bibkey: shi2026second
authors: Minjia Shi and Xuan Wang and Bouazzaoui Zakariae and Jon-Lark Kim and Patrick Sole
year: 2026
title: Second order Recurrences, quadratic number fields and cyclic codes
doi: 10.48550/arXiv.2603.25343
claim: Equality of the Pisano period modulo p and modulo p squared corresponds to a failure of p-rationality of the associated real quadratic field.
strata_touched:
  - D5/S1/Scale/Fibonacci
  - D5/S3/Arith/FibonacciRank
  - D5/S3/Arith/GoldenApparition
  - D5/S3/Arith/GoldenPrimeSplitting
license: citation-only
triage: anchor

Second order Recurrences, quadratic number fields and cyclic codes

Shi, Wang, Zakariae, Kim, and Sole relate second order recurrences to quadratic number fields and cyclic codes. The paper restates Wall’s question — whether there is a prime p with pi(p) = pi(p^2) for the Pisano period pi — records that no such prime is known below 10^14, and cites heuristics conjecturing infinitely many. For its generalized recurrences it makes the connection precise: equality of the periods modulo p and p^2 corresponds, under stated hypotheses, to failure of p-rationality of the associated real quadratic field.

This note is the literature anchor for the problem candidate Problems/wall-sun-sun-golden-unit-lift.md.

Search log

• 2026-08-18: Queried the arXiv Atom API for id_list=2603.25343. HTTP 200 with totalResults=1; the entry resolved to http://arxiv.org/abs/2603.25343v1, title Second order Recurrences, quadratic number fields and cyclic codes, authors Minjia Shi, Xuan Wang, Bouazzaoui Zakariae, Jon-Lark Kim, and Patrick Sole, published 2026-03-26, primary category math.NT. The API reported no arxiv:doi and no arxiv:journal_ref, so the arXiv-assigned DOI is used. The final author’s surname carries an acute accent in the API response; this note records it unaccented because the metadata line is restricted to ASCII.
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Wall’s question is a long-standing open problem restated by this paper rather than originated by it; no search was performed for a resolution after this arXiv version.

Verified locator

• arXiv: https://arxiv.org/abs/2603.25343
• DOI: https://doi.org/10.48550/arXiv.2603.25343

Fixed-golden arithmetic and geometric derivations

The following arguments concern the WSS golden-unit lift problem. They are repository derivations, rather than claims attributed to Shi and coauthors. Each argument retains its stated hypotheses, source roles and unresolved arithmetic boundary; none settles integer WSS existence.

PH. 固定黄金 Pell 曲线与初始 WSS 深度的高度预算

PH.1 原对象与精确秩成对分块

定义。 保留原域 $K=\mathbb{Q}(\sqrt{5})$、$\varphi =(1+\sqrt{5})/2$、$\psi =1-\varphi$，以及 $F_n=({\varphi }^n-{\psi }^n)/(\varphi -\psi )$、$L_n={\varphi }^n+{\psi }^n$。对素数 $p\ne 2,5$，令 $r(p)$ 为 Fibonacci 首现秩，并令

$$h_p=v_p(F_{r(p)})=v_p(F_{p-(5/p)}),q_p=F_{p-(5/p)}/p(\mathrm{mod}p).$$

第二个等式由经典秩界 $r(p)\mid p-(5/p)$ 和估值提升公式得到；其指标倍率与 $p$ 互素。因此 $q_p=0$ 当且仅当 $h_p\ge 2$。以下不假设 $h_p=1$。

对每个素数 $\ell \ge 7$，定义实际整数

$$A_{\ell }=F_{\ell },B_{\ell }=L_{\ell },M_{\ell }=A_{\ell }{B}_{\ell }=F_{2\ell },C_{\ell }=5A_{\ell }^2.$$

定理 PH1。 $A_{\ell },B_{\ell }$ 是互素奇数，且 $\gcd (M_{\ell },10)=1$。每个 $p\mid A_{\ell }$ 恰有 $r(p)=\ell$；每个 $p\mid B_{\ell }$ 恰有 $r(p)=2\ell$。对两种情形均有

$$v_p(M_{\ell })=h_p.$$

不同素数指标 $\ell$ 给出的 $M_{\ell }$ 两两互素。

证明。 黄金范数恒等式给出 $B_{\ell }^2-5A_{\ell }^2=-4$。Fibonacci 和 Lucas 的模二递推周期均为三，故两数为奇数。模五由 $L_n=2\cdot 3^n$ 和 $F_n=n{3}^{n-1}$ 得二者都与五互素。共同奇素因子会整除四，故两数互素。

由强整除性，$p\mid F_{2\ell }$ 蕴含 $r(p)\mid 2\ell$；秩一和二不可能，因为 $F_1=F_2=1$。所以秩为 $\ell$ 或 $2\ell$，而互素分解将两个秩准确分配给两因子。秩界还排除 $p=\ell$。若秩为 $\ell$，从 $F_{\ell }$ 到 $F_{2\ell }$ 的倍率二不改变奇素数估值；若秩为 $2\ell$，定义直接给出初始深度。最后，$\gcd (F_{2\ell },F_{2{\ell }^{\prime }})=F_2=1$ 对不同素数指标成立。

PH.2 几何等式与独立的高度假设

命题 PH2。 对全部上述指标，

$$B_{\ell }^2+4=C_{\ell }=5A_{\ell }^2,A_{\ell }<B_{\ell }<\sqrt{5}{A}_{\ell },$$

且 $(B_{\ell }^2,4,C_{\ell })$ 为两两互素的正整数加法三元组。记

$$R_{\ell }=\mathrm{rad}(4B_{\ell }^2{C}_{\ell }),$$

其中 $\mathrm{rad}$ 为不同素因子的乘积，则

$$R_{\ell }=10\mathrm{rad}(M_{\ell }),C_{\ell }>\sqrt{5}{M}_{\ell }.$$

证明。 第一式由 $\varphi \psi =-1$ 和奇指标得到。$L_{\ell }=F_{\ell }+2F_{\ell -1}>F_{\ell }$，而 Pell 等式给出右侧严格界。两数皆奇且 $B_{\ell }$ 与 $5A_{\ell }$ 互素，所以三元组两两互素；取根基得到最后的等式。由 $B_{\ell }<\sqrt{5}{A}_{\ell }$ 得高度的严格下界。

定义。 给定实数 $\kappa >1$，称这条素指标 Pell 族满足 $\mathrm{PH}(\kappa )$，若存在常数 $H>0$ 和 ${\ell }_0$，使全部素数 $\ell \ge \max (7,{\ell }_0)$ 满足

$$\boxed{C_{\ell }\le H{R}_{\ell }^{\kappa }.}$$

这是额外的统一算术假设。Pell 曲线方程本身不包含该不等式。对正互素三元组的有理数 abc 猜想会对每个 $\kappa >1$ 给出这个假设；以下条件性结论只需要所显示的限制族，而不假设一般数域 abc 或任意别的二次域结论。

PH.3 原始零集合的精确整数账目

定义。 在同一个 $M_{\ell }$ 的实际素因子上，令

$$U_{\ell }=\prod _{p\mid M_{\ell },h_p=1}p,W_{\ell }=\prod _{p\mid M_{\ell },h_p\ge 2}{p}^{h_p},E_{\ell }=\prod _{p\mid M_{\ell },h_p\ge 3}{p}^{h_p-2}.$$

空乘积为一。$W_{\ell }$ 恰好由原 WSS 初始商零集合、在精确秩 $\ell$ 和 $2\ell$ 上的实际贡献组成。它不是由后来的指标倍乘产生的平方因子。$U_{\ell }$ 只保留估值恰为一的素数，与按奇数估值定义的平方自由核不同。

定理 PH3。 无条件地有

$$\boxed{M_{\ell }=U_{\ell }{W}_{\ell },R_{\ell }^2{E}_{\ell }=100M_{\ell }{U}_{\ell },W_{\ell }{E}_{\ell }=100M_{\ell }^2/R_{\ell }^2.}$$

对任意整数 $d\ge 2$，进一步令 $W_{\ell ,\ge d}={\prod }_{p\mid M_{\ell },h_p\ge d}{p}^{h_p}$，则

$$\boxed{W_{\ell ,\ge d}^{d-1}\le {\left(M_{\ell }/\mathrm{rad}(M_{\ell })\right)}^d.}$$

证明。 逐素数比较指数。估值为一时，$\mathrm{rad}(M{)}^{2}E$ 与 $MU$ 的指数均为二；估值为二时均为二；估值 $h\ge 3$ 时均为 $h$。其余两个恒等式由分解与 PH2 得到。对于最后的不等式，若 $h\ge d$，有 $h(d-1)\le d(h-1)$；若 $h<d$，左侧该素数指数为零，右侧非负。将这些整数整除关系相乘即可。

PH.4 条件性异常深度质量界

定理 PH4。 假设 $\mathrm{PH}(\kappa )$，取其常数 $H,{\ell }_0$。令

$$K_{\kappa ,H}=100(H/\sqrt{5}{)}^{2/\kappa }.$$

则全部足够大的素数指标满足

$$\boxed{W_{\ell }\le K_{\kappa ,H}{M}_{\ell }^{2-2/\kappa }/E_{\ell },U_{\ell }\ge K_{\kappa ,H}^{-1}{E}_{\ell }{M}_{\ell }^{2/\kappa -1}.}$$

更一般地，对每个整数 $d\ge 2$，

$$\boxed{W_{\ell ,\ge d}\le {\left(10(H/\sqrt{5}{)}^{1/\kappa }\right)}^{d/(d-1)}{M}_{\ell }^{(1-1/\kappa )d/(d-1)}.}$$

证明。 高度假设给出 $R_{\ell }\ge (C_{\ell }/H{)}^{1/\kappa }$。代入 PH3，并使用 $C_{\ell }>\sqrt{5}{M}_{\ell }$，得到首个上界。用 $U_{\ell }=M_{\ell }/W_{\ell }$ 得下界。深度阈值版本使用 PH3 最后一式与同一个根基下界。常数对全部允许指标统一；未将逐点选择的常数当成统一高度界。

推论 PH5。 如果对每个 $\kappa >1$ 都有 $\mathrm{PH}(\kappa )$，则

$$\underset{\ell \to \infty ,\ell \mathrm{prime}}{\lim }\frac{{\sum }_{p\mid F_{2\ell },q_p=0}{h}_p\log p}{\log F_{2\ell }}=0.$$

证明。 分子恰为 $\log W_{\ell }\ge 0$。对每个固定 $\kappa >1$，PH4 将上极限控制为 $2-2/\kappa$；让 $\kappa$ 从上方趋于一。这里是按整数因子大小加权的质量比例，不是 WSS 素数在全部素数中的自然密度或计数密度，也未假设不同素数独立。

PH.5 每个精确秩通道的条件性非零见证

定理 PH6。 若对某个 $1<\kappa <2$ 有 $\mathrm{PH}(\kappa )$，则每个足够大的素数指标 $\ell$ 都有一个素数 $p$，满足

$$r(p)\in \{\ell ,2\ell \},q_p\ne 0.$$

若更强地 $1<\kappa <4/3$，则两个通道分别都有这样的素数：存在 $p_{\ell }\mid F_{\ell }$ 和 $s_{\ell }\mid L_{\ell }$，使

$$\boxed{r(p_{\ell })=\ell ,r(s_{\ell })=2\ell ,q_{p_{\ell }}\ne 0,q_{s_{\ell }}\ne 0.}$$

可以同时得到实际简单素因子乘积的下界

$$U(F_{\ell })\ge c_1{F}_{\ell }^{4/\kappa -3},U(L_{\ell })\ge c_2{L}_{\ell }^{4/\kappa -3},$$

其中 $U(X)={\prod }_{p\parallel X}p$，$c_1,c_2>0$ 与指标无关。

证明。 当 $\kappa <2$ 时，PH4 中 $U_{\ell }$ 的幂指数为正，所以最终 $U_{\ell }>1$。对逐通道结论，置 $\delta =2-2/\kappa$。每个通道的异常部分都不超过 $W_{\ell }\le K_{\kappa ,H}{M}_{\ell }^{\delta }$。PH2 给出 $M_{\ell }<\sqrt{5}{F}_{\ell }^2$ 及 $M_{\ell }<L_{\ell }^2$。因此分别除以异常部分后，下界幂指数均为 $1-2\delta =4/\kappa -3>0$。两个简单素因子乘积最终均大于一，再由 PH1 识别其精确秩和原商非零性。

