Selfridge attack: complete closure of the complementary f=0 cancellation locus

6 September 2026. Primary mode: ATTACK.

1. Outcome and exact scope

This packet continues the normalized primitive proper three-addition attack on Selfridge’s integer-complexity question

$$\lVert 2^n\rVert\stackrel{?}=2n.$$

In the complementary $f=0$ branch, write

$$RS+3^g=2^K, \tag{1}$$

where each factor is of one of the canonical types

$$\mathrm I:1+2^x3^v, \qquad \mathrm {II}:2^x+3^v. \tag{2}$$

The preceding attack proved that the quadratic-reciprocity gate leaves forty ordered canonical placements and that exactly twenty tied-minimum strata can cancel in their first nonzero 3-adic digit. It completely classified the two triple-minimum strata.

The new result is:

Theorem. The eighteen remaining pair-minimum cancellation strata contain exactly one solution,

$$\boxed{(4+3)(1+8\cdot3)+3^4=2^8}. \tag{3}$$

Its displayed construction costs

$$29=2\cdot8+13 \tag{4}$$

ones. Together with the preceding triple-minimum theorem, every possible lowest-digit cancellation in the $f=0$ branch is now classified: the exceptional locus consists of (3) and

$$(2+3)(1+8\cdot3)+3=2^7, \tag{5}$$

whose displayed costs exceed $2K$ by thirteen and four ones, respectively. Every other $f=0$ solution obeys the universal valuation law

$$\boxed{\min(v,y,g)=1+v_3(K-r-u)}. \tag{6}$$

Here $2^r$ and $2^u$ are the constant binary terms in the ordered expansions of $R$ and $S$.

This is an exact theorem for one restricted formula grammar. It does not classify all solutions of (1), does not prove the cost inequality for every normalized three-addition presentation, and does not resolve Selfridge’s unrestricted problem.

2. Canonical placements and the cancellation criterion

Write

$$R=2^r+2^s3^v, \qquad S=2^u+2^w3^y. \tag{7}$$

For type I, $(r,s)=(0,x)$; for type II, $(r,s)=(x,0)$, and similarly for $(u,w)$. The complete residues of the canonical factors modulo 24 and the inherited Jacobi-symbol identities leave exactly forty ordered placements.

Subtracting the constant product term from (1) gives

$$2^{r+u}(2^{K-r-u}-1) =2^{r+w}3^y+2^{s+u}3^v+2^{s+w}3^{v+y}+3^g. \tag{8}$$

The gate makes $K-r-u$ even. LTE therefore gives

$$v_3(2^{K-r-u}-1)=1+v_3(K-r-u). \tag{9}$$

Let $m=\min(v,y,g)$, and put

$$A=(-1)^{r+w},\qquad B=(-1)^{s+u}. \tag{10}$$

The coefficient of the first nonzero 3-adic digit on the right of (8) is the sum of the applicable entries among $B,A,1$. It can vanish modulo three only in the following tied masks:

$$\begin{array}{c|c} \text{minimum mask}&\text{necessary condition}\pmod3\\ \hline v=y&B+A=0\\ v=g&B+1=0\\ y=g&A+1=0\\ v=y=g&A+B+1=0. \end{array} \tag{11}$$

Exact substitution into the forty placements produces twenty cancellation strata: eighteen with exactly two coordinates at the minimum and two with all three. No $v=g<y$ pair occurs; the eighteen pair strata use either $v=y<g$ or $y=g<v$.

If (11) does not hold, the right side of (8) has valuation exactly $m$, and (6) follows from (9). Thus classifying the twenty exceptional strata is enough to establish (6) outside their actual solutions.

3. A complete four-prime quotient for the eighteen pair strata

Use the squarefree modulus

$$Q=7\cdot73\cdot5\cdot13=33215. \tag{12}$$

Its relevant exact periods are

$$\operatorname{ord}_Q(2)=36, \qquad \operatorname{ord}_Q(3)=12. \tag{13}$$

Write $x_1,x_2,x_o,x_e$ for an exponent equal to one, equal to two, odd and at least three, or even and at least four. For a pair stratum let $q$ denote the third ternary exponent, so $q>m$. A quotient state below is

$$(x_S,m,q,K)\pmod{(36,12,12,36)}. \tag{14}$$

The parity triple records $(m,q,K)$, with zero for even and one for odd. Zero in a residue state is only a congruence representative; positivity and the strict inequality $q>m$ are imposed at the next stage.

