Integer complexity · sustained attack

Restricted attacks on Selfridge's powers-of-two problem

Selfridge asked whether a power of two ever has a shorter expression than the obvious product of twos. The current FPRD attack studies a factored family with three additions. Several infinite branches are now completely excluded, including the exceptional 3-adic lift that previously survived beyond exponent107010^{70} and the complete type-II2 first-layer family. A two-modulus contradiction now closes the last type-II1 family, completing thea=1a=1 layer of this canonical branch. A valuation-informed modulus-271 quotient also closes the first serious a=2a=2family, and moduli 73 and 577 close its last canonical family. Thus both a=1a=1 anda=2a=2 are complete in this branch. At a=3a=3, exact gate quotients now exclude every entry. A paired-prime argument then eliminates every layer a≥4a\ge4at once for both canonical positive-factor types. These are exact restricted results. In the complementary positive-type-II branch, a complete quotient now closes a=3a=3, a coupled-order argument forces x≤10x\le10at every remaining low layer, and the resulting 21 periodic rows at a=1,2a=1,2 are now completely classified by the full 2-adic valuation gate and one aligned modulus. Exactly ten control presentations remain, all costing at least 2N+42N+4. In the complementary a=0<ga=0<gorientation, a quadratic-reciprocity gate now eliminates half of the canonical residue pairings and completely excludes thef=2f=2 branch where both inner sums contribute one factor of two. In the three survivingf=1f=1 rows, moduli 7 and 73 exclude every tied-minimum 3-adic cancellation, yielding one exact valuation law for all remaining solutions. The alternativef=2f=2 split now has the same structural conclusion: ten possible cancellations are eliminated by complete periods modulo 7, 73, 487, and 2593, forcing its own exact valuation law. Combining the two laws at minimum layersm=1,2m=1,2 leaves exactly 40 periodic lift graphs. The next ternary digit, two 2-adic gates, Matveev's explicit bound, and a finite modular descent now prove that those graphs contain exactly seventeen proper controls, every one costing at least2N+62N+6. In the remaining f=0f=0branch, all twenty exceptional lowest-digit cancellation strata are now classified. They contain exactly two harmless controls, and every other solution obeys one universal 3-adic valuation law. This is not a proof of Selfridge's conjecture.

FPRD-IC-O01 · open problem

Can any formula beat the product of twos?

The integer complexity ∥n∥\lVert n\rVertis the least number of occurrences of 1 in an expression fornn using only addition, multiplication, and parentheses. Since

2n=(1+1)(1+1)⋯(1+1), 2^n=(1+1)(1+1)\cdots(1+1),

one always has ∥2n∥≤2n\lVert2^n\rVert\le2n. Selfridge's question is whether equality always holds. It remains open; see the current general estimates of Konyagin and Oganesyan and the stability framework of Altman and Arias de Reyna.

FPRD-IC-D01 · restricted grammar

A three-addition fork that preserves factored cost

The active family has the form

M[A(B+C)(D+E)+F]=2n, M\bigl[A(B+C)(D+E)+F\bigr]=2^n,

where every displayed operand is(2,3)(2,3)-smooth. The paid atoms 1, 2, and 3 cost 1, 2, and 3 ones. Multiplication is free once those atoms have been built, but all three displayed additions are real operations; there is no reuse, subtraction, or division. The exact displayed cost retains the special charge for a unit operand.

Common factors of (B,C)(B,C) and(D,E)(D,E) can be transferred toAA without increasing cost. A common exterior power of two can likewise be transferred toMM. A potentially cheaper expression may therefore be assumed primitive with

gcd⁡(B,C)=gcd⁡(D,E)=1,3a(B+C)(D+E)+2f3g=2N,min⁡(a,g)=0. \gcd(B,C)=\gcd(D,E)=1, \qquad 3^a(B+C)(D+E)+2^f3^g=2^N, \qquad \min(a,g)=0.

The fork is called proper when both inner sums are nonsmooth. Smooth inner sums can be rebuilt at no greater cost and reduce to the already classified two-addition family.