推论。 在 PH6 的第二种假设下，这两族见证彼此无重复。Lucas 通道的每个见证均在原黄金域中分裂，并满足 $s_{\ell }\equiv 1(\mathrm{mod}2\ell )$。Fibonacci 通道的见证满足 $p_{\ell }\equiv 1(\mathrm{mod}4)$，但没有由此确定它分裂还是惰性。

证明。 不同指标的块互素，同一指标的两通道也互素。Lucas 素因子满足 $5F_{\ell }^2\equiv 4(\mathrm{mod}{s}_{\ell })$，其中 $F_{\ell }$ 可逆，故 $(5/s_{\ell })=1$；秩界给出显示的同余。Fibonacci 素因子满足 $L_{\ell }^2\equiv -4(\mathrm{mod}{p}_{\ell })$，故 $-1$ 为平方。存在某个简单素因子不等于存在惰性的简单素因子。

具体强度。 仅假设 $\mathrm{PH}(5/4)$，即可得到 $W_{\ell }\le K{M}_{\ell }^{2/5}$，以及两个通道各自的简单素因子乘积至少为常数乘其 $1/5$ 次幂。这些均为条件性结论。

PH.6 全异常覆盖会迫使怎样的高度反例

定义。 对上述互素三元组，令 ${\mathcal{Q}}_{\ell }=\log C_{\ell }/\log R_{\ell }$。 定理 PH7。 固定整数 $d\ge 2$。若一个无界素数指标子族的 $M_{\ell }$ 的全部素因子均有 $h_p\ge d$，则沿该子族

$$\mathrm{liminf}{\mathcal{Q}}_{\ell }\ge d.$$

若只要求 $F_{\ell }$ 的全部素因子有 $h_p\ge d$，或只要求 $L_{\ell }$ 的全部素因子有 $h_p\ge d$，则相应子族满足

$$\mathrm{liminf}{\mathcal{Q}}_{\ell }\ge \frac{2d}{d+1}.$$

证明。 全块条件给出 $\mathrm{rad}(M_{\ell })\le M_{\ell }^{1/d}$，所以 $R_{\ell }\le 10M_{\ell }^{1/d}$，而 $C_{\ell }>\sqrt{5}{M}_{\ell }$；取对数比值的下极限。

若只有 Fibonacci 通道满足条件，则 $R_{\ell }\le 10F_{\ell }^{1/d}{L}_{\ell }<10\sqrt{5}{F}_{\ell }^{1+1/d}$，而 $C_{\ell }=5F_{\ell }^2$。Lucas 通道的情形使用 $R_{\ell }\le 10F_{\ell }{L}_{\ell }^{1/d}<10L_{\ell }^{1+1/d}$ 与 $C_{\ell }=L_{\ell }^2+4$。两者均给出 $2/(1+1/d)$。

取 $d=2$，全部原始零因子覆盖成对块会迫使质量至少二，只覆盖一个通道会迫使质量至少 $4/3$。这证明为何 PH6 的两个阈值不同，但未构造任何这样的全异常子族。

PH.7 不能从这一条件证明中删除的内容

命题。 PH1–PH3 和 PH7 无需高度猜想；PH4–PH6 明确依赖 $\mathrm{PH}(\kappa )$。这些结论均允许原 WSS 零集合为空，也允许它非空。上述条件性非零见证不能推出存在某个 $q_p=0$。

证明。 零集合为空时所有 $W_{\ell }=E_{\ell }=1$，整数恒等式及质量上界仍成立；任何上界都不强制正贡献。若存在零，则它在相应块中的实际贡献受条件上界约束，但上界不指定或生成该素数。几何方程与统一高度不等式是不同命题，后者未由前者的代数恒等式或实嵌入的伸缩性推出。

边界实例。 指标换成任意合数而不移除旧素因子，会破坏 PH1 的原始深度解释：$v_{13}(F_7)=1$，$v_{13}(F_{91})=2$，而 $F_{14}/13\equiv 3(\mathrm{mod}13)$。这里的平方来自指标乘以十三，十三仍非 WSS。固定素数指标的两通道选择正是为了排除此类自动提升。

CG. 原黄金单位在连续、有限与整数几何中的同一缺陷

CG.1 固定代数环面与原初始商

定义。 令 $\mathcal{O}=\mathbb{Z}[\varphi ]$，${\varphi }^2=\varphi +1$，$\psi =1-\varphi$，$d=2\varphi -1=\sqrt{5}$，并令 $u=\varphi /\psi =-{\varphi }^2$。对 $p>5$ 素数，记 $\chi =(5/p)$、$N=p-\chi$、$h_p=v_p(F_N)$、$q_p=F_N/p(\mathrm{mod}p)$。对 $\mathcal{O}\otimes {\mathbb{Z}}_p$，用 $v_p(x)=\sup \{j:x\in p^j(\mathcal{O}\otimes {\mathbb{Z}}_p)\}$；分裂情形这是两个局部估值的最小值。

定义。 在 $\mathbb{Z}[1/10]$ 上，令 $T$ 是 $\mathcal{O}$ 的范数一群。其点以坐标写成 $a+b\varphi$，满足 $a^2+ab-b^2=1$。这是一个一维代数环面，$u\in T(\mathbb{Z}[1/10])$。它与实紧致二维环面 ${\mathbb{R}}^2/{\mathbb{Z}}^2$ 是不同对象。

命题。 $|T({\mathbb{F}}_p)|=N$，且对每个正整数 $n$，

$$u^n-1=d{F}_n{\psi }^{-n}.$$

特别地，$v_p(u^N-1)=h_p$，并有

$$\boxed{(u^N-1)/p\equiv \chi q_{p}d(\mathrm{mod}p).}$$

证明。 分裂时范数一群为 $\{(x,x^{-1}):x\in {\mathbb{F}}_p^{\times }\}$，阶为 $p-1$；惰性时为 ${\mathbb{F}}_{p^2}^{\times }$ 到 ${\mathbb{F}}_p^{\times }$ 的范数核，阶为 $p+1$。Binet 恒等式给出 $u^n-1=({\varphi }^n-{\psi }^n){\psi }^{-n}$。$d$ 和 $\psi$ 在 $p$ 处都是单位，得到估值。$N$ 偶数且黄金 Frobenius 给出 ${\varphi }^N\equiv \chi (\mathrm{mod}p)$，所以 ${\psi }^{-N}={\varphi }^N\equiv \chi$；将已整除的等式除以 $p$ 再降模即得末式。

CG.2 光滑切空间中恰有一个保周期提升

定理。 在模 $p^2$ 的范数一群中，固定 $u\mathrm{mod}p$，其全部提升恰可写成

$$u(1+pz),z\in \mathcal{O}/p\mathcal{O},\mathrm{Tr}(z)=0.$$

这样的提升有 $p$ 个，且其中恰有一个满足 $[u(1+pz){]}^N=1$。这不确定原来的 $u$ 是否就是该提升。

证明。 两个单位提升之比唯一写成 $1+pz$。其范数模 $p^2$ 为 $1+p\mathrm{Tr}(z)$。迹零空间在二维 ${\mathbb{F}}_p$ 代数中维数为一，因为 $\mathrm{Tr}(1)=2\ne 0$。写 $u^N=1+pb(\mathrm{mod}{p}^2)$，则 $b$ 迹零；二项式展开给出

$$[u(1+pz){]}^N=1+p(b+Nz)(\mathrm{mod}{p}^2).$$

因 $p\nmid N$，唯一解为 $z=-b/N$。原来的固定提升对应 $z=0$，它保周期当且仅当 $b=0$，亦即 $q_p=0$。将提升作为变量时的比例 $1/p$，不是固定黄金单位在变动素数上的频率定理。

CG.3 一维 p-进流与准确轨道闭包

定义。 令 ${\mathcal{O}}_p=\mathcal{O}\otimes {\mathbb{Z}}_p$，$T_1=\ker (T({\mathbb{Z}}_p)\to T({\mathbb{F}}_p))$。使用主单位上的收敛幂级数定义 $\log$ 与 $\exp$。

定理。 $\log$ 给出群同构 $T_1\cong p{\mathbb{Z}}_{p}d$。令 ${\beta }_p=\log (u^N)$，则 $v_p({\beta }_p)=h_p$。映射

$${\Phi }_p(t)=\exp (t{\beta }_p),t\in {\mathbb{Z}}_p$$

连续且解析，并在非负整数 $t$ 处等于 $u^{Nt}$。对 $s\ne t$，

$$\boxed{v_p({\Phi }_p(s)-{\Phi }_p(t))=h_p+v_p(s-t).}$$

此外，

$$\boxed{\overline{\{u^{Nn}:n\ge 0\}}=\exp (p^{h_p}{\mathbb{Z}}_{p}d),[T_1:\overline{\{u^{Nn}:n\ge 0\}}]=p^{h_p-1}.}$$

证明。 对 $z\in p{\mathcal{O}}_p$，对数级数首项的估值严格低于其余项；指数级数亦然。于是 $v_p(\log (1+z))=v_p(z)$，且指数与对数在主单位上互逆。范数的对数是迹，因此范数一对应迹零。${\mathcal{O}}_p$ 的迹零部分为 ${\mathbb{Z}}_{p}d$。将 ${\beta }_p$ 写为 $p^{h_p}cd$，其中 $c\in {\mathbb{Z}}_p^{\times }$，得到估值公式。非负整数在 ${\mathbb{Z}}_p$ 中稠密，所以 $\{n{\beta }_p:n\ge 0\}$ 的闭包为 $p^{h_p}{\mathbb{Z}}_{p}d$，指数映射与一维格指数给出结论。

推论。 WSS 条件等价于上述真实 p-进轨道闭包在 $T_1$ 中具有非平凡有限指数。局部流的完整描述以实际 $h_p$ 为参数，尚未决定变动素数 $p$ 上哪些 $h_p\ge 2$。该流的参数空间是 ${\mathbb{Z}}_p$，没有把它作为实连通时间区间到全不连通空间的非平凡连续路径。

CG.4 Fibonacci 多项式的单根与固定参数偏移

定义。 令 $U_0(X)=0,U_1(X)=1$、$V_0(X)=2,V_1(X)=X$，两列都满足 $P_{n+2}(X)=X{P}_{n+1}(X)+P_n(X)$。

命题。 对所有 $n\ge 0$，在 $\mathbb{Z}[X]$ 中有

$$(X^2+4)U_n^{\prime }(X)=n{V}_n(X)-X{U}_n(X).$$

证明。 在 $\mathbb{Q}(X,\sqrt{X^2+4})$ 中，取二次方程的两根 $\alpha ,\beta$，令 $D=\alpha -\beta$。有 $D^2=X^2+4$、$D^{\prime }=X/D$、${\alpha }^{\prime }=\alpha /D$、${\beta }^{\prime }=-\beta /D$。对 $U_n=({\alpha }^n-{\beta }^n)/D$ 求导，得到显示公式。两端为整数多项式，故该恒等式回到 $\mathbb{Z}[X]$。 定理。 对每个 $p>5$，

$$\boxed{U_N^{\prime }(1)\equiv -2/5\not\equiv 0(\mathrm{mod}p).}$$

在 $1+p{\mathbb{Z}}_p$ 中，$U_N$ 恰有一个根 $a_p$，并且

$$v_p(a_p-1)=h_p,(a_p-1)/p\equiv (5/2)q_p(\mathrm{mod}p).$$

证明。 $U_N(1)=F_N\equiv 0$、$V_N(1)=L_N\equiv 2\chi$，而 $N\equiv -\chi (\mathrm{mod}p)$。导数恒等式给出 $5U_N^{\prime }(1)\equiv -2$。Hensel 单根提升定理给出唯一根。对于 $x,y\in 1+p{\mathbb{Z}}_p$，多项式差商模 $p$ 恒为这个非零导数，所以 $v_p(U_N(x)-U_N(y))=v_p(x-y)$。取 $x=1,y=a_p$ 得估值等式。最后 Taylor 展开给出

$$U_N(1+pt)\equiv p(q_p-2t/5)(\mathrm{mod}{p}^2),$$

得到首位偏移。

实例。 $p=7$ 时 $N=8,q_7=3$，唯一参数提升为 $a_7\equiv 29(\mathrm{mod}49)$。这不使七成为 WSS 素数，因为原序列固定参数 $X=1$。WSS 判定是该固定参数是否已经落在根上至模 $p^2$，而不是根是否存在或是否光滑。