$$\begin{array}{c|c|c|c|c|l} j&R&S&\text{minimum}&(m,q,K)&Q\text{-states}\\ \hline 0&I,x=1&I,x_o&y=g&(0,1,0)&(3,0,1,6),(3,0,11,4)\\ 1&I,x=1&I,x_e&v=y&(1,0,0)&(2,11,10,2),(18,3,0,12)\\ 2&I,x=2&I,x_1&y=g&(0,1,0)&(1,2,1,8),(1,10,1,4)\\ 3&I,x=2&I,x_o&v=y&(1,1,0)&\varnothing\\ 4&I,x=2&I,x_o&y=g&(1,1,0)&\varnothing\\ 5&I,x=2&I,x_1&v=y&(0,0,0)&(1,0,0,4)\\ 6&I,x=2&I,x_1&y=g&(0,0,0)&(1,0,0,4)\\ 7&I,x=2&I,x_o&y=g&(1,0,0)&\varnothing\\ 8&I,x=2&I,x_o&v=y&(0,1,0)&\varnothing\\ 9&II,x=1&I,x_e&v=y&(1,1,1)&\varnothing\\ 10&II,x=1&I,x_e&y=g&(1,1,1)&(4,9,1,3),(4,11,1,5)\\ 11&II,x=1&II,x_1&y=g&(0,1,0)&(1,0,1,4),(1,2,1,6)\\ 12&II,x=2&I,x_o&v=y&(1,0,0)&(3,1,4,8),(3,11,10,4),(11,11,6,0)\\ 13&II,x=2&I,x_o&y=g&(0,1,0)&(3,0,1,6),(5,0,3,10)\\ 14&II,x=2&I,x_1&v=y&(0,0,0)&(1,0,0,4),(1,2,2,8)\\ 15&II,x=2&I,x_1&y=g&(0,0,0)&(1,0,0,4),(1,0,4,8), (1,2,2,8),(1,10,2,4)\\ 16&II,x=2&I,x_o&y=g&(1,0,0)&\varnothing\\ 17&II,x=2&I,x_o&v=y&(0,1,0)&\varnothing. \end{array} \tag{15}$$

Thus seven complete families are already empty and the other eleven contain only twenty-three quotient states.

4. The next 3-adic digit leaves one state

Modulo 27, $3^e$ is nonzero only for $e=1,2$. For a residue class $e\pmod {12}$:

• if the residue is one or two, the exponent may be that exact low value or a higher representative whose power is zero modulo 27;
• every other positive representative is necessarily high and contributes zero modulo 27.

For the ordered pair $m<q$, the feasible low/high patterns are:

$$\begin{array}{c|c} (m,q)&\text{feasibility}\\ \hline (\mathrm L,\mathrm L)&q>m\text{ at the two exact values}\\ (\mathrm L,\mathrm H)&\text{always}\\ (\mathrm H,\mathrm L)&\text{never}\\ (\mathrm H,\mathrm H)&\text{always, after choosing a sufficiently high }q. \end{array} \tag{16}$$

Substitution of every feasible pattern for every state in (15) is a complete finite calculation. Exactly one survives:

$$j=12,qquad (x,m,g,K)\equiv(3,1,4,8)\pmod{(36,12,12,36)}, \tag{17}$$

and the surviving pattern has $m$ low and $g$ high. In particular, $m=1$ exactly. The surviving equation is therefore

$$7(1+3\cdot2^x)+3^g=2^K, \tag{18}$$

with $x\ge3$ odd, $g\equiv4\pmod {12}$, and $K$ even.

5. Elementary terminal factorization

The left side of (18) is greater than $2^x$, so $K>x$. Reducing modulo $2^x$ gives

$$2^x\mid 7+3^g. \tag{19}$$

Because $4\mid g$, one has $3^g\equiv1\pmod {16}$, hence

$$v_2(7+3^g)=3. \tag{20}$$

It follows that $x\le3$, and the canonical lower bound forces $x=3$. Equation (18) becomes

$$2^K-3^g=175. \tag{21}$$

Both exponents are even, so

$$(2^{K/2}-3^{g/2})(2^{K/2}+3^{g/2})=175. \tag{22}$$

The positive factor pairs of 175 are

$$(1,175),\quad(5,35),\quad(7,25). \tag{23}$$

They give $(2^{K/2},3^{g/2})=(88,87),(20,15),(16,9)$. Only the last pair consists of the required powers. Thus $(K,g)=(8,4)$, proving the uniqueness of (3).

