:|:---| | $1$ | $\{0,2\}$ | | $16385$ | $\{0,2\}$ | | $114687$ | $\{1\}$ | | $131071$ | $\{1,2\}$ | | $131073$ | $\{1,2\}$ | | $147457$ | $\{2\}$ | | $245759$ | $\{0,1\}$ | | $262143$ | $\{0,1\}$ | The same pattern occurs for: $$ c+2^{18}. $$ The same pattern also occurs in both surviving torsion cosets. The labels: $$ 1 $$ and: $$ 2 $$ depend on the chosen orientation of the order-$3$ generator, but nonemptiness and cardinality are invariant. --- # 13. The missing explanation: a $2$-torsion symmetry Define $$ \boxed{ T_0=(0,0). } $$ For a finite point $$ P=(X,Y) $$ with $$ X\ne0, $$ the addition formula gives $$ \boxed{ X(P+T_0) = -\frac{AB}{X(P)}. } $$ This extends projectively across the exceptional points. Let $$ X' = -\frac{AB}{X}. $$ R12 now studies the action of this translation on the two gate squareclasses. --- # 14. First gate after $T_0$ translation Substitute $$ X'=-\frac{AB}{X}. $$ Then $$ \mathcal R_1(P+T_0) = \frac{ MX-2AB }{ LX }. $$ Meanwhile $$ \mathcal R_2(P) = LX(MX-2AB). $$ Their quotient is $$ \boxed{ \frac{ \mathcal R_2(P) }{ \mathcal R_1(P+T_0) } = L^2X^2. } $$ Since the right side is a rational square, $$ \boxed{ [ \mathcal R_1(P+T_0) ] = [ \mathcal R_2(P) ]. } $$ --- # 15. Second gate after $T_0$ translation Again use $$ X'=-\frac{AB}{X}. $$ Then $$ \mathcal R_2(P+T_0) = \frac{ LA^2B^2(M+2X) }{ X^2 }. $$ But $$ \mathcal R_1(P) = \frac{ M+2X }{ L }. $$ Therefore $$ \boxed{ \frac{ \mathcal R_2(P+T_0) }{ \mathcal R_1(P) } = \left( \frac{ LAB }{ X } \right)^2. } $$ Hence $$ \boxed{ [ \mathcal R_2(P+T_0) ] = [ \mathcal R_1(P) ]. } $$ --- # 16. Gate-Swap Theorem ## Theorem Translation by $$ T_0=(0,0) $$ exchanges the two R8 gate squareclasses: $$ \boxed{ \left( [\mathcal R_1], [\mathcal R_2] \right) \longmapsto \left( [\mathcal R_2], [\mathcal R_1] \right). } $$ Therefore the simultaneous predicate $$ \boxed{ \mathcal R_1\in\mathbb Q^2 \quad\text{and}\quad \mathcal R_2\in\mathbb Q^2 } $$ is invariant under $$ \boxed{ P\mapsto P+T_0. } $$ This is an exact global algebraic theorem. It is not specific to one reduction prime. --- # 17. Why the two R9/R11 torsion cosets always match The three nonzero rational $2$-torsion points satisfy $$ T_A+T_{-B}=T_0. $$ Therefore $$ \boxed{ T_A+T_0=T_{-B}, } $$ and $$ \boxed{ T_{-B}+T_0=T_A. } $$ Because double-gate squarehood is invariant under translation by $$ T_0, $$ the two surviving R9/R11 torsion cosets must always have identical double-gate behavior. Thus they are not two genuinely independent arithmetic states. They form one gate-symmetry orbit. --- # 18. Corrected effective torsion count R11 reported two surviving torsion cosets: $$ T_A, \qquad T_{-B}. $$ R12 refines the interpretation. Modulo the exact gate-swap symmetry $$ \langle T_0\rangle, $$ these two classes are one orbit: $$ \boxed{ \{T_A,T_{-B}\}. } $$ Likewise the two eliminated classes form the other orbit: $$ \boxed{ \{O,T_0\}. } $$ Therefore the effective gate-state torsion count is: $$ \boxed{ 1 } $$ surviving orbit, not two unrelated surviving classes. --- # 19. Why the new prime cannot see the next coefficient bit At $$ p=6{,}291{,}881, $$ we proved $$ \boxed{ [2^{18}]P_2=T_0. } $$ Therefore $$ [c+2^{18}]P_2+T = [c]P_2+(T+T_0). $$ Thus changing the top coefficient bit is exactly equivalent to applying the gate-swap torsion translation. Because the simultaneous gate predicate is invariant under this translation, $$ \boxed{ c \quad\text{and}\quad c+2^{18} } $$ must have identical survival behavior. Hence the failure to cut the new bit is forced algebraically. It is not a weakness of the implementation. --- # 20. Prime-selection correction R11 proposed maximizing $$ \operatorname{ord}([m]G_0). $$ R12 shows that this is not sufficient. Suppose $$ \operatorname{ord}([m]G_0)=2^e. $$ If $$ \boxed{ [2^{e-1}][m]G_0=T_0, } $$ then the highest coefficient bit is invisible to the simultaneous double-gate predicate. Therefore a better prime-selection score must record both: $$ \boxed{ e, } $$ and the half-order torsion point $$ \boxed{ H_p = [2^{e-1}][m]G_0. } $$ A prime with