CG.5 真实三维黄金映射环面的同调接口

定义。 令

$$A=\left(\begin{array}{cc}1& 1\\ 1& 0\end{array}\right),B=A^2=\left(\begin{array}{cc}2& 1\\ 1& 1\end{array}\right),S=2A-I=\left(\begin{array}{cc}1& 2\\ 2& -1\end{array}\right).$$

$B$ 在实紧致环面 ${\mathbb{R}}^2/{\mathbb{Z}}^2$ 上定义自同构。对正偶数 $n$，令 $M_n$ 为 $B^n$ 的映射环面。

定理。 对每个正偶数 $n$，

$$B^n-I=F_{n}S{A}^n,$$

$$\boxed{H_1(M_n;\mathbb{Z})\cong \mathbb{Z}\oplus \mathbb{Z}/F_n\oplus \mathbb{Z}/(5F_n).}$$

因此，对 $n=p-(5/p)$、$p>5$，

$$\boxed{\mathrm{Tor}{H}_1(M_n;\mathbb{Z})[p^{\infty }]\cong (\mathbb{Z}/p^{h_p}{)}^2.}$$

证明。 $A^n=F_{n}A+F_{n-1}I$，且 $I-A=-A^{-1}$。偶数 $n$ 时，将两者相减得到 $A^n-A^{-n}=F_n(2A-I)$，再乘以 $A^n$。矩阵 $A^n$ 整数可逆，而 $S$ 的元素最大公约数为一、行列式为 $-5$，所以 Smith 对角为 $(1,5)$，返回矩阵的 Smith 对角为 $(F_n,5F_n)$。映射环面的基本群为 ${\mathbb{Z}}^2{⋊}_{B^n}\mathbb{Z}$，阿贝尔化为时间方向 $\mathbb{Z}$ 与 $B^n-I$ 的余核之直和。取 $p$-初等部分得到末式。

推论。 原 WSS 条件等价于这个指定三维覆盖的一阶同调中存在阶为 $p^2$ 的元素。等价式依赖随 $p$ 变化的同一个覆盖次数 $N$；它没有从实双曲伸缩或三维拓扑分类中得到额外的素数深度约束。

CG.6 算术微分与被迹删除的一阶信息

定义。 在 ${\mathcal{O}}_p$ 上令 ${\sigma }_p$ 为 Frobenius 提升：$\chi =1$ 时取恒等，$\chi =-1$ 时取黄金共轭。定义

$${\delta }_p(x)=\frac{{\sigma }_p(x)-x^p}{p}.$$

分子在 $p{\mathcal{O}}_p$ 中，所以此定义在除以 $p$ 后仍取值于 ${\mathcal{O}}_p$。

命题。 对任意 $x,y\in {\mathcal{O}}_p$，

$${\delta }_p(xy)=x^p{\delta }_p(y)+y^p{\delta }_p(x)+p{\delta }_p(x){\delta }_p(y).$$

并有

$$\boxed{{\delta }_p(\varphi )\equiv -(d/2){\varphi }^{\chi }{q}_p(\mathrm{mod}p).}$$

证明。 将 ${\sigma }_p(x)=x^p+p{\delta }_p(x)$ 代入乘法保持关系，展开即得第一式。由 $(L_N-2\chi )(L_N+2\chi )=5F_N^2$ 且第二因子模 $p$ 为单位，得到 $L_N\equiv 2\chi (\mathrm{mod}{p}^2)$。因此 ${\varphi }^N\equiv \chi +(d/2)p{q}_p(\mathrm{mod}{p}^2)$。若 $\chi =1$，乘以 $\varphi$；若 $\chi =-1$，乘以 ${\varphi }^{-1}$，并使用 $-{\varphi }^{-1}=\psi$。两种情形均得到显示公式。

推论。 $q_p=0$ 当且仅当 ${\delta }_p(\varphi )=0(\mathrm{mod}p)$。另一方面，

$$v_p(L_N-2\chi )=2h_p.$$

所以普通返回迹在模 $p^2$ 上对所有这些素数都返回，已经删除了原缺陷的一阶信息。算术微分的乘积律与实导数不同，不能未证明地移入实几何流的微分方程或估计。

CG.7 切空间读数、圆周等分布与精确零点

定义。 令 $c_p\in \{0,\dots ,p-1\}$ 为 $\chi q_p$ 的整数代表，$w_p=c_p/p(\mathrm{mod}\mathbb{Z})\in \mathbb{R}/\mathbb{Z}$。以 $\mathbb{Z}d$ 为 $T$ 的迹零 Lie 格，CG.1 表明这正是由 $u^N\mathrm{mod}{p}^2$ 的切空间缺陷除以 $p^2$ 得到的圆周读数。$u$ 不是扭元素，因此在一维环面的几何泛纤维中生成 Zariski 稠密子群。

命题。 $w_p=0$ 当且仅当 $p$ 为原 WSS 素数。圆周上 Haar 等分布的性质本身，不推出这种零点至少出现一次。

证明。 第一项由 $\chi$ 是单位直接得到。为说明后一项，按递增顺序记素数为 $p_j$，取任意无理实数 $\theta$，令

$$b_{p_j}=1+\lfloor (p_j-1)\{j\theta \}\rfloor .$$

有 $1\le b_{p_j}\le p_j-1$，且 $|b_{p_j}/p_j-\{j\theta \}|\le 2/p_j\to 0$。无理旋转的等分布和连续函数在紧圆周上的一致连续性，说明 $b_{p_j}/p_j$ 仍等分布，然而它永不为零。这只是对一般推理的反例，没有把这列人工余数当成 Fibonacci 余数。

恒等式。 令 $Z(X)=\#\{5<p\le X:p\text{ 素且 }{q}_p=0\}$。字符正交性给出

$$Z(X)=\sum _{5<p\le X}\frac{1}{p}+\sum _{5<p\le X}\frac{1}{p}\sum _{a=1}^{p-1}\exp (2\pi ia{c}_p/p).$$

若第二项能被独立证明为 $o(\log \log X)$，则 Mertens 素数调和和公式将给出 $Z(X)\sim \log \log X$。本节没有证明这个误差估计。固定频率的宏观等分布只控制每个固定 $a$ 的平均，不能直接替代同时到 $a=p-1$ 的精确零点尺度。

CG.8 不同接口的共同假设边界

命题。 CG.1–CG.7 保留原黄金单位，但没有强制任何变动素数满足 $h_p\ge 2$。其中局部流、唯一提升、根位置和三维同调都容许 $h_p=1$ 与 $h_p\ge 2$；将它们互相代换不产生新的素数存在结论。

证明。 CG.2 在同一个约化点上同时构造一个保周期提升及 $p-1$ 个不保周期提升。CG.3 的公式以原 $h_p$ 为输入，CG.4 将同一个深度记为固定参数到唯一根的距离，CG.5 将它记为同调挠阶，CG.6 将它记为算术微分的零阶。上述等价关系均不排除原固定点始终选中非零的一阶缺陷。CG.7 又排除了仅由宏观均匀性推出精确零点的推理。因此跨素数的深度控制仍需独立命题。

FPD. Recurrence-level square transport and maximal-prime descent

FPD.1 Original objects and the square-factor threshold

Use the original Fibonacci recurrence F_0=0, F_1=1 and F_(n+2)=F_n+F_(n+1). A positive integer M is powerful when every prime p dividing M also satisfies p^2|M. Thus one is powerful. Put E={1,2,6,12}. For an odd prime p different from five, write chi(p)=(p/5)=(5/p), and N(p)=p-chi(p). The standard WSS square condition is p^2|F_(N(p)). Index primes and value-primes are different variables.

Theorem FPD1. （文献：T. Lengyel, Fibonacci Quart. 33 (1995) 234–239；L. A. Medina, E. Rowland, Fibonacci Quart. 53 (2015) 265–271, Theorem 1.4） For every prime p, positive n, and natural k, if p|F_n, then

$$\boxed{p^2\mid F_{nk}⟺p^2\mid F_n\text{or}p\mid k.}$$

This includes p=2 and k=0. It does not assume any initial valuation.

Proof from the recurrence. In R=Z/(p^2), put f=F_n, g=F_(n+1) and c=F_(n-1). Then g=c+f and f^2=0. The Fibonacci addition formulas imply, by simultaneous induction on k>=0,

$$F_{n(k+1)}=(k+1)f{g}^k,F_{n(k+1)+1}=g^{k+1}\text{in }R.$$

For the first induction step the new coordinate is (k+1)fg^k c+g^(k+1)f. Replacing c by g-f gives (k+2)fg^(k+1)-(k+1)g^k f^2, as required. The second new coordinate is (k+1)f^2g^k+g^(k+2), also as required. The initial pair is (f,g). Consecutive Fibonacci coprimality shows gcd(p,g)=1. Consequently, for positive k, p^2|F_(nk) exactly when p^2|k F_n. Write F_n=p u and cancel one p in the integers. Primality gives p|ku exactly when p|k or p|u. The latter is equivalent to p^2|F_n. At k=0 both sides are true.

In particular, if p does not divide k, square divisibility is exactly preserved between F_n and F_(nk). If p divides F_n simply and p does not divide k, it still divides F_(nk) simply. A square produced solely by multiplying the index by p is not an initial WSS exception. For example F_7=13, 13^2|F_91, but F_14/13=29=3 modulo thirteen.

FPD.2 The small-support case is completely eliminated

Lemma FPD2. If m>0 has no prime divisor greater than five, then

$$\boxed{F_m\text{ powerful}⟺m\in E.}$$

Proof. The exact small values are

$$F_5=5,F_8=3\cdot 7,F_9=2\cdot 17,F_{25}=5^2\cdot 3001.$$

The number 3001 is prime, by trial division through its square root, which is less than55. If 25|m, the factor3001 remains simple in F_m by FPD1, since it cannot divide this small-support index. Thus 25 does not divide m. If 5|m, write m=5k. The preceding exclusion makes 5 not divide k; FPD1 at F_5 then gives another simple factor, a contradiction. Likewise 9|m preserves the simple prime17 from F_9, and 8|m preserves seven from F_8. Hence a powerful value forces 5 not dividing m, 9 not dividing m and 8 not dividing m. Unique factorization now gives m|12. The six positive divisors of12 are1,2,3,4,6,12. Indices3 and4 give the simple values2 and3, and the four remaining values are1,1,8,144, all powerful. This proves both directions without an external valuation formula or a classification of perfect powers.

FPD.3 Prime-index factors escape the index support

Lemma FPD3. Suppose ell>=7 is prime and p is a prime divisor of F_ell. Then the least positive Fibonacci zero index for p is ell, and p>ell.

Proof. If p|F_j, strong divisibility gives p|gcd(F_ell,F_j)=F_(gcd(ell,j)). Unless ell|j, primality of ell makes that greatest common divisor one, impossible for p. Thus every zero index is a multiple of ell. The initial zero at ell proves exact rank.

The cases p=2,5 are excluded by their zeros at indices3,5 respectively. The existing golden Frobenius/rank theorem gives ell|p-1 or ell|p+1. In the first case p>ell immediately. In the second, if p<=ell, the only positive multiple of ell that can equal p+1 is ell itself. This gives p+1=ell, impossible because both primes are odd. Thus p>ell in both cases. No lower bound on the value-primes is assumed as an input.