The first factor $4+3$ costs seven ones, the second factor $1+8\cdot3$ costs ten, and $3^4$ costs twelve. Their sum is 29, proving (4).

6. Synthesis: a universal f=0 valuation law with two controls

The preceding triple-minimum certificate proves that one of the two triple families is empty and the other contains only (5). Sections 3–5 prove that seventeen of the pair families are empty and the last contains only (3). Consequently every possible cancellation described by (11) is now resolved.

For any other solution, the lowest coefficient on the right of (8) is nonzero modulo three. Its valuation is therefore $m$, and comparison with (9) proves (6). This is the exact $f=0$ analogue of the earlier universal valuation laws for the complementary $f=1$ and $f=2$ branches, except for the two explicit harmless controls.

7. Independent evidence

The reproducible archive contains:

1. a primary Python certificate that reconstructs the forty placements, twenty cancellation strata, eighteen pair families, complete quotient table (15), low/high calculation (16), terminal factorization, and cost;
2. a separately written C++20 implementation that independently reconstructs the same canonical enumeration and all exact conclusions;
3. a diagnostic of the most obvious attempted continuation to the full $f=0$ remainder.

Both exact certificates report:

$$40\longrightarrow20\longrightarrow18,qquad 18\longrightarrow 7\text{ empty}+23\text{ states} \longrightarrow1\text{ control}. \tag{24}$$

No bounded search is used for completeness.

8. Useful failure and stopping audit

After proving (6), the same fixed quotient was applied to all forty $f=0$ placements, without assuming a cancelling minimum. It leaves 149 residue states. Lifting the binary coordinates to the period 972 required by modulo $3^6=729$, and treating each ternary coordinate exactly as low $(1\le e\le5)$ or high $(e\ge6)$, leaves 2,032 modular patterns. Imposing the new valuation law (6), while retaining the two exceptions, still leaves 1,953 patterns across twenty placements.

This diagnostic is not a theorem that the remaining equations have solutions. It proves only that the successful next-digit mechanism does not extend as a uniform finite closure: outside the exceptional cancellation locus it branches instead of freezing. A higher digit or another prime would enlarge coordinate periods before any known height bound makes the quotient finite.

The completed attack therefore wraps up the exceptional-cancellation phase, but not the normalized complementary family. The unresolved obligations are:

1. the nonexceptional $f=0$ strata governed by (6);
2. the higher minimum layers $m\ge3$ in the valuation-controlled $f=1,2$ branches;
3. most importantly, any reduction from arbitrary integer-complexity formulas to the normalized three-addition grammar.

Evertse–Schlickewei–Schmidt gives finiteness of the normalized complementary family but no effective height cutoff. Bajpai and Bennett give effective height bounds for homogeneous five-term S-unit equations when the set of places has size at most three; the expanded equation here has six homogeneous terms (five positive monomials and the power of two on the other side), so that theorem does not directly supply the missing cutoff.

The evidence does not support continuing layer-by-layer merely to enlarge periods. A further Selfridge attack is warranted only if it targets a genuine structural bridge: an effective reduction of this six-term product-structured S-unit equation, a uniform dominance/freezing lemma, or a minimal-counterexample normal-form theorem connecting the restricted grammar to arbitrary formulas.

9. Status and recommendation

This is an exact, independently reconstructed computer-assisted restricted theorem. It has no new external theorem dependency beyond the inherited quadratic gate and LTE. It has not received external review, and novelty has not been assessed.

The promised eighteen-stratum target is complete, but neither the normalized three-addition family nor Selfridge’s conjecture is closed. The decisive current attack route has reached a clean stopping boundary.

RECOMMENDED NEXT MODE: HUNT. Preserve Selfridge as a paused program and search broadly for a new target. Reopen it only for a high-level structural idea of the kind listed above, not for another minimum layer or a larger bounded quotient.

10. Sources and dependency boundary

• The quadratic-reciprocity gate and canonical modulo-24 table are inherited FPRD results in this restricted program.
• LTE supplies (9).
• Evertse, Schlickewei, and Schmidt, Linear equations in variables which lie in a multiplicative group, Annals of Mathematics 155 (2002), 807–836, supplies the previously recorded ineffective solution-count bound.
• Prajeet Bajpai and Michael A. Bennett, Effective S-unit Equations Beyond 3 Terms: Newman’s Conjecture (2023), supplies the five-homogeneous-term comparator discussed in Section 8; it is not a proof dependency for the new theorem.