bibkey: ribenboim2005squareclasses authors: Paulo Ribenboim year: 2005 title: “FFF: (Favorite Fibonacci Flowers)” doi: null url: https://www.fq.math.ca/Papers1/43-1/paper43-1-1.pdf claim: “Statement (3.4) records the complete positive Fibonacci square classes; GSE uses its prime-power and ratio-25 consequences, not a newly assumed rigidity theorem.” strata_touched: [] license: citation-only triage: anchor

Square-class rigidity and explicit original-depth support bounds

Exact classical source and scope

Paulo Ribenboim, FFF: (Favorite Fibonacci Flowers), Fibonacci Quarterly 43(1) (2005), 3-14. Statement (3.4), printed page 8, says that the only nonsingleton positive-index square classes of Fibonacci values are {1,2,12} and {3,6}. Reference [20] attributes this to the author’s Square Classes of Fibonacci and Lucas Numbers, Portugaliae Mathematica 46 (1989), 159-175. The 1989 paper’s bibliographic record was verified; its full proof was not independently read in this pass. The actual 2005 statement was read in parsed primary text. Requested images of its pages 8 and 9 failed; no successful visual-page check is claimed.

Statements (3.7)-(3.8), printed page 9, record a classical effective fixed-squareclass bound and a finite closure under prime factors of the entry ranks. The closure operation and general effective finiteness are therefore prior work. Neither is counted as a new invention of GSE. No unproved abc statement from the same paper is used.

The valuation input is Lengyel’s classical theorem, restated in Medina-Rowland, p-regularity of the p-adic valuation of the Fibonacci sequence, Theorem 1.4, arXiv:0910.2907v4. Its original first-zero depth is retained; its formulas for two and five are used separately. The prime number theorem is used only for the final asymptotic corollary, not for the exact divisor bound or finite procedure. The standard prime-counting form is recorded by NIST DLMF Section 27.12: https://dlmf.nist.gov/27.12 . Partial summation and removal of higher prime powers give the equivalent Chebyshev form psi(x)~x.

Consumer and actual addition

Section 10 (GSE) of docs/develop/theory/GOLDEN_CUBIC_BLOCK_PRIME_PERIODS.md proves:

• for a finite rank-closed prime set H containing {2,3,5}, the condition that the squarefree kernel of F_n is supported in H forces n | 5*lcm_(p in H) rho(p);
• the same explicit divisor bound for the original external odd-depth support U(n) contained in a prescribed finite set S, using the existing OSE support descent in the migrated Andrejic companion;
• a finite divisor-enumeration procedure using exact square roots of the residual after removing H-prime factors, without full factorization of every large Fibonacci value;
• the uniform escape condition n not dividing 10*lcm(1,…,Y), where Y=max(5,floor((Q+1)/2)), forces an external odd-original-depth prime greater than Q. It is an odd-depth WSS prime only with the additional powerful-Fibonacci antecedent.

The proof combines nonsquare prime-power quotients with parity cancellation in Q_positive^times modulo squares at the prime five. The latter requires preserving rank-divisibility thresholds and uses square-class rigidity at the ratio 25. The map n -> [F_n] is not treated as a group homomorphism. PBC.3 already contains the nonsquare quotient mechanism for primes at least seven; GSE retains that attribution and checks two, three and five separately to obtain the explicit cutoff. Global first-discovery priority for this bound has not been established.

Recent Erdos methods: inspiration, not imported hypotheses

The curator’s Problem 936 concerns finiteness of powerful values among 2^n+/-1 and n!+/-1: https://www.erdosproblems.com/936 . The retrieved page is marked open, last edited 31 October 2025, and records abc-conditional results. The indexed copy is not represented as a new September 2026 status audit. The exponential branch shares the multiplicity/first-lift issue with Fibonacci WSS. No reduction solving one problem through the other is asserted.

The primary repository tadamcz/erdos126, README blob ef92c4278100cf0063424710b4c352e5c2e31df6, describes recently verified proofs for the number of prime divisors of pairwise sums. Its accounts use prime-power residues, negation orbits, signed kernels, and matching or collision budgets to produce polynomial bounds. The README and its statement/provenance were read; those Lean files were not recompiled or imported here. The useful methodological comparison is to turn local prime data into an actual global bound, beyond merely counting abstract characters. No specific inequality from Erdos126 is a GSE premise.

Nat Sothanaphan, Resolution of Erdos Problem #728: a writeup of Aristotle’s Lean proof, arXiv:2601.07421v2: https://arxiv.org/html/2601.07421v2 . Sections 3-4 reduce factorial quotients to prime-by-prime valuations and use carry bounds and spike control. This suggests preserving the exact bad valuation event rather than replacing it with a coarse density analogy. That theorem is also context only, with no factorial-to-WSS reduction.