large $$ e $$ but $$ H_p=T_0 $$ may add no information at the top bit. --- # 21. Desirable future primes The most promising primes are those with: $$ \operatorname{ord}([m]G_0)\ge2^{19} $$ and half-order torsion: $$ H_p $$ equal to $$ T_A $$ or $$ T_{-B}, $$ rather than $$ T_0. $$ Alternatively one wants projected order at least $$ 2^{20} $$ so that even after quotienting by the gate-swap symmetry there remains a genuinely new binary digit to test. Thus the corrected search target is: $$ \boxed{ \text{large projected order modulo gate symmetry}. } $$ --- # 22. Quotient by the gate-swap torsion The involution $$ P\mapsto P+T_0 $$ is the translation action associated with the kernel $$ \{O,T_0\}. $$ The quotient elliptic curve is the standard degree-$2$ isogenous curve. Define the invariant coordinate $$ \boxed{ U = \frac{Y^2}{X^2}. } $$ Using the elliptic equation, $$ \boxed{ U = X+M-\frac{AB}{X}. } $$ Under $$ X\mapsto-\frac{AB}{X}, $$ the value of $$ U $$ is unchanged. Thus $$ U $$ is a natural coordinate on the quotient. --- # 23. The quotient elliptic curve Define $$ \boxed{ V = - \frac{ Y(X^2+AB) }{ X^2 }. } $$ Then $$ (U,V) $$ lies on $$ \boxed{ \widehat E: \qquad V^2 = U^3 - 2MU^2 + L^2U. } $$ This is the standard degree-$2$ isogenous curve with kernel: $$ \boxed{ \{O,T_0\}. } $$ After rational scaling, it is the historical normalized: $$ E_0 $$ family already encountered in R5-R8. Therefore the gate-swap quotient is not a new elliptic object. It is precisely the existing $E_0$ direction. --- # 24. Product gate on the quotient Multiply the two R8 gate functions: $$ \mathcal R_1\mathcal R_2 = X(2X+M)(MX-2AB). $$ Use: $$ U = X+M-\frac{AB}{X}. $$ Exact algebra gives: $$ \boxed{ X(2X+M)(MX-2AB) = X^2 \left( 2MU-L^2 \right). } $$ Therefore: $$ \boxed{ [ \mathcal R_1\mathcal R_2 ] = [ 2MU-L^2 ]. } $$ The square factor: $$ X^2 $$ disappears from the squareclass. --- # 25. Linear Square-Gate Theorem ## Theorem On the quotient elliptic curve $$ \widehat E, $$ the historical genus-$3$ product-square condition is equivalent to $$ \boxed{ 2MU-L^2 \in \mathbb Q^2. } $$ Explicitly: $$ \boxed{ 1{,}901{,}250\,U - 1352033^2 \in \mathbb Q^2. } $$ Thus a cubic-looking square-value function on $$ E $$ becomes a linear square-value function after quotienting by the gate-swap torsion. This is a major FDRS simplification. --- # 26. Interpretation of the individual gates The quotient remembers the unordered pair of gate squareclasses: $$ \boxed{ \{ [\mathcal R_1], [\mathcal R_2] \}. } $$ Translation by $$ T_0 $$ swaps the two members. The product squareclass is quotient-invariant and becomes the linear function: $$ 2MU-L^2. $$ To require both individual classes to be trivial, one still needs one additional quadratic lift above the quotient. Thus the complete double-gate problem may be reorganized as: $$ \boxed{ \text{linear square gate on }\widehat E } $$ plus $$ \boxed{ \text{one residual gate-label lift}. } $$ This is potentially simpler than carrying both original gates independently on: $$ E. $$ --- # 27. State-count compression R11 described: $$ 8 $$ coefficient residues and: $$ 2 $$ surviving torsion cosets. R12 shows that the two torsion cosets are one gate-symmetry orbit. Therefore the symmetry-reduced state count is: $$ \boxed{ 8 } $$ rather than: $$ 16. $$ The eight R11 coefficient classes remain the true unresolved binary states at the current precision. --- # 28. Search result beyond the new prime The continuous prime scan was extended beyond $$ p=6{,}291{,}881 $$ through approximately $$ 8.26\times10^6. $$ No second prime was found in the scanned interval for which the projected known Mordell-Weil direction had order at least $$ 2^{19}. $$ This is only a finite computational observation. It is not a density or nonexistence theorem for such primes. --- # 29. Why brute-force prime extension is no longer the best next step The first successful arbitrary-prime hit already demonstrates two facts: 1. large projected order primes exist; 2. large projected order does not automatically refine the