FPRD-IC-T01 · normalization theorem

Only three final valuations survive

For a coprime smooth pair, the 2-adic valuation of its sum is 0, 1, or 2. Equality of the two summands' 2-adic valuations in a sum equal to a power of two first givesf≤4f\le4. A complete periodic analysis of the cases f=3,4f=3,4leaves six residue rows, all contradicted modulo 9, 27, 64, 81, or 271. Hence every normalized primitive proper fork satisfies

f≤2. \boxed{f\le2}.

There is also a purely elementary mod-eight obstruction. For oddxx, putχ(x)=+1\chi(x)=+1 on residues 1 and 3 and χ(x)=−1\chi(x)=-1 on residues 5 and 7. If P0,Q0P_0,Q_0 are the odd parts of the two nonsmooth inner sums, then

χ(P0)χ(Q0)=−1. \boxed{\chi(P_0)\chi(Q_0)=-1}.

Thus equal nonsmooth factors and every same-character family are impossible before any size estimate is used.

FPRD-IC-T02 · closed infinite branch

The cyclotomic ladder has no nonsmooth rung

One f=0f=0 branch reduces to

3aP=22rx−12x+1,r≥2. 3^aP=\frac{2^{2rx}-1}{2^x+1}, \qquad r\ge2.

Exact valuation and character constraints reduce the ladder to five fixed exponential equations and one periodic family. The fixed equations fail modulo 163, 37, 7, 757, and 271. In the last family, a complete residue certificate modulo 7681 forces its parameter to be odd, while a complete certificate modulo 8641 forces it to be even. Therefore no canonical nonsmooth solution remains for r≥2r\ge2.

The boundary reductions inherit one use of Bennett's theorem on Pillai-type equations. The final six modular contradictions are independently checkable finite quotients, not bounded searches in the exponents.

FPRD-IC-T03 · closed odd branch

The odd-exponent branch has one harmless control

After the cyclotomic boundary, the two nonsmooth canonical shapes at odd NN can be classified completely. Their unique solution is

7⋅73+1=29. 7\cdot73+1=2^9.

The two relevant factored presentations cost 20 and 21 ones, respectively 2N+22N+2 and2N+32N+3. Neither threatens the conjectured 2N2N minimum. The proof combines multiplicative-order certificates modulo 7, 19, 27, and 73 with Zsigmondy's primitive-divisor theorem on one boundary. Consequently the complete odd-NN portion of this normalized branch satisfies the desired cost inequality.

FPRD-IC-T04 · intermediate even-layer reduction

A zero-exponent identity first appeared to lift remotely

In the first even layer, a=1a=1, the remaining exceptional equation is

3(4+3v)(1+16⋅3y)+1=2N. 3(4+3^v)(1+16\cdot3^y)+1=2^N.

It is a lift of the genuine boundary identity15⋅17+1=2815\cdot17+1=2^8. A complete cover modulo 5, 7, 13, 17, 19, 37, and 41 leaves exactly

(v,y,N)≡(0,0,8)(mod(576,144,360)). (v,y,N)\equiv(0,0,8)\pmod{(576,144,360)}.

Expanding the equation gives the exact identity

v3(2N−13)=min⁡(v,y)+1. v_3(2^N-13)=\min(v,y)+1.

The resulting simple Hensel root modulo31403^{140} proves that any positive solution must satisfy

N≥19403145444478990487147845757777065330721025558761886758514814368314768,v+y>1070. N\ge 19403145444478990487147845757777065330721025558761886758514814368314768, \qquad v+y>10^{70}.

This enormous lower bound was not a contradiction. It is retained because it records the exact input to the next theorem; the lift is no longer an active survivor.

FPRD-IC-T05 · closed exceptional lift

The first Hensel digit and nine residues close x=4

Put N=8+3240zN=8+3240z. Dividing the inherited valuation identity by353^5 and reducing its first Hensel digit modulo three gives

z≡1(mod3),N≡3248(mod9720). z\equiv1\pmod3, \qquad N\equiv3248\pmod{9720}.