FPD.4 Constructive descent to a maximal prime-index WSS block

Theorem FPD4. （方法先例：N. Robbins, Fibonacci Quart. 21 (1983) 215–218；V. Andrejić, Univ. Beograd Publ. Elektrotehn. Fak. Ser. Mat. 17 (2006) 38–44） If m>0, m is outside E, and F_m is powerful, then there exists a prime ell>=7 such that

$$\boxed{\ell \mid m,\forall q\text{ prime},q\mid m⟹q\le \ell ,F_{\ell }\text{ is powerful}.}$$

Moreover every prime p dividing F_ell satisfies

$$\boxed{p>\ell ,p^2\mid F_{N(p)}.}$$

Proof. By FPD2, m has a prime factor greater than five. Choose its largest prime factor ell. It is at least seven. For each prime p|F_ell, FPD3 gives p>ell, so p does not divide m. Write m=ell k. Fibonacci divisibility gives p|F_m; powerfulness gives p^2|F_m. In FPD1 the alternative p|k is impossible because k|m. Therefore p^2|F_ell. This proves powerfulness of the entire prime-index block.

FPD3 gives its actual entry point ell. Apply the golden rank bound to obtain ell|N(p), and then F_ell|F_(N(p)). Thus the square divisibility also holds at p’s own signed Frobenius index. Since p>ell>=7, the small and ramified primes are automatically absent from this endpoint. The theorem does not assert that the index-prime ell is WSS.

FPD.5 Consequences and the remaining open arithmetic

Corollary. The following two assertions are equivalent:

$$\begin{array}{rl}& \forall m>0,F_m\text{ powerful}⟹m\in E;\\ & \forall \ell \ge 7\text{ prime},F_{\ell }\text{ is not powerful}.\end{array}$$

If the first assertion fails, its least counterexample index is prime and at least seven.

Proof. The forward implication specializes to prime indices. For the reverse, a nonclassical counterexample would descend by FPD4 to a prime-index counterexample. For the least one, the descending prime ell satisfies ell<=m and is itself a counterexample, so minimality forces ell=m. Neither assertion is established by this equivalence.

For a prime index ell>=7, the existence of a simple prime divisor of F_ell is equivalent to the existence of a non-WSS prime of exact rank ell. Indeed, FPD3 supplies the exact rank, and ell|N(p) with p not dividing N(p)/ell allows FPD1 to identify the two square-divisibility conditions. Thus a remaining target is an independent simple-factor existence theorem at prime indices. A primitive-prime theorem only supplies a prime with that rank and does not assert exponent one.

Absence of all WSS primes would imply the first classification above, since every nontrivial prime-index Fibonacci value has a prime divisor. The classification’s failure would produce WSS primes by FPD4. The converse implication from a single WSS prime to failure of the powerful classification is not asserted: the other factors of its rank block may still be simple.

A prospective stronger lifting theorem is preservation of the entire p-valuation under a multiplier coprime to p. The square-zero argument already identifies its mechanism: at an actual initial depth e>=1, F_n^2 vanishes modulo p^(e+1). Proving the corresponding depth theorem requires retaining the nondivisibility at p^(e+1), the nonzero index and the coprime multiplier. The present public transfer theorem certifies the square threshold only; it does not silently assert that extension.

FPD.6 Source roles and mathematical scope

FPD1 is a recurrence proof of a classical special case of the Fibonacci valuation theory of Lengyel, recorded in Medina and Rowland, p-regularity of the p-adic valuation of the Fibonacci sequence, The Fibonacci Quarterly53(2015),265-271, Theorem1.4, https://arxiv.org/abs/0910.2907 . The source proof here does not use that theorem as an axiom or a lifting hypothesis.

Bates, Jesubalan, Lee, Lu and Shim, Powerful Fibonacci polynomials over finite fields, arXiv:2601.02664v1 (2026), Theorem1.2 and Section2.1, https://arxiv.org/html/2601.02664v1 , classify a polynomial problem over finite fields and explicitly distinguish the conditional integer powerful-number problem. A finite-field polynomial multiplicity does not determine divisibility of an evaluated integer by p squared. Their theorem is contextual prior work, not an input to FPD1-FPD4.

The full classification and WSS existence remain open targets of this line, not conclusions of FPD4.

SJC. Predictive completion of the original golden orbit at square precision

SJC.1 What is being completed

Definition. Retain K=Q(sqrt(5)), O=Z[phi], phi^2=phi+1, conjugation bar(phi)=1-phi, and delta=2phi-1. Put v=phi^2=1+phi, so N(v)=1 and Tr(v)=3. For a prime p>5 write chi=(5/p), N=p-chi, q_p=F_N/p modulo p, and h_p=v_p(F_N). Division by p in a residue formula is performed before reduction; denominators prime to p are then inverted in the residue ring.

The prediction task consists of the actual update x mapping to vx and specified observations of its state. Coarse reduction modulo p, full state modulo p^2, and delayed scalar traces are different observation maps. Completeness of an observation means sufficiency for the specified outputs and their future updates; it is not an assertion about incompleteness of arithmetic or of set theory.

Proposition SJC1. Z with its ordinary absolute-value metric is complete, whereas its p-adic completion is Z_p. The projection O/p^2O -> O/pO is a noninjective ring map and cannot have a unital ring section. In particular its source cannot be replaced as a ring by (O/pO)[epsilon]/(epsilon^2).

Proof. An ordinary Cauchy sequence of integers is eventually constant, by applying the Cauchy condition with distance less than one. In the p-adic metric, compatible residues modulo p^a define Cauchy classes and conversely, giving the inverse-limit completion Z_p. For example the partial geometric sums 1+p+…+p^n converge p-adically to 1/(1-p), which is not an integer for p>2. The square-precision ring has characteristic p^2, while the residue ring and its dual numbers have characteristic p. A unital section would send p1=0 to p1!=0. This proves the last two assertions. A multiplicative group section constructed below does not contradict the absence of a ring section.

The repository’s CONTEXTUAL_SPACETIME_ARITHMETIC_ML.md, Sections 1-3, makes the relevant distinction precise: a quotient may have closed updates yet fail a required query; exact task descent requires that all such queries and successor labels be constant on its fibers. The current draft SYMPLECTIC_PREDICTIVE_COMPLETION.md, Section 2, treats related real linear dynamics, with additional hypotheses for positive-energy and symplectic conclusions. These frameworks provide a method of selecting state, not an independent WSS prime-distribution theorem.

SJC.2 The exact square-precision state and drift

Definition. For a>=1 let T^(a) be the group of norm-one units of O/p^aO. Put r=ord_(T^(1))(v mod p), C=, and X_2 the inverse image of C in T^(2). The symbol r here is the period of the norm-one unit phi^2, not the Fibonacci zero rank rho(p). One has r|N, p not dividing r, and r>=3. The first fact follows from the split/inert order N of T^(1). The cases r=1,2 are excluded by v-1=phi and v+1=2+phi, whose norms are -1 and 5.

Theorem SJC2. There is a unique multiplicative section s:C -> X_2 whose image consists of r-torsion. Put xi=s(v mod p). Every state of X_2 has a unique expression

$$x={\xi }^j(1+pz\delta ),(j,z)\in \mathbb{Z}/r\mathbb{Z}\times {\mathbb{F}}_p.$$

In these coordinates the actual update by v is

$$\boxed{(j,z)⟼(j+1,z-q_p).}\text{\hspace{1em}(SJC1)}$$

Thus X_2 has p disjoint cycles of length r if q_p=0, and one cycle of length rp otherwise. The orbit of the specified initial state one has length r or rp respectively.

Proof. The kernel of norm-one reduction consists of 1+pw with Tr(w)=0 modulo p. Since Tr(1)=2 and delta^2=5, its elements are exactly 1+pz*delta with z in F_p. Multiplication adds z. Reduction is surjective: for any unit lift of a norm-one residue, its norm is 1+pc, and multiplication by the scalar 1-pc/2 corrects the norm modulo p^2.

For a lift x of an element of C, write x^r=1+pbdelta. Multiplication by 1-pr^(-1)b*delta produces an r-torsion lift. It is unique, because taking the rth power is invertible on the order-p kernel. Products of unique torsion lifts are the unique torsion lift of the product, proving the group section. This also gives the unique displayed coordinates.

The golden identity L_N^2-5F_N^2=4, together with L_N=2chi modulo p, gives L_N=2chi modulo p^2: the other factor L_N+2chi is a unit. Therefore

$${\varphi }^N\equiv \chi +\frac{1}{2}p{q}_p\delta (\mathrm{mod}{p}^{2}O),v^N\equiv 1+p\chi q_p\delta (\mathrm{mod}{p}^{2}O).$$

Write v=xi(1+pcdelta). Since xi^N=1 and N=-chi modulo p, comparison of v^N gives Nc=chiq_p, hence c=-q_p. This proves SJC1. A return after t steps requires r|t and tq_p=0 modulo p. Since p is coprime to r, the asserted cycle lengths and counts follow.

This is the fixed-golden specialization of the square-level Lie-kernel mechanism in Katz, Wieferich past and future, Section 4, the exact sequence on printed page 3. The explicit coordinates and task interpretation above do not make Katz’s cross-prime equidistribution conjecture a theorem.

SJC.3 A cycle obstruction that rephasing cannot remove

Theorem SJC3. For any prescribed gamma_j in F_p on j in Z/rZ, consider T_gamma(j,z)=(j+1,z+gamma_j). A fiberwise translation z’=z+f(j) changes the drift to

$${\gamma }_j^{\prime }={\gamma }_j+f(j+1)-f(j).$$

The sum H=sum_j gamma_j is invariant under every such change. There is a rephasing with all gamma’_j=0 if and only if H=0; when it exists, f is unique up to an additive constant. For the actual golden update, H=-r*q_p, so complete removal of the drift is equivalent to q_p=0.

Proof. Substitute z=z’-f(j) in the update. Summing cancels every f term. If H=0, choose f(0) arbitrarily and define f(j+1)=f(j)-gamma_j; the closing equation at r holds exactly because the sum is zero. Conversely zero drift has zero sum. Two solutions have zero successive difference and thus differ by a constant. Since p does not divide r, the actual sum vanishes precisely with q_p.

The obstruction is the accumulated drift of a Z-action over a finite phase cycle, or equivalently the circulation of a labeled cycle graph. It is not being identified with H^1 of the finite group C_r acting trivially on F_p. In that finite-group cohomology, p not dividing r would give a different vanishing statement.

Proposition SJC4. All lifts v_c=xi(1+pc*delta), c in F_p, have the same complete history after reduction modulo p. Their refined return behaviors differ. The original integer recurrence selects exactly one of them by the fixed polynomial equation v_c^2-3v_c+1=0, and

$$\boxed{q_p=\frac{\mathrm{Tr}(\xi )-3}{5p}(\mathrm{mod}p).}\text{\hspace{1em}(SJC2)}$$

Here Tr(xi)-3 is divisible by p in Z/p^2Z. Divide by p first, then multiply by the inverse of five modulo p; no integer divisibility by five is asserted for an arbitrary representative of the trace.

Proof. Coarse equality persists under every power because all lifts reduce to v. For norm-one xi, its inverse is its conjugate and xi^2-Tr(xi)xi+1=0. In the kernel direction, direct first-order expansion gives

$$v_c^2-3v_c+1\equiv \xi (\mathrm{Tr}(\xi )-3+5pc)(\mathrm{mod}{p}^{2}O).$$

Indeed modulo p, 2xi-3=delta and delta^2=5. Since xi and five are units, this has exactly one solution c. SJC2 already identifies the original solution as c=-q_p, proving SJC2. The uniqueness shows why choosing the torsion lift c=0 does not construct a WSS example: it must also satisfy the fixed golden polynomial. The other lifts are comparison states, not alternative realizations of that same fixed recurrence.

SJC.4 Two exact delayed traces recover the fine state

Theorem SJC5. Let R=Z/p^aZ for a>=1 and p>5. For x=A+Bphi in O/p^aO, write y_0=Tr(x) and y_1=Tr(vx). Then

$$\left(\begin{array}{c}{y}_0\\ y_1\end{array}\right)=\left(\begin{array}{cc}2& 1\\ 3& 4\end{array}\right)\left(\begin{array}{c}A\\ B\end{array}\right),\boxed{A=(4y_0-y_1)/5,B=(-3y_0+2y_1)/5.}\text{\hspace{1em}(SJC3)}$$

Consequently, equality of all future exact trace observations under multiplication by v is already decided by these two observations and is equality of the state itself. On X_2 every one of its rp states is distinguishable by this two-observation experiment, independently of which drift branch the fixed v occupies.