Formalization handoff and evidence boundary

The actual PR source already has FiniteFibonacciRankClosure, FibonacciPrimeToIndexValuation and OriginalOddDepthSupport. The last source displays its prime-index odd-factor premise explicitly. GSE’s square-class rigidity is a separate classical prerequisite, not already proved by those Lean endpoints. A useful order of further formal work is: prime-power quotient cancellation and its odd witness; the ratio-25 parity transport with rank thresholds; the exponent bounds n|5R_H; then the uniform lcm and finite-set consequences.

No Lean source, frozen marker, atom coverage, or CI claim is added by this note. Existing parallel scalar-contraction work in theory Section 9 is preserved. The new ordinary deductions do not produce a WSS prime, decide a previously unknown WSS prime family, or exclude the entire P^2Q^3 branch. Finite arithmetic certificates are supplemental to the written general proof; synthetic higher depths are never WSS examples.

GPF. Independent square-class packets and an original-depth forest

GPF.0 Fixed arithmetic, source inputs, and the new question

Retain F_0=0, F_1=1, L_0=2, L_1=1 and their original recurrences. For primes p>5 put rho(p)=min{r>=1:p|F_r} and h_p=v_p(F_rho(p))=v_p(F_(p-(5/p))). Define

$$U(n)=\{p>5:p\mid F_n,p\nmid n,h_p\text{ odd}\}(n\ge 1).$$

The equality v_p(F_n)=h_p for a prime-to-index zero is already a repository prerequisite. When F_n is powerful, every member of U(n) has original odd h_p>=3. Without powerfulness, h_p=1 remains possible. The result below counts distinct such factors inside ONE actual value, including when some primitive factors have been absorbed into its index.

Use the classical Fibonacci valuation formulas stated in Section 10 of GOLDEN_CUBIC_BLOCK_PRIME_PERIODS and the full square-class statements (3.4)-(3.5) of Ribenboim’s cited paper. The exceptional positive-index Fibonacci classes are {1,2,12} and {3,6}; the exceptional Lucas classes are {1,3}, and {0,6} when index zero is included. The latter creates no exception among distinct positive odd prime powers other than 1,3. These are substantive classical premises, not newly proved Lean results.

Write G=Q_positive^times/(Q_positive^times)^2 additively. Its coordinate at p is v_p modulo two. For a positive integer A, [A] denotes its class. Nothing here treats the map n -> [F_n] as a homomorphism.

GPF.1 Actual, pairwise disjoint rank packets

For a fixed n>=1 construct the following positive integer packets:

• A_(q,s)=F_(q^s)/F_(q^(s-1)), whenever q^s|n, with q>=7 prime and s>=1, or q=3 and s>=2, or q=2 and s>=3.
• B_(q,s)=L_(q^s)/L_(q^(s-1)), whenever 2q^s|n, with odd prime q>=5 and s>=1, or q=3 and s>=2.
• D_r=F_(5^(2r))/(25 F_(5^(2r-2))), whenever 5^(2r)|n and r>=1.

The letters A,B here are packet labels, not the earlier cubic orders or the earlier ternary B_j. Their associated possible-rank sets are, respectively, {q^s}, {2q^s}, and {5^(2r-1),5^(2r)}.

Lemma GPF1. Each packet is an integer greater than one, divides F_n, and is not a square. Its prime factors exceed five. Every prime factor p has rank in its specified rank set and occurs in that packet with exponent exactly h_p. Every such p exceeds the source prime q, where the source of D_r is five. Distinct packets have disjoint prime supports.

Proof for A. This is the prime-power cancellation in Section 10.3, with the initial packets containing primes two and three now omitted. An old factor p of F_(q^(s-1)) is different from q, because rho(q) is prime to q for q!=5 (and rho(2)=3). Its valuation is unchanged when multiplying that index by q. At p=2, the only possible odd base is q=3; both old and new odd multiples of three have valuation one. Five is absent. Thus the quotient is coprime to the preceding term, and every new factor has exact rank q^s and original exponent h_p. The two Fibonacci indices are not one of the exceptional square-class pairs, so the quotient is not square. The omitted small packets are the only ones with rank three or four. For odd q, q^s divides p-1 or p+1 and is odd, giving p>=2q^s-1>q. For q=2,s>=3 the rank bound gives p>=2^s-1>2. Positivity and divisibility follow from the recurrence.