Write v=576rv=576r andy=144sy=144s. Modulo 19441,

3576≡3144≡19199,191993≡1,24860≡1,23248≡15812. 3^{576}\equiv3^{144}\equiv19199, \quad19199^3\equiv1, \quad2^{4860}\equiv1, \quad2^{3248}\equiv15812.

Hence each power of three has only three possible residues. The nine possible left sides of the exceptional equation are

(25625918974730432736725124961264315951)(mod19441). \begin{pmatrix} 256&259&18974\\ 7304&3273&6725\\ 12496&12643&15951 \end{pmatrix}\pmod{19441}.

None is 15812. Therefore the entire positivex=4x=4 lift is empty. Primality of 19441 is not required; the displayed period identities suffice. Moreover, every hypothetical point on this branch would already have cost at most 2N−12N-1, so this is a genuine restricted counterexample family that has been ruled out, not merely a formal Diophantine curiosity.

FPRD-IC-T06 · intermediate large-x reduction

The large-x gates make every surviving point cost-threatening

Two families remain at the first even layer. Their 2-adic gates are

3v+1≡−7(mod2x)(II1),3v+1≡−13(mod2x)(II2). 3^{v+1}\equiv-7\pmod{2^x}\quad\text{(II1)}, \qquad 3^{v+1}\equiv-13\pmod{2^x}\quad\text{(II2)}.

Exact compatible lifts at the least surviving precisions give

v≥55313581in II1,v≥60254788in II2. v\ge55313581\quad\text{in II1}, \qquad v\ge60254788\quad\text{in II2}.

These are unbounded consequences of the gates, not search cutoffs. The defining equations also imply the desired strict cost saving automatically at such sizes. Thus every remaining equation solution would be an actual counterexample inside the normalized grammar; the problem is now existence rather than cost accounting.

There is a first exact reduction in II2. Ify<vy<v andy≥5y\ge5, thenv3(2N−13)=y+1v_3(2^N-13)=y+1 forcesN≡332(mod486)N\equiv332\pmod{486}. Combining that class with the periods modulo 19441 eliminates 17 of the 28 inherited classes forx(mod540)x\pmod{540}. The surviving necessary classes are

31,67,88,112,196,211,223,247,340,379,472. 31,67,88,112,196,211,223,247,340,379,472.

These eleven classes were only an intermediate quotient. The next theorem closes all of them, the other seventeen classes, the small values y=2,3,4y=2,3,4, and both relative orders of v,yv,y.

FPRD-IC-T07 · complete type-II2 closure

The entire first-layer type-II2 family is empty

Consider the remaining large-xx equation

3(4+3v)(1+2x3y)+1=2N, 3(4+3^v)(1+2^x3^y)+1=2^N,

under the inherited type-II2 conditions. Every allowedxx is 1 modulo 3, whileN≡8(mod18)N\equiv8\pmod{18}. A complete reduction modulo 7 forces 3∣y3\mid y, so y=2,4y=2,4 are impossible.

If y=3y=3, the exact 3-adic valuation fixes N≡8(mod9)N\equiv8\pmod9, and the 2-adic gate forces v≡0(mod4)v\equiv0\pmod4. Modulo 73, the nine possible left sides are

(23722610406152349), \begin{pmatrix} 23&72&26\\10&40&61\\52&3&49 \end{pmatrix},

whereas the right side is 37. Thusy=3y=3 is impossible as well.

For y≥5y\ge5, expanding gives

2N−13=3v+1+2x3y+1(4+3v). 2^N-13=3^{v+1}+2^x3^{y+1}(4+3^v).

Both terms are divisible by 363^6, independently of whether y<vy<v,y=vy=v, ory>vy>v. HenceN≡332(mod486)N\equiv332\pmod{486}. Exact finite quotients reduce the 28 allowedx(mod540)x\pmod{540} classes as

28→ mod 194414→ mod 64811→ mod 130. 28\xrightarrow{\bmod 19441}4 \xrightarrow{\bmod 6481}1 \xrightarrow{\bmod 13}0.