Proof. Multiply (1+phi)(A+Bphi)=(A+B)+(A+2B)phi and use Tr(C+Dphi)=2C+D. The determinant is five, a unit in R, proving the inverse and the claim about future observations. This is a fixed linear observability computation over the actual finite ring; replacing it by an approximate real trace is not part of the assertion.

Corollary. At the phase return r,

$$\mathrm{Tr}(v^r)=2(\mathrm{mod}{p}^2),\boxed{\frac{L_{2r+2}-3}{p}=-5r{q}_p(\mathrm{mod}p).}\text{\hspace{1em}(SJC4)}$$

Thus the return trace alone deletes the first-order defect, while the next exact trace recovers it.

Proof. SJC1 gives v^r=1-prq_pdelta modulo p^2. Its trace is two since Tr(delta)=0. Multiplying by v and using Tr(v)=3 and Tr(v delta)=5 gives the second formula. In the integers Tr(v^(r+1))=L_(2r+2), so the displayed division is legitimate. This does not supply q_p independently of the exact integer traces; it specifies which observation retains it.

The task-quotient principle in CONTEXTUAL_SPACETIME_ARITHMETIC_ML.md, Theorems 2.2 and 3.2, applies literally: coarse reduction is closed for the coarse task but fails this refined task. This is a different kind of information loss from HCR’s non-generator auxiliary Hecke observation, where the exact kernel is a relative field trace. Both require retaining the actual joint observation map.

SJC.5 All p-adic levels retain a single drift parameter

Theorem SJC6. In O_p=O tensor Z_p there is a unique norm-one torsion lift xi of v modulo p with order r. There is a unique a_p in Z_p such that

$$v=\xi \exp (p{a}_p\delta ),v_p(a_p)=h_p-1,a_p=-q_p(\mathrm{mod}p).$$

At precision p^s the inverse image of C has coordinates

$$\mathbb{Z}/r\mathbb{Z}\times \mathbb{Z}/p^{s-1}\mathbb{Z},(j,z)⟼(j+1,z+a_p).$$

It has exactly p^min(s-1,h_p-1) cycles, each of length r*p^max(s-h_p,0). At s=1 the second coordinate is the singleton group.

Proof. Simple-root lifting constructs the compatible torsion lift; the norm is one by uniqueness. The convergent logarithm and exponential are inverse valuation-preserving maps between 1+pO_p and pO_p. This follows from their power series for odd p, since the linear term has strictly smaller valuation than the higher terms. Norm one corresponds to trace zero, which is the free line Z_pdelta. Hence log(v/xi)=pa_pdelta with a unique a_p.

Using phi^N=(L_N+F_N delta)/2 and the norm identity, one has v_p(phi^N-chi)=h_p in every completion above p. The factor phi^N+chi is a unit, giving v_p(v^N-1)=h_p. Since N is a p-adic unit and xi^N=1, this is the valuation of log(v/xi), proving the depth of a_p. Reduction to the first layer recovers -q_p. Additivity of logarithms gives the translation action. A return time must be divisible by r and make t*a_p zero modulo p^(s-1); these coprime requirements give the lengths and counts.

The integer time is common to all selected primes. Combining their local states gives the image of that one time parameter, not every point of their Cartesian product. The exact gcd correction in ROC4-ROC5 is an existing arithmetic instance of this joint-compatibility requirement. Neither local cycle counts nor their product imply that the actual fixed drift vanishes at a new prime.

SJC.6 The real symplectic realization has indefinite energy

Theorem SJC7. In the coefficient basis (1,phi), multiplication by phi and v has matrices

$$A=\left(\begin{array}{cc}0& 1\\ 1& 1\end{array}\right),B=A^2=\left(\begin{array}{cc}1& 1\\ 1& 2\end{array}\right).$$

Put S=2A-I, J=((0,1),(-1,0)), and c=2log(phi)/sqrt(5). Then

$$S^2=5I,B^{T}JB=J,B=\exp (cS).$$

The real flow exp(tcS) is Hamiltonian with quadratic Hessian -JcS, which is indefinite. No positive-definite quadratic energy is preserved by B.

Proof. The matrix identities follow by multiplication. The eigenvalues of S are sqrt(5) and -sqrt(5), and exp(cS) has eigenvalues phi^2 and phi^(-2) on the same eigenlines as B, proving the exponential identity. The matrix cS has trace zero and is Hamiltonian in dimension two; -JcS is symmetric with eigenvalues 2log(phi) and -2log(phi), so the energy is indefinite. If a positive-definite Q satisfied B^TQB=Q, its value on a real eigenvector of B with eigenvalue phi^2>1 would both remain fixed and multiply by phi^4, a contradiction.

The positive-energy hypothesis in the draft SYMPLECTIC_PREDICTIVE_COMPLETION.md, Section 2.2, is therefore unavailable for this real golden flow. Its automatic symplectic compression and normalized Gibbs conclusions cannot be applied merely because B is symplectic. The basis here is the coordinate swap of CG.5, with the same arithmetic map. Real interpolation adds a legitimate geometric realization but does not alter any p-adic return.

SJC.7 A global cyclotomic norm of the canonical torsion trace

Theorem SJC8. Let r be the norm-one period in SJC.2. For a primitive rth root of unity zeta_r let Psi_r be the minimal polynomial of zeta_r+zeta_r^(-1), and set G_r=Psi_r(3). Then G_r is a positive integer and

$$\boxed{v_p(G_r)=h_p,1<G_r<5^{\phi (r)/2}.}\text{\hspace{1em}(SJC5)}$$

In particular WSS at p forces p^2|G_r and p^2<5^(varphi(r)/2). This r divides p-chi and is prime to p; it is distinct from the p-power character order used in HCR and SGN.

Proof. Because r>=3, every real conjugate 2cos(2pi j/r) lies strictly between -2 and 2. Thus every factor of Psi_r(3) lies strictly between one and five, proving positivity and the bound.

Since p=+/-1 modulo r and p does not divide r, p splits completely in the real cyclotomic field, and the conjugate torsion traces are distinct modulo p. Distinctness follows directly: two rth roots with the same sum with their inverse must agree up to inversion, since both solve X^2-tX+1. The canonical torsion lift xi gives one such trace alpha=xi+xi^(-1) with alpha=3 modulo p. It is the unique conjugate with this residue. All other factors of Psi_r(3) are p-adic units.

Finally,

$$3-\alpha =(v-\xi )(1-(v\xi {)}^{-1}).$$

The second factor is a unit because the residue of xi has order r>2. The first factor has valuation h_p by SJC6. Thus the valuation of the entire norm is exactly h_p. At a split p the two golden embeddings exchange xi and xi^(-1), which have the same alpha; they do not create a second norm factor.

Corollary. Conversely, for any r>=3 and prime p>5 with p not dividing r, p^2|Psi_r(3) implies that the order of v modulo p is r and that p is WSS.

Proof. A zero Psi_r(3) modulo p comes from a primitive rth root whose trace is three, since reduction of rth roots is separable. Its two roots solve X^2-3X+1, the same roots as v and v^(-1) in the quadratic residue algebra. Hence v has order r. Apply SJC8 to identify the full valuation with h_p.

The use of cyclotomic norms and their exact entry-point valuations is classical for regular Lucas sequences. Ross, Shen and Cai, The p-adic Valuations of Mobius Duals of Lucas Sequences, arXiv:2512.03481v1, Introduction and Theorem 2.2(c), explicitly credit Carmichael’s regular case and extend it to irregular sequences. SJC8 is proved above using the canonical torsion lift; neither the norm scale nor its valuation role is claimed as a new arithmetic discovery. The source does not prove WSS existence in the fixed golden field.

SJC.8 Research consequence and limits

Proposition. The useful transfer from the repository’s spacetime model is task-relative completion: retain phase, the square-level Lie coordinate, the fixed update, and their joint compatibility. It identifies the original WSS defect with a rephasing-invariant cycle displacement and gives a sufficient two-trace observation. It does not make the integer formulation inadequate or establish a new actual WSS prime family.

Proof. SJC1-SJC4 construct the extra coordinate and prove the fixed polynomial determines its drift. SJC3 excludes removing a nonzero cycle displacement by a change of coordinates. SJC5 gives exact reconstruction from two required observations, while SJC7 shows that a positive real energy cannot be assumed. SJC8 replaces the same unknown valuation by a specified global norm without independently bounding it. Thus a new prime-family decision still needs an independent constraint on the fixed torsion trace, on G_r, or on their cross-prime distribution. Arbitrarily choosing another lift, treating local states as independent worlds, or substituting a dual-number ring for characteristic p^2 changes the problem instead of proving that constraint.

TBN. A fixed-golden split tower and simultaneous initial-depth balances

TBN.1 Calibration of a quantum-information-derived field construction

Definition. Keep the original field K=Q(sqrt(5)), O=Z[phi], phi^2=phi+1, v=phi^2, and the original Fibonacci and Lucas sequences. For an integer d>=4 let D(d) be the squarefree part of (d+1)(d-3). The dimensions belonging to the fixed golden field are exactly

$$d=L_{2n}+1=v^n+v^{-n}+1,n\ge 1.$$

Proof. D(d)=5 means (d-1)^2-5b^2=4 for an integer b>0. The two coordinates have the same parity, and ((d-1)+b sqrt(5))/2 is a totally positive norm-one unit greater than one in O. The golden unit classification gives v^n for a unique positive n. Conversely the Pell identity for L_(2n),F_(2n) gives the displayed square class.

Theorem TBN1. Among square integers d>=4, the equation D(d)=5 holds only at d=4. In particular D(p^(2a))!=5 for every odd prime p and positive a.

Proof. Write d=y^2=L_(2n)+1. If n is even, L_(2n)=L_n^2-2, so L_n^2-y^2=1. The positive factors L_n-y and L_n+y would both be one, forcing y=0, a contradiction. If n is odd, y^2-L_n^2=3, so (y-L_n)(y+L_n)=3. The only positive factorization gives y=2,L_n=1, hence n=1. The value d=4 indeed has square class five. The theorem only excludes square seeds; it makes no claim for all nonsquare perfect powers or all mixed seeds.

Gary McConnell, Some new infinite families of non-p-rational real quadratic fields, arXiv:2406.14632v1, Theorems 2.1-2.2, constructs fields while D varies, using ideas originating in SIC-POVM quantum information. Its definitions (2.1), (3.2), and Lemma 3.3 describe the associated dimension towers. The argument above enforces the fixed-field condition. In particular the pure p^2 seed in that construction cannot yield an original golden WSS example. This is a restriction on that construction, not an exclusion of p from the original WSS set.

TBN.2 A companion to the existing GP3 blocks

Definition. For j>=1 put n_j=3^j, r_j=3^(j+1), x_j=L_(n_j), and

$$B_j=L_{2n_j}+1=x_j^2+3.$$

The notation B_j is an integer block, not the cyclotomic coefficient field B of SGN. The existing PERIODIC_TREE.md, GP3, uses the different blocks C_j=x_j^2+1. Its results are retained and are not recounted as new theorems.

Theorem TBN2. The following exact identities hold:

$$L_{3n_j}=L_{n_j}{B}_j,B_{j+1}=B_j^3-3B_j^2+3,B_1=19,$$

$$B_j={\Psi }_{r_j}(3),B_j\equiv 19(\mathrm{mod}80),v_3(B_j-1)=j+1,$$

$$\frac{B_j-1}{3^{j+1}}\equiv (-1{)}^j(\mathrm{mod}3).\text{\hspace{1em}(TBN1)}$$

Here Psi_r is the real cyclotomic polynomial from SJC. Every B_j is odd, greater than one, and nonsquare.

Proof. Since n_j is odd, the Lucas triple identity is L_(3n)=L_n^3+3L_n; division gives the first formula. Substituting x_(j+1)=x_j^3+3x_j into x_(j+1)^2+3 gives the displayed cubic in B_j. For r=3^(j+1), the cyclotomic polynomial is 1+Z^(3^j)+Z^(2*3^j). Dividing by Z^(3^j) and evaluating at v, with v+v^(-1)=3, gives B_j=Psi_r(3).