Proof for B. Odd-index Lucas divisibility gives an integer quotient. The source prime q does not divide either Lucas term. For q!=5, if it did, its rank would divide 2q^s and would be prime to q, hence at most two, impossible. For q=5 use L_m^2-5F_m^2=4(-1)^m. An old odd factor p of L_(q^(s-1)) is therefore prime to q. Since gcd(F_m,L_m) divides two, the factorization F_(2m)=F_m L_m and the Fibonacci valuation law show that its Lucas valuation is unchanged under this odd multiplier. The possible old prime two occurs only for q=3; both Lucas valuations are two, by v_2(L_m)=v_2(F_(2m))-v_2(F_m)=3-1 at odd multiples of three. Thus all old factors cancel. A new odd factor has rank 2q^s: its rank divides 2q^s, cannot divide q^s, and rank 2q^t with t<s would put it in the old Lucas term. The original exponent follows from F_(2q^s). Primes two, three and five are absent in the stated packet ranges. At such a factor, the odd-index norm identity gives 5F_(q^s)^2=4 modulo p. Thus (5/p)=1 and 2q^s divides p-1, so p>=2q^s+1>q. The only exceptional positive odd Lucas square-class pair is 1,3, which was excluded. Hence B_(q,s) is not a square. Finally it divides L_(q^s), hence F_(2q^s), hence F_n.

Proof for D. Put a=5^(2r-2), so the numerator index is 25a. The valuation at five in F_(25a)/F_a is exactly two. Every other old prime factor has unchanged valuation under the multiplier 25. Two and three never divide a Fibonacci value at a power-of-five index. Removing 25 therefore leaves an integer coprime to F_a and to thirty. Its factors have ranks 5^(2r-1) or 5^(2r), and their exponents are their original h_p because they are not five. Such primes are at least 2*5^(2r-1)-1>5. If D_r were square then F_(25a)/F_a=25D_r would be square. This would put indices a and 25a in the same Fibonacci square class, contrary to (3.4). Thus D_r>1 and is nonsquare. It divides F_(25a), hence F_n.

For a prime p>5 the converses also hold: if its rank belongs to a packet’s specified set, it divides that packet. For A this follows from dividing the new Fibonacci value but not the old one. For B it follows from F_(2q^s)=F_(q^s)L_(q^s), absence from F_(q^s), and absence from the old Lucas term. For D it follows from absence in F_(5^(2r-2)) and from p not dividing 25. This also justifies using rank membership alone, rather than factoring a packet, to compute the capacity below.

All displayed rank sets are disjoint. A prime has only one first-zero rank, proving disjointness of the full prime supports. This proves GPF1. In particular D_1=3001. The factor five is removed by pairing TWO successive layers, not by falsely claiming each individually normalized fifth-power quotient is nonsquare.

GPF.2 A group-rank lower bound that subtracts internal absorption

Let P_n be this finite packet collection, E(n)=|P_n|, and let R(n) be the union of its possible-rank sets. Define the finite internal capacity

$$I(n)=\{p>5:p\mid n,\rho (p)\in R(n)\},t(n)=|I(n)|.$$

This I(n) is a SET of index primes, not the earlier normalization index I_j. It is computable by factoring n and computing ranks only at those index primes; no factorization of the large value F_n enters t(n).

Theorem GPF2. In G let V_n=span_F2{[C]:C in P_n}, and project to the coordinates at primes not dividing n, writing this map pi_n. Then

$$\boxed{\dim V_n=E(n),E(n)-t(n)\le \dim {\pi }_n(V_n)\le |U(n)|.}\text{\hspace{1em}(GPF1)}$$

Proof. Every nonsquare packet has a nonzero square class. Their prime supports are disjoint, so any nonempty sum of their classes has a surviving odd coordinate. This proves independence and dim V_n=E. A vector in ker(pi_n|V_n) is supported on primes dividing n and lying in some packet support. Those primes belong to I(n), by GPF1. Thus this kernel embeds in F2^I(n) and has dimension at most t(n). Rank-nullity gives the lower bound. Every retained coordinate lies at a prime p>5, p not dividing n, in a packet with odd original h_p, and that packet divides F_n. The image is therefore contained in F2^U(n), giving the upper bound. Distinct classes alone would not prove independence; the disjoint actual rank supports do that work here.

Forest interpretation. Choose one odd-exponent prime p_C from each packet C and draw an edge from its source prime q to p_C. The targets are distinct. Every edge increases the prime, so the graph is a finite directed forest, with indegree at most one. Any target dividing n is an index prime greater than five and has at least one outgoing A packet. Its terminal targets consequently lie in U(n). At most t(n) edges can end internally, leaving at least E(n)-t(n) distinct terminal primes. This is a second constructive proof of the numerical bound, not an assumption of independent or uniformly distributed prime factors.

GPF.3 A factorization-only budget for any positive index

Put a_q=v_q(n), epsilon=1 if n is even and zero otherwise, and define

$$A(n)=\sum _{q>5\mathrm{prime}}{a}_q,k(n)=\#\{q>5:q\mid n\},$$ $$s(n)=(a_2-2{)}_{+}+(1+ϵ)(a_3-1{)}_{+}+\lfloor a_5/2\rfloor +ϵa_5.$$

Here x_+=max(x,0). Then E(n)=s(n)+(1+epsilon)A(n).