Together with the earlier x=4x=4theorem, this closes the complete a=1a=1type-II2 branch. The next theorem closes type-II1; exterior layers a≥2a\ge2 remain open.

FPRD-IC-T08 · complete first-layer closure

Two small moduli close the last a=1 family

The sole remaining equation was

3(2+3v)(1+2x3y)+1=2N, 3(2+3^v)(1+2^x3^y)+1=2^N,

with odd vv,N≡16(mod18)N\equiv16\pmod{18}, and sixteen inherited xxclasses modulo 270, all 2 modulo 6. Modulo 7 the complete quotient is

y mod 6v≡1v≡3v≡5062110512301356144315111(mod7). \begin{array}{c|ccc} y\bmod6&v\equiv1&v\equiv3&v\equiv5\\ \hline 0&6&2&1\\1&0&5&1\\2&3&0&1\\ 3&5&6&1\\4&4&3&1\\5&1&1&1 \end{array}\pmod7.

The required residue 2 occurs only at(v,y)≡(3,0)(mod6)(v,y)\equiv(3,0)\pmod6. Hence 3v≡3y≡1(mod13)3^v\equiv3^y\equiv1\pmod{13}. The left side modulo 13 is then in{0,7}\{0,7\}, while the allowed powers 2N2^N are{3,10}\{3,10\}. The sets are disjoint.

Therefore type-II1 is empty. With the earlier canonical closures, the complete a=1a=1layer of the active branch is closed. The unused 2-adic gate independently raises the hypothetical lower bound tov≥137494267053v\ge137494267053, but the modulus-13 contradiction makes that bound only a control.

FPRD-IC-B01 · next-layer method boundary

The coarse a=2 frontier defeats coprime-modulus covers

The first serious next-layer survivor is

9(1+4⋅3v)(1+2⋅3y)+1=2N. 9(1+4\cdot3^v)(1+2\cdot3^y)+1=2^N.

Modulo 7 and 13 first force the coarse classes

v≡5(mod6),y≡0(mod6),N≡6(mod36). v\equiv5\pmod6,\qquad y\equiv0\pmod6, \qquad N\equiv6\pmod{36}.

The equation also has a formal rational boundary at(v,y,N)=(−1,0,6)(v,y,N)=(-1,0,6). For any finite collection of moduli coprime to 6, take these three exponents modulo the corresponding multiplicative orders. The formal boundary then satisfies every chosen coarse congruence simultaneously. This blocks a direct repetition of the first-layer prime-cover method. It does not block the sharper 3-adic refinement used next.

FPRD-IC-T09 · first a=2 closure

A 3-adic lift breaks the formal boundary

Subtracting the boundary value 64 gives

2N−64=18(−3+3y+2⋅3v+4⋅3v+y). 2^N-64=18(-3+3^y+2\cdot3^v+4\cdot3^{v+y}).

On the coarse classes, the bracket has 3-adic valuation one. LTE therefore gives v3(N−6)=2v_3(N-6)=2, or equivalently

N≡42 or 78(mod108). N\equiv42\text{ or }78\pmod{108}.

This excludes the formal boundary from the admissible progression. Finally,ord⁡271(2)=135\operatorname{ord}_{271}(2)=135and ord⁡271(3)=30\operatorname{ord}_{271}(3)=30. The allowed exponent classes produce 25 distinct left residues and 10 right residues modulo 271, and those sets are disjoint. Hence the first serious a=2a=2type-I2 family has no solution.

L={0,13,17,32,64,100,103,105,117,138,140,151,157,160,161,{165,169,171,185,187,194,210,243,256,264},R={34,35,41,69,79,125,139,148,166,248},L∩R=∅. \begin{aligned} L={}&\{0,13,17,32,64,100,103,105,117,138,140,151,157,160,161,\\ &\phantom{\{}165,169,171,185,187,194,210,243,256,264\},\\ R={}&\{34,35,41,69,79,125,139,148,166,248\}, \qquad L\cap R=\varnothing. \end{aligned}

FPRD-IC-T10 · complete second-layer closure

Moduli 73 and 577 close the last a=2 family

The remaining equation was

9(4+3v)(1+2x3y)+1=2N,x=v2(37+3v+2), 9(4+3^v)(1+2^x3^y)+1=2^N, \qquad x=v_2(37+3^{v+2}),

with positive yy and oddvv. The second exact gate is

37+3v+2≡2x3y(mod22x). 37+3^{v+2}\equiv2^x3^y\pmod{2^{2x}}.