The recursion for x_j sends 4 to -4 and -4 to 4 modulo40. Thus x_j=+/-4 modulo40, giving B_j=19 modulo80. For the ternary claim extend the same definition to B_0=L_2+1=4. The exact factorization

$$B_{j+1}-1=(B_j-1)((B_j-1{)}^2-3)$$

raises the 3-adic valuation by exactly one and changes the normalized quotient by a minus sign modulo3. Starting with B_0-1=3 proves both ternary statements. Finally x_j>=4, so x_j^2<B_j<(x_j+1)^2. This gives nonsquareness without any factorization.

TBN.3 Each factor has a specified original rank and actual depth

Theorem TBN3. If p is a prime divisor of B_j, then

$$p>5,(5/p)=1,{\mathrm{ord}}_p(v)=r_j,\rho (p)=2r_j,p\equiv 1(\mathrm{mod}2r_j),$$

$$\boxed{v_p(B_j)=h_p.}\text{\hspace{1em}(TBN2)}$$

Here rho(p) is the Fibonacci zero rank, and h_p=v_p(F_(p-(5/p))) is the original initial depth. Different B_j are pairwise coprime.

Proof. The congruences in TBN2 exclude 2,3,5. In O/pO let w=v^(n_j). The equation B_j=0 gives w^2+w+1=0. Since p!=3, w!=1 and has order three. The order of v divides 3^(j+1) but not 3^j, hence is r_j. The identity F_a=(phi^a-psi^a)/sqrt(5) shows that rho(p) is the order of phi/psi=-v. The order r_j is odd, so -1 does not belong to the group generated by v; consequently -v has order 2r_j.

The Lucas identity gives p|L_(r_j). The odd-index Pell identity L_(r_j)^2-5F_(r_j)^2=-4 implies 5F_(r_j)^2=4 modulo p. Thus F_(r_j) is a unit modulo p and five is a square. The standard golden Frobenius rank bound now gives 2r_j|p-1.

Also L_(r_j)=L_(n_j)B_j, and L_(n_j)^2=-3 modulo p, so L_(n_j) is a unit. Therefore F_(2r_j)=F_(r_j)L_(r_j) has p-valuation exactly v_p(B_j). This is the valuation at the first Fibonacci zero, so it is h_p. Equality with the valuation at p-(5/p) uses the classical prime-to-p multiplier formula, with multiplier (p-1)/(2r_j) prime to p. Distinct j give distinct exact ranks, so no prime can divide two blocks.

The rank and valuation inputs are classical Lucas arithmetic. They are stated, for example, in Medina-Rowland, p-regularity of the p-adic valuation of the Fibonacci sequence, Fibonacci Quarterly53(2015), Theorem1.4, arXiv:0910.2907v4. McConnell’s dimension-tower Lemma3.3 has the same order-three mechanism. Ross-Shen-Cai, arXiv:2512.03481v1, explicitly credit the regular cyclotomic valuation case to classical results. TBN does not claim a new general valuation theorem.

TBN.4 Three simultaneous balances on the actual initial depths

Theorem TBN4. The prime factors of each B_j satisfy

$$\boxed{\sum _{\begin{array}{c}p\mid B_j\\ p\equiv 3(4)\end{array}}{h}_p\equiv 1(\mathrm{mod}2),}\text{\hspace{1em}(TBN3)}$$

$$\boxed{\sum _{\begin{array}{c}p\mid B_j\\ p\equiv 4(5)\end{array}}{h}_p\equiv 1(\mathrm{mod}2),}\text{\hspace{1em}(TBN4)}$$

$$\boxed{\sum _{p\mid B_j}{h}_p\frac{p-1}{r_j}\equiv (-1{)}^j(\mathrm{mod}3).}\text{\hspace{1em}(TBN5)}$$

In particular each block has a split prime p with odd initial depth and

$$p\equiv 2r_j+1(\mathrm{mod}4r_j).$$

Each block also has a prime p with 3 not dividing h_p and v_3(p-1)=j+1. These two existential witnesses are not asserted to be the same prime.

Proof. The exact factorization is B_j=product_(p|B_j)p^(h_p) by TBN3. Reduce it modulo4. Since B_j=3 modulo4, the exponents of the prime factors congruent to3 have odd sum. This proves TBN3. All factors split in the golden field, hence are1 or4 modulo5; reducing the same factorization modulo5 and using B_j=4 proves TBN4.

For TBN5 write p=1+r_j a_p, which is legitimate for every factor. Since r_j^2 is divisible by3r_j, expansion of the finite product modulo3r_j gives

$$B_j\equiv 1+r_j\sum _{p\mid B_j}{h}_p{a}_p(\mathrm{mod}3r_j).$$

The last congruence in TBN2 identifies the sum modulo3. Its nonzero right side forces a term with both h_p and a_p nonzero modulo3, proving the second witness claim. The first witness has h_p odd and p=3 modulo4. Combining this with p=1 modulo2r_j, and using r_j odd, gives its displayed class modulo4r_j.

These balances genuinely concern the original h_p, because TBN3 removes index-multiplier contributions. They are elementary joint constraints on one actual finite prime support. They do not specify which factor has depth one.

TBN.5 A fifth-power budget for an all-WSS block

Definition. For r=3^(j+1), define

$$b(r)=\{\begin{array}{ll}8r+1,& r\equiv 1(\mathrm{mod}5),\\ 4r+1,& r\equiv 2(\mathrm{mod}5),\\ 6r+1,& r\equiv 3(\mathrm{mod}5),\\ 2r+1,& r\equiv 4(\mathrm{mod}5).\end{array}$$

Theorem TBN5. If every prime factor of B_j is WSS, then

$$\sum _{p\mid B_j}{h}_p\ge 5,\boxed{B_j\ge b(r_j{)}^5.}\text{\hspace{1em}(TBN6)}$$

Consequently B_j<b(r_j)^5 is a sufficient, independently checkable condition for B_j to have a non-WSS factor. It does not construct a WSS prime.

Proof. Under the premise all h_p>=2. The first parity balance excludes total exponent two and every possibility of total exponent four, because all exponents would be even. A total exponent three would consist of a single exponent three; the ternary balance excludes that case because every summand would be zero modulo3. Totals zero and one are impossible as B_j>1 and each exponent is at least two. Hence the total is at least five.

Each prime factor is p=1+2r_j t with integer t>=1, and p is1 or4 modulo5. The smallest positive allowed t is respectively4,2,3,1 in the four residue classes of r shown. This gives p>=b(r_j), without asserting that b(r_j) itself is prime. Multiplying these lower bounds with the actual exponents proves TBN6.

For the fixed indices j=1,2,3 the respective pairs (B_j,b(r_j)^5) are

$$(19,2476099),(5779,15386239549),(192900153619,115139273278249).$$

All satisfy the strict inequality, establishing a non-WSS factor in each of these three known finite blocks without factoring the third one. These bounded checks are not previously unknown WSS-family decisions. At j=4 this same test fails: B_4=7177905237579946589743592924684179 exceeds1459^5. No conclusion about that block follows from this inequality.

TBN.6 The infinite consequence and the unclosed branch

Corollary. There are infinitely many distinct split primes p with odd h_p, p=3 modulo4, and rho(p)=2*3^s for some s>=2. Within this restricted set, either infinitely many have h_p=1 or infinitely many have odd h_p>=3, or both. At least J distinct primes of those two types combined lie below B_J.

Proof. Choose a parity witness from each of the J pairwise-coprime blocks B_1,…,B_J. TBN3-TBN4 give all stated properties and ensure distinctness. The blocks increase strictly, so each selected prime is at most B_J. Every positive odd h_p is one or at least three. If the union is infinite at least one of these two subsets is infinite.

This is a specialization of the existing GP3 nonsquare-block strategy to the complementary split Lucas tower, with the additional ternary balance displayed above. It is not a proof that either designated subset by itself is infinite. In particular an initial depth of three remains allowed and is itself WSS. The fifth-power lower budget grows polynomially in r_j, whereas B_j grows exponentially in3^j, so the bound does not exclude the all-WSS branch uniformly. No new actual WSS prime-family existence or exclusion, or resolution of WSS, is claimed.

DCE. Primitive depth vectors from a solved exponential equation

DCE.1 The exact classical input

Theorem (Lebesgue-Nagell, D=3). There are no integers x,y and integer e>=3 satisfying x^2+3=y^e. In particular this holds for positive x and y. This is a cited theorem, not a newly proved exponential Diophantine result in this dossier. A precise published locator is Y. Bugeaud, M. Mignotte and S. Siksek, Classical and modular approaches to exponential Diophantine equations II. The Lebesgue-Nagell equation, Compositio Math. 142 (2006), 31-62, DOI 10.1112/S0010437X05001739, Theorem 1 and Section 16, the empty D=3 row. The primary author preprint is arXiv:math/0405220; the table is on its printed page 47. The historical D=3 case is attributed to Nagell in Section 2. We use the classification as an external mathematical input and do not claim to reproduce its proof or certify it in Lean.

Lemma DCE1. For every integer x>=2, x^2+3 is not a perfect power y^e with y>=2 and e>=2.

Proof. Exponents e>=3 are excluded by the displayed classical theorem. At e=2, positive factorization gives (y-x)(y+x)=3, so y-x=1 and y+x=3. This forces x=1, contrary to x>=2. This lemma does not say that x^2+3 is never powerful. Mixed exponents are not excluded by the cited equation.

DCE.2 No common divisor of the actual initial depths

Theorem DCE2. For every j>=1, retain the actual block B_j=L_(3^j)^2+3 and its original depths from TBN. Then

$$\boxed{\gcd \{h_p:p\text{ prime},p\mid B_j\}=1.}\text{\hspace{1em}(DCE1)}$$

Equivalently, for every integer e>=2 there is a prime p|B_j such that e does not divide h_p. In particular B_j cannot be a power of a single WSS prime. If B_j has only one distinct prime factor, that factor has h_p=1 and B_j itself is prime.

Proof. The support is finite and nonempty because B_j>1. TBN3 gives the exact integer factorization B_j=product_(p|B_j) p^(h_p), without an index-multiplier contribution. If the gcd of these positive exponents were g>1, the integer Y=product p^(h_p/g) would satisfy B_j=Y^g. But L_(3^j)>=4, so DCE1 prohibits this equality. This proves the gcd. If every exponent were divisible by a chosen e>=2, the gcd would also be divisible by e, proving the witness statement. Conversely the witness statement for every e rules out any gcd greater than one. For a one-prime support the gcd equals its only exponent, giving the last conclusion.

This uses a solved global exponential equation to constrain the actual fixed-golden depth vector. It is stronger than the two parity balances and single modulo-three balance of TBN4: it excludes a common exponent divisor of any size. The implication is a repository specialization of the cited theorem, not a new solution of the Lebesgue-Nagell problem or an independent proof of WSS existence.

Corollary DCE3. Fix any real bound X and any integer e>=2. There is a split prime p>X with

$$\rho (p)=2\cdot 3^s\text{for some }s\ge 2,e\nmid h_p.$$

There are infinitely many such primes for every fixed e. More strongly, the positive integers h_p for the primes p>X in the union of these block supports have gcd one.

Proof. Choose j so large that 2r_j+1>X. Every prime dividing B_j exceeds X by TBN3. DCE2 gives a witness for e and shows that the depths in this finite subset already have gcd one. Different blocks have disjoint prime support, so repeating with arbitrarily large j gives infinitely many witnesses. A common divisor of the depths in the entire tail would divide the gcd in this one finite block, hence is one. These primes may have depth one or larger depths not divisible by e; neither possibility is selected by this argument.

DCE.3 An all-depth covering budget and the exact smallest pattern

Theorem DCE4. Let H>=2 be an integer. If every prime divisor of B_j has h_p>=H, then B_j has at least two distinct prime factors and

$$\boxed{\Omega (B_j):=\sum _{p\mid B_j}{h}_p\ge 2H+1,B_j\ge b(r_j{)}^{2H+1}.}\text{\hspace{1em}(DCE2)}$$

Here b(r_j) is the explicit residue-class lower bound defined in TBN5. If equality Omega(B_j)=2H+1 holds, there are exactly two distinct prime factors and their depths are exactly H and H+1. In particular, under all-WSS covering, total multiplicity five can only have the form P^2 Q^3 with distinct primes P,Q.