Corollary GPF3. For every n>=1,

$$\boxed{|U(n)|\ge E(n)-t(n)\ge E(n)-k(n)+1_{k(n)>0,s(n)=0}.}\text{\hspace{1em}(GPF2)}$$

Proof. The first equality of counts follows by listing the packets. Always t(n)<=k(n). If s(n)=0 and k(n)>0, every active source prime is greater than five. The smallest such prime cannot lie in I(n), since GPF1 would require it to exceed another active source prime. Hence then t(n)<=k(n)-1. Apply GPF2. The right side is nonnegative: A>=k, and if s>0 its integer value is at least one when needed.

This bound concerns prime factors of a single actual F_n. It retains all original depths even when index multiplication changes a factor’s valuation inside F_n. It specializes to zero for some small indices; no positive witness is asserted when there are no packets.

GPF.4 Two channels force many distinct original exceptional primes

Let m=product_(q>5)q^v_q(n), assume m>1, and let ell be its smallest prime factor. Write Omega(m)=sum_q v_q(m) and omega(m)=#Supp(m). Let epsilon=1 for even n and zero for odd n.

Theorem GPF4 (large-index-prime channels). One has

$$\boxed{|U(n)|\ge (1+ϵ)\Omega (m)-\omega (m)+1.}\text{\hspace{1em}(GPF3)}$$

Moreover the distinct-prime mass satisfies the rational inequality

$$\boxed{\prod _{p\in U(n)}p\ge \frac{\ell }{\mathrm{rad}(m)}\prod _{q^a\parallel m}\prod _{j=1}^a(2q^j-1)(2q^j+1{)}^{ϵ}.}\text{\hspace{1em}(GPF4)}$$

Proof. Use only A packets with source q>5 and, if n is even, the B packets with these sources. There are (1+epsilon)Omega(m) disjoint packets. For each choose an odd-exponent prime. At most omega(m)-1 of these targets divide n, since all exceed their source and ell cannot be an internal target. Every other target belongs to U(n), proving GPF3. At rank q^j the chosen A factor is at least 2q^j-1; at the B rank 2q^j it is at least 2q^j+1 by splitness. Thus the product of all chosen primes is at least the double product in GPF4. Their internal product is at most rad(m)/ell. Divide to obtain a lower bound on the selected external product, which is at most product_(p in U(n))p. This proves GPF4 without a probabilistic factor-size premise.

For an EVEN index this becomes

$$|U(n)|\ge 2\Omega (m)-\omega (m)+1,\prod _{p\in U(n)}p\ge \frac{\ell }{\mathrm{rad}(m)}\prod _{q^a\parallel m}\prod _{j=1}^a(4q^{2j}-1).$$

Original WSS consequence. If F_n is powerful, all these external odd-depth witnesses satisfy h_p>=3 odd, in the ORIGINAL Fibonacci sequence. For an even n, let b(n) count all primes p>5 dividing F_n with odd h_p>=3, including any internal ones. If m>1, then

$$\boxed{b(n)\ge 2\Omega (m)-\omega (m)+1\ge \Omega (m)+1.}\text{\hspace{1em}(GPF5)}$$

Thus b(n)<=r forces Omega(m)<=r-1. In particular, if m is squarefree with k prime factors greater than five, a powerful F_(2m) requires at least k+1 distinct odd-original-depth WSS primes. For m=q^a, q>=7, the required count is at least 2a. These are necessary consequences, not assertions that such a powerful value or exceptional prime exists.

The earlier PBC argument selected layers of a largest index prime; OSE also used two channels at a prime index. GPF uses all index primes simultaneously and subtracts exactly the possible absorption into n. One internal prime cannot absorb two independent rank packets. Merely selecting one witness per packet without this subtraction would be false.

GPF.5 Group extension and finite-support consumers

For clarity, the group conclusion also has its usual multiquadratic interpretation. Let K_n=Q(sqrt(C):C in P_n), and let J_n=Q(sqrt(p):p divides n). Square-class independence gives Gal(K_n/Q)=(Z/2)^E(n). In the composite with J_n the relative degree is

$$[K_n{J}_n:J_n]=2^{\dim {\pi }_n(V_n)}\ge 2^{E(n)-t(n)}.$$

Indeed adjoining the square roots of the index primes removes exactly their coordinate span in the rational square-class group. This is the standard multiquadratic Kummer correspondence. The quantitative input here is the actual packet rank computation GPF1. No conclusion about a p-adic first-lift zero follows from an abstract group order alone.

For the previous fixed-support set P(S), all n in P(S) are already known by GSE to divide 5R_(H(S)). Every such candidate must additionally satisfy E(n)-t(n)<=|S|. If n is even with m>1 it must also satisfy 2Omega(m)-omega(m)+1<=|S|. These are finite candidate filters involving n and its own prime ranks, prior to factoring F_n. They do not assume that S is the complete set of exceptional primes.