coefficient sieve. The second fact is explained exactly by the gate-swap torsion symmetry. Therefore simply extending the prime search farther without accounting for $$ H_p = [2^{e-1}][m]G_0 $$ is inefficient. Future prime searches should be symmetry-aware. --- # 30. R12 theorem package ## Theorem A — New general prime $$ \boxed{ p=6{,}291{,}881 } $$ satisfies $$ \boxed{ \#E(\mathbb F_p)=3\cdot2^{21}. } $$ ## Theorem B — Projected order $$ \boxed{ \operatorname{ord}([3]G_0)=2^{19}. } $$ ## Theorem C — Half-order torsion $$ \boxed{ [2^{18}][3]G_0=T_0. } $$ ## Theorem D — No R11 class elimination All lifts of the R11 eight classes to modulo $$ 2^{19} $$ have at least one odd-primary preimage satisfying both reduced square gates. ## Theorem E — Gate-swap torsion symmetry $$ \boxed{ [\mathcal R_1(P+T_0)] = [\mathcal R_2(P)], } $$ $$ \boxed{ [\mathcal R_2(P+T_0)] = [\mathcal R_1(P)]. } $$ ## Theorem F — Quotient product gate On the $2$-isogenous quotient: $$ \boxed{ [ \mathcal R_1\mathcal R_2 ] = [ 2MU-L^2 ]. } $$ --- # 31. Proof-safety boundary R12 does not claim: - that the eight R11 classes contain global nonboundary points; - that no stronger prime exists beyond the scanned interval; - that the quotient linear square gate alone is equivalent to full Perfect Cuboid closure; - that the known odd-index point is globally primitive; - that the new prime gives no useful information in every other quotient representation. The exact claims are restricted to the displayed group order, projected order, finite gate enumeration, torsion symmetry, and quotient identities. --- # 32. Proposed R13 program R13 should exploit the gate-swap quotient before continuing the prime search. ## R13-A — Rewrite the double cover over $\widehat E$ Use $$ \widehat E: \qquad V^2 = U^3-2MU^2+L^2U. $$ Impose the first quotient square gate: $$ \boxed{ Z^2 = 2MU-L^2. } $$ Determine the genus and an explicit model of this cover. ## R13-B — Recover the residual individual-gate lift Express one of: $$ [\mathcal R_1], \qquad [\mathcal R_2] $$ as a rational function on the first cover. Because the two are exchanged by the kernel involution, only one additional square condition is necessary. ## R13-C — Compare with historical $E_0$ square-value formulation R7 and G2E.5 already produced square-value conditions on normalized: $$ E_0. $$ Identify whether: $$ 2MU-L^2 $$ is birationally equivalent to one of those historical simple functions. This may collapse R12 back into an already studied rank-one elliptic square-value problem. ## R13-D — Symmetry-aware prime search For every prime candidate record: $$ \left( e, H_p \right), $$ where: $$ 2^e = \operatorname{ord}([m]G_0), $$ and: $$ H_p = [2^{e-1}][m]G_0. $$ Prioritize primes for which: $$ H_p\ne T_0. $$ ## R13-E — Quotient-level Mordell-Weil sieve Rather than tracking the two torsion cosets separately on: $$ E, $$ perform the finite-prime sieve directly on: $$ \widehat E. $$ This removes the exact gate-swap redundancy before enumeration. --- # 33. Final conclusion R12 finds the first arbitrary good prime beyond the R11 modulus threshold: $$ \boxed{ p=6{,}291{,}881. } $$ Its Sylow-$2$ projection gives a Mordell-Weil direction of order: $$ 2^{19}. $$ Yet it does not reduce the R11 coefficient set. R12 proves why. The half-order point is exactly: $$ T_0=(0,0), $$ and translation by: $$ T_0 $$ exchanges the two square gates. Therefore the extra coefficient bit is a gate-label symmetry bit rather than new arithmetic information. This simultaneously explains: - the persistent pairing of the two surviving torsion cosets; - the identical residue patterns under: $$ c\mapsto c+2^{18}; $$ - the failure of the new high-order prime to refine R11. Quotienting by this symmetry lands on the familiar $2$-isogenous: $$ E_0 $$ direction. There, the product-square obstruction becomes the linear gate: $$ \boxed{ 2MU-L^2 \in\mathbb Q^2. } $$ Thus the next reduction should be performed after the $T_0$ quotient, not by blindly extending the original coefficient sieve. This is the canonical starting point for R13.