Modulo 27 first gives N≡6(mod18)N\equiv6\pmod{18}. The complete modulus-73 quotient has six rows; the primary gate removes three and leaves

v≡5(mod12),(x,y)≡(1,3),(4,11),(7,7)(mod(9,12)). v\equiv5\pmod{12},\qquad (x,y)\equiv(1,3),(4,11),(7,7)\pmod{(9,12)}.

These classes also givev3(2N−37)=min⁡(v,y)+2v_3(2^N-37)=\min(v,y)+2and N≡96(mod162)N\equiv96\pmod{162}.

For the final quotient, modulo 577 one has3=21053=2^{105} andord⁡577(2)=144\operatorname{ord}_{577}(2)=144. Set t=x+105yt=x+105y. The three surviving classes require t≡1,4,7(mod9)t\equiv1,4,7\pmod9. Solving the remaining equation for 2t2^tgives the complete logarithm table

N\v51729416−92−−24−−−8742−59−−60−11078378−−−−96−−−−114−−−−1325−−−. \begin{array}{c|rrrr} N\backslash v&5&17&29&41\\ \hline 6&-&92&-&-\\24&-&-&-&87\\42&-&59&-&-\\ 60&-&110&78&3\\78&-&-&-&-\\96&-&-&-&-\\ 114&-&-&-&-\\132&5&-&-&- \end{array}.

A dash means the quotient is not a power of two modulo 577. The seven existing logarithms occupy only{2,3,5,6}(mod9)\{2,3,5,6\}\pmod9, disjoint from the required classes. Therefore this family is empty. Since the preceding results exhaust the other canonical forms, the complete a=2a=2 layer of the active branch is closed.

FPRD-IC-T11 · third-layer entry theorem

The complete a=3 entry table is empty through x=20

The third exterior layer has the four canonical equations

27P(1+2x3y)+1=2N,P∈{1+2⋅3v,1+4⋅3v,2+3v,4+3v}. 27P(1+2^x3^y)+1=2^N, \qquad P\in\{1+2\cdot3^v,1+4\cdot3^v,2+3^v,4+3^v\}.

Modulo 81 gives the exact progressionsN≡18(mod54)N\equiv18\pmod{54} whenP≡1(mod3)P\equiv1\pmod3, andN≡36(mod54)N\equiv36\pmod{54} whenP≡2(mod3)P\equiv2\pmod3. The twox=1x=1 families fall modulo 13 and 73. Every entry with x≥3x\ge3obeys the exact second gate

27P+1≡2x3y(mod22x). 27P+1\equiv2^x3^y\pmod{2^{2x}}.

For the remaining low-bit roots, none of the possible values ofPP vanishes modulo 73. Since 2N≡1(mod73)2^N\equiv1\pmod{73}, the complete quotient forces

(x,y)≡(0,6),(3,2),(6,10)(mod(9,12)). (x,y)\equiv(0,6),(3,2),(6,10)\pmod{(9,12)}.