A concrete arithmetic check is n=182=2713. The four high-source packets are F_7=13, L_7=29, F_13=233, L_13=521. Thirteen is absorbed into n and has original depth ONE; its exponent in F_182 is two. The other three are distinct external odd-depth primes. Here t(n)=1 and dim pi_n(V_n)=3, attaining the packet-deficiency bound. These are non-WSS verification examples, not exceptional samples. At n=1274, the six high-source packets have internal capacity one, forcing at least five external odd-depth primes.

GPF.6 Source roles, formalization handoff, and scope

Ribenboim (2005), printed page 8, statements (3.4) and (3.5), supplies Fibonacci AND Lucas square-class rigidity. The Fibonacci statement was already used by GSE; the Lucas statement is additionally needed here. The source’s printed pages 5-6 give divisibility, gcd and rank identities. Its parsed statements were inspected in this pass. Requests for page-8 and page-9 screenshots failed; no successful visual-page check is claimed. The full square-class theorems are used as explicit literature inputs, not reproved in this continuation. No analytic averaging result is used.

The optional field interpretation in GPF.5 uses the classical Kummer correspondence: J. S. Milne, Fields and Galois Theory, version 5.10 (2022), Theorem 5.30, printed page 75, https://www.jmilne.org/math/CourseNotes/FT.pdf . The author-hosted text and that page image were inspected successfully. The theorem is used only at exponent two over Q. The packet-rank and absorption estimates are proved here; they are not assertions attributed to Milne.

Suggested local formalization order: (i) Lucas prime-power cancellation and exact rank; (ii) the two-level fifth packet and its nonsquare premise; (iii) pairwise disjoint possible-rank sets and exact packet count; (iv) finite-support coordinate independence and kernel-dimension bound; (v) the smallest-source exclusion and distinct-prime product inequality. The finite forest proof offers an alternative to first formalizing the whole rational square-class quotient. Record nonsquare inputs explicitly, as OriginalOddDepthSupport already does for its prime-index input.

This is a continuation of the same owner and GSE square-class method. No new Lean declaration, freeze, atom coverage, CI run, or independent model review is claimed. Global priority for the packet/forest bounds is unestablished. No new WSS prime, previously undecided prime-family resolution, or elimination of a single B_j=P^2Q^3 pattern is obtained. The assumptions of GPF5 remain visible. GPF1-GPF4 give unconditional statements about actual original odd-depth supports, and GPF5 gives a new simultaneous necessary budget for any proposed powerful value.

GDR. Higher dyadic rank channels and source-prime collision repair

GDR.0 Fixed original depths and the additional obstruction

Retain the original Fibonacci and Lucas sequences and the definitions rho(p), h_p and U(n) from GPF. In particular

$$U(n)=\{p>5:p\mid F_n,p\nmid n,h_p\text{ odd}\}.$$

Let a=v_2(n), let m=product_(q>5)q^v_q(n), and assume m>1. Set b_q=v_q(m), k=omega(m), Omega(m)=sum_q b_q, and ell=min Supp(m). All q in the notation below are primes greater than five. No initial depth is set equal to one. GPF used ranks q^s and, for even n, 2q^s. The new construction retains all available ranks 2^t q^s, 1<=t<=a.

A genuine new obstruction appears in those higher rows: the source prime q itself can divide the old Lucas value. For example

$$L_{28}/L_4=101521=7\cdot 14503,\rho (7)=8.$$

Seven here has old rank eight, not the putative new rank 56. Thus the unmodified quotient cannot be called a packet of pure rank 56. This refutes that cancellation argument, not every possible stronger count. Pairing two successive source-prime steps will remove this obstruction.

The exact valuation formulas and the full Fibonacci/Lucas square-class classifications used in GPF remain classical inputs. In particular the only nonsingleton Lucas index classes are {1,3} and {0,6}; none of the Lucas index pairs used below is exceptional. The source statement is Ribenboim (2005), (3.5), with the same attribution as in this note.

GDR.1 The local order calculation identifies every colliding row

For t>=1 and s>=1 put

$$Q_{t,q,s}=\frac{L_{2^{t-1}{q}^s}}{L_{2^{t-1}{q}^{s-1}}},{\delta }_t(q)=1_{\rho (q)=2^t}.$$

Lemma GDR1 (collision classification). The quotient is a positive integer greater than one. Its factors have exactly this description:

$$\boxed{v_q(Q_{t,q,s})={\delta }_t(q),}\text{\hspace{1em}(GDR1)}$$

and every other prime p dividing it is greater than five, has rho(p)=2^t q^s, and occurs with exponent h_p. No other old prime factor survives. Conversely every p>5 of rank 2^t q^s divides the quotient. There is at most one t with delta_t(q)=1.