Proof. A single factor would make the depth gcd at least H, contrary to DCE2. With at least three factors the total is at least 3H>2H+1, since H>=2. With two factors the total is at least 2H, but equality would give the exponent vector (H,H), again contrary to its gcd being one. Hence the total is at least 2H+1. Equality leaves two integers at least H whose sum is 2H+1, necessarily H,H+1. Every actual prime factor is at least b(r_j) by TBN5, so multiplying its powers gives the size bound. No primality of b(r_j) is required.

Corollary. For every H>=2, the exact inequality B_j<b(r_j)^(2H+1) guarantees some actual prime p|B_j with h_p<H. For example the fixed fourth block satisfies

$$1459^{10}<B_4=7177905237579946589743592924684179<1459^{11},$$

so it has a prime factor of original depth at most four. This does not decide whether that factor is WSS, and uses no factorization of B_4.

Proof. Take the contrapositive of DCE4. In the fourth block r_4=243 and b(r_4)=1459; H=5 gives the conclusion. The two displayed inequalities are exact comparisons of the written integers. The bound permits depths one, two, three or four.

Theorem DCE5. Suppose all prime factors of B_j are WSS and Omega(B_j)=5. With the forced factorization B_j=P^2 Q^3, one has

$$\boxed{Q\equiv 19(\mathrm{mod}40),P\equiv 1+2(-1{)}^j{r}_j(\mathrm{mod}6r_j),v_3(P-1)=j+1.}\text{\hspace{1em}(DCE3)}$$

Both primes also have the split exact rank 2r_j from TBN3. If Q=19 modulo80 then P^2=9 modulo16; if Q=59 modulo80 then P^2=1 modulo16.

Proof. DCE4 forces two distinct factors with the indicated depths. Since P is odd, P^2=1 modulo8. The congruence B_j=19 modulo80 gives Q^3=3 modulo8; cubing is the identity on odd residues modulo8, so Q=3 modulo8. Every factor is 1 or4 modulo5, and hence P^2=1 modulo5. The same product gives Q^3=4 modulo5, so Q=4 modulo5. Combining gives Q=19 modulo40.

Write P=1+2r_j u. In TBN5’s depth-balance equation modulo3, the contribution from Q has its factor h_Q=3 and vanishes. The remaining contribution is 2*(P-1)/r_j=4u=u modulo3. Thus u=(-1)^j modulo3, proving the displayed class and exact ternary valuation. Finally B_j=3 modulo16 and Q=19 or59 modulo80 give Q^3=11 or3 modulo16 respectively. Multiplication by P^2 yields the two stated square classes. These necessary conditions do not prove existence or impossibility of this two-prime pattern.

DCE.4 Pure prime-power seeds cannot reach the fixed golden field

Theorem DCE6. For every odd prime p and every integer a>=2, McConnell’s dimension parameter satisfies

$$\boxed{\mathbb{D}(p^a)\ne 5.}\text{\hspace{1em}(DCE4)}$$

This extends TBN1 from square seeds to every pure odd-prime-power seed, including nonsquare exponents.

Proof. If D(p^a)=5, TBN1’s dimension description gives p^a=L_(2n)+1 for a positive n. If n is odd and n>=3, this is L_n^2+3, prohibited by DCE1. At n=1 the value is four, not an odd prime power. If n is even then n>=2, L_n>=3 and

$$p^a=(L_n-1)(L_n+1).$$

Since p^a is odd, both factors are odd and their gcd, which divides two, is one. Both factors are greater than one. They cannot both be powers of the same prime p. This is a contradiction. No Catalan theorem and no classification of all mixed seed values are used.

The conclusion excludes this unbounded family of parameter inputs from one varying-field construction. It does not exclude any p from being WSS by another mechanism. The fixed-field condition and the prime-power-seed condition are both essential to the stated interpretation.

DCE.5 What the new restriction still permits

Proposition. DCE2 rules out every common-divisor depth pattern on each actual B_j, but does not prove a simple prime factor in every B_j. The vector (2,3) has gcd one and all entries at least two. DCE5 imposes additional residue conditions on a realization of that vector, without eliminating it. No all-WSS block or WSS prime is constructed in DCE.

Proof. The gcd computation on the abstract vector is immediate. The arguments of DCE2-DCE6 exclude common-divisor and pure prime-power branches, whereas none derives a contradiction for two distinct actual primes meeting all conditions of DCE5, nor constructs such primes. DCE4 is a lower size budget. Since B_j grows exponentially in 3^j while b(r_j) grows linearly in r_j, the budget at fixed H cannot exclude covering at every j. Hence a new zero-set decision still requires an independent treatment of the heterogeneous-depth branch, or of a different actual initial-rank block. The published input and its application are not counted as a newly solved external problem.

GPC. Golden prime clocks, exact interlevel jets, and quadratic reciprocity

GPC.1 The same golden matrix and three different return times

Definition. Retain the actual TBN blocks

$$x_j=L_{3^j},r_j=3^{j+1},B_j=x_j^2+3(j\ge 1),$$

and the original golden unit phi, v=phi^2 and initial depths h_p. Write rho(M) for the least positive Fibonacci zero index modulo M, and pi(M) for the least positive n with (F_n,F_(n+1))=(0,1) modulo M. Thus pi(M) is the order modulo M of the integer matrix

$$U=\left(\begin{array}{cc}0& 1\\ 1& 1\end{array}\right),U^n=\left(\begin{array}{cc}{F}_{n-1}& F_n\\ F_n& F_{n+1}\end{array}\right).$$

The matrix Q in PERIODIC_TREE.md, Section 8.1, is obtained by interchanging these two coordinates; it has the same orders. That section’s real golden-return matrices are powers of Q. The real fractional-part clock, the norm-one residue order of v, and the exact finite-ring pair return remain different observations of this common integer matrix.

Theorem GPC1. If p is any prime divisor of B_j, then

$$\boxed{{\mathrm{ord}}_p(v)=r_j,\rho (p)=\pi (p)=2r_j,\pi (p^a)=2r_j{p}^{\max (a-h_p,0)}(a\ge 1).}\text{\hspace{1em}(GPC1)}$$

Every divisor d>1 of B_j has pi(d)=rho(d)=2r_j. More generally, for every a>=1,

$$\boxed{\pi (B_j^a)=2r_j{B}_j^{a-1}.}\text{\hspace{1em}(GPC2)}$$

In particular pi(B_j^2)=B_j pi(B_j), independently of whether a prime factor has h_p=1 or h_p>=2.

Proof. TBN3 proves the first two orders except for pi(p), and proves v_p(B_j)=h_p. Since phi^(2r_j)=v^(r_j)=1 modulo p, pi(p) divides 2r_j. The zero rank rho(p)=2r_j divides every pair-return time, so equality follows.

Put T=2r_j. Its Lucas trace satisfies L_T^2-5F_T^2=4, L_T=2 modulo p and v_p(F_T)=h_p. Since L_T+2 is a unit at p, v_p(L_T-2)=2h_p. Therefore phi^T-1=(L_T-2+sqrt(5)F_T)/2 has valuation h_p in each golden completion above p. For a unit 1+y with v_p(y)=h>=1 and odd p, binomial expansion gives v_p((1+y)^b-1)=h for p not dividing b, and raises the valuation by exactly one on taking a pth power. Iteration gives h+v_p(b). Every lift return is a multiple of T, so its minimum multiple modulo p^a is T p^max(a-h_p,0). This proves the full prime-power formula without assuming h_p=1.

Write B_j=product p^(h_p), the actual factorization from TBN3. A prime-power divisor p^e of B_j has 1<=e<=h_p and hence pair period T; its zero rank also remains T by the classical Fibonacci valuation formula. The Chinese remainder theorem takes least common multiples of the prime-power periods and ranks, proving the divisor assertion. At modulus B_j^a the p component instead has period T p^((a-1)h_p). Every such p is prime to T, so their least common multiple is T times their product of p-parts, namely T B_j^(a-1). This proves GPC2.

The prime-power and CRT mechanisms are classical period theory, as stated in Marc Renault, The Period, Rank, and Order of the (a,b)-Fibonacci Sequence Mod m, Mathematics Magazine 86 (2013), 372-380, Theorems 1-2, DOI 10.4169/math.mag.86.5.372, and in the valuation theorem cited in TBN3. The displayed block specialization is proved here. Regular growth of a composite block’s period therefore cannot, by itself, distinguish its factors’ initial WSS depths.

GPC.2 Exact propagation from a zero block to the fixed point three

Theorem GPC2. For every j>=1 and every 1<=i<j,

$$\boxed{x_j=4\prod _{1\le t<j}{B}_t,B_j=16{\left(\prod _{1\le t<j}{B}_t\right)}^2+3,}\text{\hspace{1em}(GPC3)}$$

$$\boxed{B_j\equiv 3-3^{2(j-i)-1}{B}_i^2(\mathrm{mod}{B}_i^3).}\text{\hspace{1em}(GPC4)}$$

Consequently B_j=3 modulo B_i^2, and for each prime p|B_i,

$$\boxed{v_p(B_j-3)=2h_p.}\text{\hspace{1em}(GPC5)}$$

Thus for these fixed i,j and p, p is WSS if and only if p^4 divides B_j-3. This is an exact transport of the original depth, not an independent assertion that the last divisibility occurs.

Proof. TBN2 gives x_(j+1)=x_j B_j and x_1=4, proving the product by induction. The second identity follows from B_j=x_j^2+3. For i<j, divide that product at i to get the exact integer identity

$$\frac{B_j-3}{B_i^2}=x_i^2\prod _{i<t<j}{B}_t^2.$$

Each intermediate block is three modulo B_i by the same product formula, while x_i^2=B_i-3. Reducing this quotient modulo B_i gives -3*9^(j-i-1), proving GPC4. Its residue is a unit at every prime of B_i, since 3 does not divide B_i. Thus the exact p-valuation of B_j-3 is twice v_p(B_i)=h_p. Equivalently the scalar cubic map f(Y)=Y^3-3Y^2+3 sends zero to the fixed point three with order-two contact, after which its derivative f’(3)=9 is a unit at these primes. This description retains the fixed recurrence and does not insert a freely chosen local lift.

GPC.3 Directed reciprocity and primewise depth constraints

Definition. For coprime odd positive integers A,B use the usual Jacobi symbol (A/B), including composite B. A value +1 for a composite denominator is not assumed to mean that A is a square modulo B. For distinct odd primes q,p let epsilon(q,p) be zero or one according as (q/p) is +1 or -1.

Theorem GPC3. For every 1<=i<j,

$$\boxed{\left(\frac{B_j}{B_i}\right)=-1,\left(\frac{B_i}{B_j}\right)=+1.}\text{\hspace{1em}(GPC6)}$$

For every individual prime p|B_i one also has the stronger primewise identities

$$\boxed{\prod _{q\mid B_j}{\left(\frac{q}{p}\right)}^{h_q}=\left(\frac{3}{p}\right)=(-1{)}^{(p-1)/2},\prod _{q\mid B_j}{\left(\frac{p}{q}\right)}^{h_q}=1.}\text{\hspace{1em}(GPC7)}$$

In particular the actual depth vectors of two blocks satisfy

$$\boxed{\sum _{p\mid B_i}\sum _{q\mid B_j}{h}_p{h}_qϵ(q,p)=1(\mathrm{mod}2).}\text{\hspace{1em}(GPC8)}$$

Proof. GPC4 gives (B_j/B_i)=(3/B_i). All B_i are three modulo four and one modulo three, so Jacobi reciprocity with three gives (3/B_i)=-1. The blocks are coprime and both three modulo four; reciprocity between them gives the opposite symbol +1. The extension of reciprocity and the prime-factor definition of the Jacobi symbol are the classical laws in NIST DLMF 27.9.1 and 27.9.3.

For a prime p|B_i, GPC4 gives (B_j/p)=(3/p). TBN3 gives p=1 modulo three, so (3/p)=(-1)^((p-1)/2). Insert B_j=product_(q|B_j)q^(h_q) to obtain the first primewise identity. Reciprocity for p and each q contributes sign (-1)^((p-1)/2) raised to sum h_q*(q-1)/2. This sum is odd because B_j=3 modulo four. The sign cancels (3/p), proving the second identity. Finally expand the Jacobi symbol (B_j/B_i)=-1 using the actual factorizations of BOTH blocks to obtain GPC8. These formulas keep original initial depths, not multiplicities introduced by multiplying a Fibonacci index.