Thus 3∣x3\mid x. Exact gate-residue covers at the remaining values give

xI2II1II237193193641,7341,19419257,513193,1712257,193,17257,5,17257,41,715257,5193257,19318257,12289,5257,193257,17. \begin{array}{c|c|c|c} x&\mathrm{I2}&\mathrm{II1}&\mathrm{II2}\\ \hline 3&7&193&193\\ 6&41,73&41,19&41\\ 9&257,5&13&193,17\\ 12&257,193,17&257,5,17&257,41,7\\ 15&257,5&193&257,193\\ 18&257,12289,5&257,193&257,17. \end{array}

Each cell lists a complete successive cover, not a bounded exponent search. At x=18x=18, the I2 gate quotient contracts262144→8192→512→0262144\to8192\to512\to0; each II quotient contracts65536→2048→065536\to2048\to0. Consequently every survivor satisfies

x≥21,3∣x, x\ge21,\qquad3\mid x,

and the first 2-adic gate forcesv≥132267v\ge132267 in I2,v≥457519v\ge457519 in II1, andv≥736154v\ge736154 in II2. These values exceed the exact cost-saving thresholds, so every remaining solution would genuinely beat2N2N within this grammar. The next theorem closes all of these remote classes.

FPRD-IC-T12 · complete third layer

The remote a=3 tail is empty

Let p=262657p=262657. It is prime, and its relevant orders are

ord⁡p(2)=27,ord⁡p(3)=14592. \operatorname{ord}_p(2)=27, \qquad \operatorname{ord}_p(3)=14592.

Using the inherited classes for x,y,v,Nx,y,v,N, the complete quotient modulo ppchecks 623,808 residue triples for each surviving negative-factor shape. It leaves only

Px mod 27v mod 14592y mod 14592I23101559446II1949115682185167139821107997658248158122II266810124669370535024132105818. \begin{array}{c|c|c|c} P&x\bmod27&v\bmod14592&y\bmod14592\\ \hline \mathrm{I2}&3&10155&9446\\ \mathrm{II1}&9&4911&5682\\ &18&5167&1398\\ &21&10799&7658\\ &24&815&8122\\ \mathrm{II2}&6&6810&1246\\ &6&9370&5350\\ &24&13210&5818. \end{array}

Six rows fail modulo 7. For the other two, modulo 13 the possible left residues are {0,4,7,11}\{0,4,7,11\}and {3,5,6,8}\{3,5,6,8\}, while the right side lies in {1,12}\{1,12\}. Hence no remote row survives, and the completea=3a=3 positive-type-I layer is empty.

FPRD-IC-T13 · uniform exterior obstruction

Two primes eliminate every layer a at least four

An odd NN would forcea=0a=0. In the even branch,a=1+v3(N)a=1+v_3(N). Thusa≥4a\ge4 implies27∣N27\mid N. Modulo 73 and 262657 the right side is therefore one. Every canonical negative factor is nonzero at both primes, so each prime forces the positive factor QQ to vanish.

QQ=0(mod73)Q=0(mod262657)1+2x3y(0,6),(3,2),(6,10)(0,7296),(9,2432),(18,12160)2x+3y(0,6),(3,10),(6,2)(0,7296),(9,12160),(18,2432). \begin{array}{c|c|c} Q&Q=0\pmod{73}&Q=0\pmod{262657}\\ \hline 1+2^x3^y&(0,6),(3,2),(6,10)&(0,7296),(9,2432),(18,12160)\\ 2^x+3^y&(0,6),(3,10),(6,2)&(0,7296),(9,12160),(18,2432). \end{array}

The exponent moduli in the two columns are(9,12)(9,12) and(27,14592)(27,14592). Projecting the second column to the first gives no common row for either form ofQQ. This excludes alla≥4a\ge4 at once; no exterior-layer induction remains.

FPRD-IC-T14 · branch synthesis

The complete g=0 positive-type-I branch is closed

The cyclotomic and odd-exponent theorems handlea=0a=0. The complete first, second, and third-layer theorems handlea=1,2,3a=1,2,3, and the paired-prime obstruction handles every a≥4a\ge4. Therefore no normalized primitive proper fork in this entire positive-type-I branch has displayed cost below2N2N. This completes one infinite branch of the three-addition grammar, not the grammar itself.

FPRD-IC-T15 · complementary third-layer closure

The positive-type-II a=3 layer is empty

For the complementary positive factor, the remaining low-layer equation is

3aP(v)(2x+3y)+1=2N,a∈{1,2,3}. 3^aP(v)(2^x+3^y)+1=2^N, \qquad a\in\{1,2,3\}.