Proof. Odd-multiplier Lucas divisibility gives integrality. If an odd p divides L_d, then p does not divide F_d, since their common odd factor would divide four. From F_(2d)=F_d L_d, the rank of p divides 2d but not d; hence v_2(rho(p))=v_2(2d). This also proves the converse: if rho(p)|2d but rho(p) does not divide d, then p|L_d.

For d=2^(t-1)q^(s-1), an old odd p divides L_(dq) and not F_(dq), because multiplication by odd q does not change the two-part of d. The original Fibonacci valuation law therefore gives

$$v_p(L_{dq})-v_p(L_d)=v_p(q).$$

So old factors cancel except possibly q, which contributes exactly one. Since rho(q) is prime to q and divides q-(5/q), membership of q in L_(2^(t-1)q^s) is equivalent to rho(q)=2^t. When that equality fails q is absent at all s, and when it holds q is already present at s=0. This proves GDR1. A new rank divides 2^t q^s, has two-part 2^t, and cannot have smaller q-exponent, since then it would occur in the old term. Thus its rank is exactly 2^t q^s. The corresponding index is prime to p: the only possible index primes are two and q, both excluded. Its exponent is consequently h_p. Two is absent from all Lucas terms here since their indices are prime to three. The old factor three can occur only when t=2 and cancels; five divides no Lucas number. This accounts for all small primes. A prime has only one rank, proving the last assertion.

Lemma GDR2 (order and residue type). For any new factor of rank r=2^t q^s, the Fibonacci pair period is r for t=1 and 2r for t>=2. At t=1 the prime splits in Q(sqrt(5)) and p=1 modulo r. At t>=2,

$$\boxed{\{\begin{array}{ll}p\equiv 1(\mathrm{mod}2r),& (5/p)=1,\\ p\equiv r-1(\mathrm{mod}2r),& (5/p)=-1.\end{array}}\text{\hspace{1em}(GDR2)}$$

Proof. Write d=r/2. The golden element phi^d has trace zero and norm (-1)^d modulo p. Its quadratic identity gives phi^r=-(-1)^d. Any return exponent is divisible by the rank r. For t=1, d is odd, so phi^r=1 and the period is r; the identity L_d^2-5F_d^2=-4 gives splitness and r|p-1. For t>=2, d is even, phi^r=-1 and the period is 2r. In the split case it divides p-1. In the inert case phi^(p+1)=-1 and r|p+1, so (p+1)/r is odd. This gives both residue classes. The faithful matrix bridge transfers the golden element’s order to the pair period. These are classical rank/order consequences, not new general prime-period theorems.

In particular, a new factor is at least 2q^s+1 for t=1, and at least 2^t q^s-1 for t>=2. Each is strictly greater than its source q.

GDR.2 Repair by an even valuation shift and construct disjoint packets

For each q^b_q||m include the original Fibonacci packets

$$A_{q,s}=F_{q^s}/F_{q^{s-1}},1\le s\le b_q.$$

For each 1<=t<=a with delta_t(q)=0, include every Q_(t,q,s), 1<=s<=b_q. For the possible colliding row delta_t(q)=1, include instead

$$P_{t,q,j}=\frac{L_{2^{t-1}{q}^{2j}}}{q^2{L}_{2^{t-1}{q}^{2j-2}}},1\le j\le \lfloor b_q/2\rfloor .$$

Their possible-rank sets are respectively {q^s}, {2^t q^s}, and {2^t q^(2j-1),2^t q^(2j)}. All these sets are disjoint. Let the collection be D(n). Write

$$c_a(q)=1_{\rho (q)=2^t\text{ for some }1\le t\le a},C_a(m)=\sum _{q\mid m}{c}_a(q)\lceil \frac{b_q}{2}\rceil .$$

Theorem GDR3 (complete packet construction). Every member of D(n) is a nonsquare positive integer greater than one dividing F_n. Its prime factors exceed five and its prime exponents are their original h_p. A prime divides it exactly when its rank belongs to its displayed rank set. Distinct members are coprime, and

$$\boxed{|D(n)|=E_a(m)=(a+1)\Omega (m)-C_a(m).}\text{\hspace{1em}(GDR3)}$$

Proof. The A statements are the previous GPF lemma. In a noncolliding Lucas row GDR1 gives the exact new factors. If the quotient were square, its two distinct Lucas indices would lie in the same square class; they are not {1,3} or {0,6}, a contradiction.