Corollary GPC4. For every i<j there are primes p|B_i and q|B_j with both h_p,h_q odd and (q/p)=-1. For every fixed i, at least one such p has infinitely many distinct later witnesses q. In particular there are infinitely many split primes q with odd h_q, (q/19)=-1, and rho(q)=pi(q)=2*3^s for some s>=3.

Proof. The odd sum in GPC8 contains a summand with all three factors odd. As j varies, the later supports are disjoint. The fixed finite set of primes in B_i therefore contains one p occurring for infinitely many j. At i=1, B_1=19 is prime and h_19=1 by TBN3, so the first coordinate must be nineteen. The remaining period and splitting assertions follow from GPC1 and TBN3. The odd depth of q may be one or at least three; the corollary does not select one of these alternatives.

GPC.4 Odd-depth kernels and the full earlier-prime splitting condition

Definition. Write each actual block as B_j=s_j a_j^2, where

$$s_j=\prod _{q\mid B_j,h_q\text{ odd}}q,a_j=\prod _{q\mid B_j}{q}^{\lfloor h_q/2\rfloor }.$$

For j>=2 put A_j=product_(i<j)rad(B_i), let t_j be the number of distinct prime factors of A_j, and define the multiquadratic field

$${\mathcal{E}}_j=\mathbb{Q}(\sqrt{p}:p\mid A_j).$$

Here rad uses all distinct primes, whereas s_j retains only odd depths. The field E_j is totally real and exists independently of any WSS hypothesis.

Theorem GPC5. Each s_j is a nontrivial squarefree integer, the s_j are pairwise coprime, and

$$s_j=19(\mathrm{mod}120),s_j=1(\mathrm{mod}2r_j).$$

Removing square factors preserves GPC6:

$$\left(\frac{s_j}{s_i}\right)=-1,\left(\frac{s_i}{s_j}\right)=1(i<j).$$

The degree of E_j is 2^(t_j), and its Frobenius elements at the primes q|B_j obey

$$\boxed{\prod _{q\mid B_j}{\mathrm{Frob}}_q^{h_q}=1\text{in }\mathrm{Gal}({\mathcal{E}}_j/\mathbb{Q}).}\text{\hspace{1em}(GPC9)}$$

Proof. Nonsquareness of B_j from TBN2 gives s_j>1. Disjoint block supports give pairwise coprimality. The square a_j^2 is one modulo eight. Each of its prime factors is one or four modulo five, so its square is also one modulo five. Thus s_j is three modulo eight and four modulo five. Every prime in the block is one modulo 2r_j and hence one modulo three; these facts give the two displayed congruences. Squares in either slot of a coprime Jacobi symbol contribute one, proving the reduced symbol identities.

Distinct rational primes have independent square classes in Q^/Q^()^2: taking the valuation at any selected prime forces its exponent in a square product to be even. The classical degree formula for multiquadratic extensions therefore gives 2^(t_j), with independent sign changes on the chosen square roots. The extension is unramified at every q|B_j. On the coordinate sqrt(p), its Frobenius acts by the sign (p/q). GPC7 makes the product of these signs, with multiplicities h_q, equal one for EVERY p|A_j. This proves GPC9 coordinatewise. It uses all earlier prime factors, including those of even depth, not only the squarefree kernels s_i.

Corollary GPC6. If B_j has exactly one prime Q of odd initial depth, then Q splits completely in E_j. In particular, if B_j is an all-WSS block of total multiplicity five, DCE4 forces B_j=P^2 Q^3 with distinct primes and

$$\boxed{\left(\frac{p}{Q}\right)=1\text{for every }p\mid A_j,\left(\frac{Q}{B_i}\right)=-1(1\le i<j).}\text{\hspace{1em}(GPC10)}$$

The same Q satisfies Q=19 modulo40 and Q=1 modulo2r_j. For j>=3, in particular, (Q/19)=(Q/5779)=-1.

Proof. In the group of exponent two, all even h_q terms of GPC9 disappear. The only remaining term is Frob_Q to an odd power, so it must be the identity. This is equivalent to the primewise splitting conditions. Also (B_j/B_i)=(Q/B_i) because all the other exponents are even, giving the second condition. The congruence for Q follows from the s_j congruences when there is only one odd-depth prime, or directly from DCE5 in the two-prime case. The exact period gives Q=1 modulo2r_j. The first two earlier blocks are19 and5779, so the stated Jacobi tests follow without factoring any later block. No existence or impossibility of the two-prime pattern is asserted by these necessary conditions.

GPC.5 The finite reciprocity tests are compatible, but omit the actual clock

Theorem GPC7. Fix j>=2 and its actual earlier supports. The set of rational primes Q satisfying

$$Q=19(\mathrm{mod}40),Q=1(\mathrm{mod}{r}_j),(p/Q)=1\text{ for every }p\mid A_j$$

has Dirichlet density

$$\boxed{\frac{1}{2^{t_j}\phi (40r_j)}=\frac{1}{2^{t_j+5}{3}^j}>0.}\text{\hspace{1em}(GPC11)}$$

This set does not impose Q|B_j, pi(Q)=2r_j, or h_Q>=2. For fixed j, the primes with pi(Q)=2r_j and Q>5 form a finite set, in fact exactly the prime divisors of B_j. Thus the positive density in GPC11 supplies no prime of the required exact clock or a WSS prime.

Proof. The two displayed congruences specify a single unit class modulo 40r_j, by CRT. In this class Q=3 modulo four. Reciprocity rewrites each condition (p/Q)=1 as (Q/p)=(-1)^((p-1)/2). Every such nontrivial quadratic character modulo p takes each sign on half its nonzero residues. All primes p|A_j are distinct and coprime to 40r_j. CRT therefore leaves product_(p|A_j)(p-1)/2 admissible unit classes modulo 40r_j A_j. Dirichlet’s theorem on prime progressions assigns each class density 1/phi(40r_j A_j); summing gives GPC11. Equivalently this is the abelian cyclotomic case of the Chebotarev density theorem, A. V. Sutherland, MIT 18.785 Lecture 28 (2021), Theorem 28.9. Since r_j=3^(j+1), phi(40r_j)=32*3^j.

If Q>5 has pi(Q)=2r_j, the order of phi^2 modulo Q is r_j, because the square of an element of exact even order 2r_j has order r_j. Then w=(phi^2)^(3^j) is a nontrivial cube root of unity in the norm-one group. Its components are reciprocal in the split case, so neither equals one; in the inert case it is a nonidentity field element. Thus w-1 is a unit and w^2+w+1=0, which makes its trace minus one, giving L_(2*3^j)+1=B_j=0 modulo Q. Conversely GPC1 gives pi(Q)=2r_j for every Q|B_j. Thus the exact-period primes are precisely that finite support. Removing the exact-order requirement before applying a density theorem changes the question; it is not a proof of fixed-block or WSS existence.

GPC.6 Arithmetic scope of the golden-clock transfer

Proposition. GPC2-GPC6 give exact relations between different actual prime supports, while GPC1 shows that the normal composite-period ratio is insensitive to their individual initial WSS status. The quadratic identities constrain only the parity of the depths in the later support. They allow h=1 and also odd h>=3, and the complete-splitting conditions in GPC10 have not been proved to contradict the required factorization or exact clock.

Proof. Equation (GPC5) transports each original depth exactly, but requires evaluation on the actual later block. The Jacobi and quadratic Frobenius groups have exponent two, so replacing an exponent h by h+2 does not change their character values. GPC11 shows that the necessary finite splitting tests are consistent for unrestricted comparison primes, whereas its last paragraph retains the missing exact-period restriction. None of these statements chooses a new prime with h=1 or h>=2. They therefore provide a period-aware cross-prime constraint and an explicit remaining obligation for heterogeneous-depth covering, rather than a resolution of WSS or a new prime-family WSS decision.

PCL. Golden-clock sampling, prime-period coupling, and native block periods

PCL.1 Three clocks and the retained arithmetic object

Definition. Keep the original Fibonacci sequence, the golden integer ring O=Z[phi], phi^2=phi+1, and the matrix

$$Q=\left(\begin{array}{cc}1& 1\\ 1& 0\end{array}\right),\pi (m)=\mathrm{ord}(Q\mathrm{mod}m),\pi (1)=1.$$

For an integer stride s>=1, put Pi_s(m)=ord(Q^s modulo m). The full two-coordinate matrix of golden-clock layer L in PERIODIC_TREE.md, Section 8, is Q^(L+2), so its stride is s=L+2. The scalar floor map T_L and the irrational circle rotation are different observables. In particular the former has the parity-carry correction in GC11, while the latter has no nonzero exact integer return. Neither is identified here with the matrix reduced modulo m.

Lemma PCL1. For every m,s>=1,

$$\boxed{{\Pi }_s(m)=\frac{\pi (m)}{\gcd (\pi (m),s)}.}\text{\hspace{1em}(PCL1)}$$

For coprime positive m,n, pi(mn)=lcm(pi(m),pi(n)); the same identity holds for Pi_s.

Proof. If an element has order t, its sth power returns after a steps exactly when t divides sa, whose smallest positive solution is t/gcd(t,s). CRT makes simultaneous matrix return modulo mn equivalent to return modulo m and n. These are statements about the fixed invertible matrix, with no choice of a different lift or recurrence coefficient.

The period/rank/lift interfaces are already retained in PERIODIC_TREE.md, Appendix R. The classical CRT and lift theorems are also stated in B. Benfield and O. Lippard, Fixed Points of K-Fibonacci Sequences, arXiv:2404.08194v2, Theorems 2.1-2.2; the K=1 fixed-point classification and eventual convergence are attributed there to Fulton-Morris (1969). This appendix uses that pinned source only for the stated classical inputs and supplies its specialized arguments below. The paper’s separate generalized (a,-1) conjecture is not used.

PCL.2 The original lift depth and the exact combined masking formula

Definition. For a prime p>5 write tau_p=pi(p), chi_p=(5/p), rho(p) for the first Fibonacci zero, and h_p=v_p(F_(rho(p)))=v_p(F_(p-chi_p)). The equality of depths uses the prime-to-p multiplier law; it does not set h_p to one.

Lemma PCL2. For all a>=1,

$$\boxed{p\nmid {\tau }_p,\pi (p^a)={\tau }_p{p}^{\max (a-h_p,0)}.}\text{\hspace{1em}(PCL2)}$$

Every prime factor of tau_p is strictly smaller than p. If rho(p) is odd, tau_p=4rho(p).

Proof. Frobenius in O/pO gives tau_p dividing p-1 in the split case and 2(p+1) in the inert case. Thus p does not divide tau_p. A prime divisor of p+1 is at most (p+1)/2 because p+1 is even and greater than two; this and the split bound give strict prime descent.

The matrix identity Q^n=F_n Q+F_(n-1)I shows that at n=rho(p) the matrix is a nonzero scalar c. Cassini gives c^2=(-1)^rho(p) modulo p. Thus tau_p/rho(p) is 1,2 or4 and is prime to p; if rho(p) is odd, that scalar has exact order four. The classical valuation law v_p(F_(rho(p)u))=h_p+v_p(u) implies v_p(F_(tau_p))=h_p. This law is recorded in the existing GP3.3 and, as a classical input, in Ross-Shen-Cai, arXiv:2512.03481v1, Proposition 3.2 specialized to U(1,-1).

For completeness, the matrix lift has exactly this same depth. Put f=F_(tau_p) and c=F_(tau_p-1). The return matrix is cI+fQ, c=1 modulo p, and tau_p is even. Its determinant gives (c-1)(c+1)=f(f-c). Both c+1 and f-c are p-adic units, so v_p(c-1)=h_p. Hence Q^(tau_p)=I+p^(h_p)A with A nonzero modulo p. Raising such a matrix to an exponent prime to p preserves its first nonzero depth; raising it to p increases the depth by exactly one. This follows by binomial expansion: the first term has depth h_p+1 and all higher terms have depth at least h_p+2 for p>2. Any return exponent is a multiple of tau_p, so the least return modulo p^a is exactly PCL2.

Theorem PCL3. Let S be a nonempty finite set of primes greater than five, let a_p>=1, and set