In a colliding row, multiplying the quotients at s=2j-1 and s=2j contributes exactly q^2. Division by q^2 removes the only old factor. Thus P is an integer with precisely the two displayed new rank sets, with exponents h_p. If P were square then

$$\frac{L_{2^{t-1}{q}^{2j}}}{L_{2^{t-1}{q}^{2j-2}}}=q^{2}P$$

would be square. Its two Lucas indices differ by the factor q^2 and are not an exceptional pair. This proves nonsquareness and positivity. The integer P divides its numerator, which divides F_(2^t q^(2j)), which divides F_n. The converse rank statements follow from GDR1. Disjoint rank sets give coprime full supports. A noncolliding source contributes (a+1)b_q packets; a colliding one replaces b_q packets in one row by floor(b_q/2), losing ceil(b_q/2). Summing proves GDR3.

The repair is cancellation in the group of positive rational numbers modulo squares: two forced q factors vanish together. The argument does NOT prove that Q_(t,q,s)/q is individually nonsquare in a colliding row. A final unpaired layer is deliberately omitted.

GDR.3 Project away the index and count the remaining independent directions

Let R(n) be the union of the packet rank sets, and define

$$I(n)=\{p:p\mid m,\rho (p)\in R(n)\},t(n)=|I(n)|.$$

To keep multiple internal primes in the same packet from being counted twice, also set

$$b(n)=\#\{D\in D(n):\text{some }p\mid m\text{ has }\rho (p)\in R(D)\}.$$

Both capacities need only the factorization of n and ranks at its index primes. No factorization of F_n is needed. They satisfy b(n)<=t(n). These symbols refer to finite capacities, not GIR’s normalization index.

Theorem GDR4 (simultaneous original-depth bound). In the positive rational square-class group G, let V be spanned by {[D]:D in D(n)} and let pi delete coordinates at primes dividing n. Then

$$\boxed{\dim V=E_a(m),E_a(m)-b(n)\le \dim \pi (V)\le |U(n)|,}\text{\hspace{1em}(GDR4)}$$

and in particular

$$\boxed{|U(n)|\ge (a+1)\Omega (m)-C_a(m)-\omega (m)+1.}\text{\hspace{1em}(GDR5)}$$

Proof. Every packet is nonsquare and the full prime supports are disjoint, so their square classes are independent. Projection preserves disjointness of the remaining supports. A packet untouched by every index prime has a nonzero projected class. There are at least E-b such packets, and their projected classes are still independent. Every nonzero retained coordinate is at p>5, p not dividing n, with odd original h_p and p|F_n, hence lies in U(n). This proves GDR4.

There are k index primes greater than five. The least one ell cannot belong to I(n): all packet factors strictly exceed their source, which is itself at least ell. Thus t(n)<=k-1. Combining b<=t with GDR4 proves GDR5. This capacity subtraction concerns one common actual F_n; it does not assume statistical independence or choose different indices for different witnesses.

Equivalently, form the multiquadratic extension generated by the square roots of the packets. Its degree is 2^E. After adjoining square roots of the index primes, its relative degree is 2^dim(pi(V)), hence at least 2^(E-b). This is the standard square-class/Kummer interpretation of the proved independent directions; it is not itself a WSS existence theorem.

GDR.4 A sharp three-channel theorem and the first repaired higher row

No prime q>5 has rank two or four. Hence C_a(m)=0 for a<=2. Specializing GDR5 to a=2 gives

$$\boxed{v_2(n)=2,m>1⟹|U(n)|\ge 3\Omega (m)-\omega (m)+1.}\text{\hspace{1em}(GDR6)}$$

If merely 4|n, apply this bound to the packets already dividing F_(4m), keeping projection relative to the ORIGINAL n. Their factors exceed five, so the extra factors two, three and five in n cannot absorb them. The same bound GDR6 thus holds for every n divisible by four. It is attained at n=28: F_28=31329*281, and its external odd-original- depth set is exactly {13,29,281}. This is an equality example for the count, not a WSS example or an asymptotic-optimality assertion.

For a=3 the only new colliding source is seven: a prime of rank eight must divide F_8=3*7, and only seven has rank eight. Therefore

$$\boxed{v_2(n)=3⟹|U(n)|\ge 4\Omega (m)-\lceil v_7(m)/2\rceil -\omega (m)+1.}\text{\hspace{1em}(GDR7)}$$

For instance n=392=8*7^2 has seven repaired packets, with no internal packet factor possible because seven is its only large index prime. Thus |U(392)|>=7, whereas the preceding two-channel GPF bound gives four. This deduction uses exact rank and nonsquare inputs, not a complete factorization of F_392.

At n=364=4713 the six packets are

$$13,29,281,233,521,90481.$$

Thirteen is absorbed by the index, while the other five displayed primes are external, with original depth one. The packet projection has rank five. The actual F_364 was not completely factored in this verification; these five primes are not claimed to be its full external support.

GDR.5 Distinct-prime mass and the direct powerful-Fibonacci consequence

Attach a lower bound w(D) to each packet as follows:

$$w(A_{q,s})=2q^s-1,w(Q_{1,q,s})=2q^s+1,$$

$$w(Q_{t,q,s})=2^t{q}^s-1(t\ge 2),w(P_{t,q,j})=2^t{q}^{2j-1}-1.$$

A colliding row always has t>=3, so the last formula uses the correct higher-row bound. By GDR2 and the odd-rank bound for A, every prime factor of a packet is at least its indicated w(D). Consequently

$$\boxed{\prod _{p\in U(n)}p\ge \frac{\ell }{\mathrm{rad}(m)}\prod _{D\in D(n)}w(D).}\text{\hspace{1em}(GDR8)}$$

Proof. Choose one odd-exponent prime from each packet. These primes are distinct. Their product is at least the product of w(D). The product of those dividing n is at most rad(m)/ell, by the smallest-source argument. Dividing gives the stated bound on the selected external product, which is at most the product over all of U(n).

In particular, the three-channel subcollection at every 4|n gives

$$\boxed{\prod _{p\in U(n)}p\ge \frac{\ell }{\mathrm{rad}(m)}\prod _{q^e\parallel m}\prod _{s=1}^e(4q^{2s}-1)(4q^s-1).}\text{\hspace{1em}(GDR9)}$$

Let B_o(n) count all p>5 dividing F_n with odd original h_p>=3, including those dividing n. If F_n is powerful, then U(n) is a subset of that set: v_p(F_n)=h_p for p not dividing n, and an odd exponent at least two is at least three. Thus all count and mass bounds apply with these genuinely original exceptional witnesses. In particular

$$\boxed{4\mid n,m>1,F_n\text{ powerful}⟹B_o(n)\ge 3\Omega (m)-\omega (m)+1\ge 2\Omega (m)+1.}\text{\hspace{1em}(GDR10)}$$

Therefore B_o(n)<=r forces Omega(m)<=floor((r-1)/2). A powerful F_(4m) with m squarefree and k primes all greater than five requires at least 2k+1 different odd-depth WSS primes. If m=q^e, it requires at least 3e. The old five-smooth classification also yields: any positive powerful Fibonacci index divisible by four, other than twelve, requires at least three such primes. None of these implications supplies a powerful counterexample or asserts that even one WSS prime exists.

For the earlier fixed-support set P(S), its candidates still lie among divisors of 5R_(H(S)). They must now additionally satisfy E_a(m)-b(n)<=|S|, and at 4|n must satisfy 3Omega(m)-omega(m)+1<=|S|. These are factorization- of-index filters before computing or factoring the large Fibonacci value.

GDR.6 Repository, source, formalization and verification boundaries

The inspected latest branch already contains GPF and the local matrix- period, prime-to-index-depth and cubic-order work. The separate repository atom 0641c337c4e10a26e833dfb22c40d0e8cbf85fa13ce1513271780c97955ea5cd records gcd(L_(2^i H),L_(2^j H))|2. That older dyadic disjointness statement is compatible with this proof; it does not remove the source-prime factor when an odd multiplier q is introduced within ONE such row. GDR proves that collision classification, its paired repair, and the original-depth budget for the enlarged collection. Earlier GPF bounds remain valid.

Primary source roles: Ribenboim, FFF (2005), (3.4)-(3.5), supplies the Fibonacci and Lucas square-class classifications. The parsed statements were re-read in this pass; the requested image of page 8 failed. The classifications remain explicit prerequisites, not new theorem claims. Medina-Rowland’s Theorem 1.4 supplies the classical original valuation formula already used in GPF. Ballot-Elia, Rank and period of primes in the Fibonacci sequence. A trichotomy, Fibonacci Quarterly 45 (2007), 56-63, DOI 10.1080/00150517.2007.12428243, is related classical context; only its publisher abstract was read here, and no unread numbered theorem is imported. GDR2 is proved directly above from trace, norm and the existing golden Frobenius identities. No first general order theorem or established global priority for the particular new bounds is claimed.

For concurrent formalization, the useful order is: the precise Lucas zero-rank two-part; the source-prime valuation increment; paired removal of q^2; the explicit disjoint rank sets; the finite witness injection and smallest-source exclusion; then GDR6 and GDR10. The forest/injection proof can precede construction of the full rational square-class quotient. Keep the Lucas nonsquare premise explicit until its classical proof is available. Pairing does not require cubic reciprocity or the maximal order of the auxiliary pure-cubic field.

The companion verifier checks the new packets with original ranks and valuations, source-prime leakage, disjointness, independent finite root- of-unity order checks and the count/mass formulas. Complete factorizations of F_n are restricted to n<=60. Larger chosen cases factor only the small packets; paired high-rank cases use exact quotient, gcd and integer square-root checks without complete factorization. No finite computation substitutes for the general proof. This continuation changes no Lean or Scribe theorem, registration, freeze or CI status. It decides no new WSS prime family and does not exclude the entire P^2Q^3 golden-block branch.