At a=3a=3, the exact 2-adic gate quotient modulo 2112^{11}contains 524,288 admissible states. Solution-preserving filters give the complete chain

524288→648116512→2571268→73390→7124→1930. 524288\xrightarrow{6481}16512\xrightarrow{257}1268 \xrightarrow{73}390\xrightarrow7 124\xrightarrow{193}0.

Each filter deliberately allows independent exponent lifts, so it only enlarges the genuine solution set. The empty intersection therefore proves that the complete complementarya=3a=3 layer has no solution.

FPRD-IC-T16 · uniform high-x obstruction

Every remaining low-layer solution has x at most ten

The exact gates are

a=1+v3(N),x=v2 ⁣(3a+yP(v)+1). a=1+v_3(N),\qquad x=v_2\!\left(3^{a+y}P(v)+1\right).

Independent prime filters leave false high-xx states because they may choose different exponent lifts. Coupling the samex,Nx,N residues at 17, 193, 257, and 12289 leaves only

(1,I2,222,60,1061,3394),(1,II2,260,64,2724,4192)(mod(512,512,6144,6144)). (1,\mathrm{I2},222,60,1061,3394),\qquad (1,\mathrm{II2},260,64,2724,4192) \pmod{(512,512,6144,6144)}.

Here the coordinates after the shape are(v,y,x,N)(v,y,x,N). Each row has 81 simultaneous lifts to periods(1536,1536,18432,18432)(1536,1536,18432,18432). None satisfies the equation modulo all of 7, 13, 73, 97, and 577. Hence every solution ata=1,2,3a=1,2,3 hasx≤10x\le10.

FPRD-IC-T17 · exact residual quotient

The complementary low layers reduce to twenty-one lift classes

A coupled exact quotient leaves onlyx∈{1,3,4,5,6,8}x\in\{1,3,4,5,6,8\} and the following rows. The columns v,y,Nv,y,Nare residues modulo 1536, 1536, and 18432.

aPvyxN
1I11535014
1I20048
1I20238
1II11048
1II11238
1II20048
1II20238
1II231310
1II232110
1II240816
2I110612
2I203612
2I211512
2I213312
2I21535016
2II113612
2II11535016
2II203612
2II210612
2II221512
2II223312

These are necessary lift classes, not solutions. A zero residue also allows positive exponents such asy=1536y=1536. A bounded search through v,y≤384v,y\le384 finds ten control presentations, representing four factorizations, all with displayed cost at least2N+42N+4. The table was the exact finite input to the later second-gate classification below; its nonzero lifts are no longer open.

Download the proof code and exact certificates.

FPRD-IC-T18 · shared modular lift

Five rows lock to two boundary factorizations

Five rows in the residual table sharea=2, x=6, N≡12(mod18432)a=2,\ x=6,\ N\equiv12\pmod{18432}. Applying simultaneous exponent periods at the twelve primes

19,41,61,151,101,251,271,541,631,29,751,811 19,41,61,151,101,251,271,541,631,29,751,811

leaves exactly one lift of each row. Every solution therefore has

v=v0+36,288,000r,y=y0+36,288,000s,N=12+48,384,000t. v=v_0+36{,}288{,}000r,\qquad y=y_0+36{,}288{,}000s,\qquad N=12+48{,}384{,}000t.

The base residues encode only9⋅7⋅65+1=2129\cdot7\cdot65+1=2^{12} and9⋅5⋅91+1=2129\cdot5\cdot91+1=2^{12}. No single-prime search through five million eliminated any row; the useful obstruction is the shared lift, not an isolated lucky modulus.

This lock remains an exact intermediate theorem. The full valuation gate in FPRD-IC-T21 now excludes every nonboundary lift in all five rows.

FPRD-IC-T19 · certified height barrier

Every non-boundary lift begins beyond 26 quadrillion

Put k=r+sk=r+s. Exact logarithmic normalization turns every non-boundary solution into the one-sided approximation

0<34log⁡2(3) k−⌊34log⁡2(3) k⌋<7.2548,384,000. 0<\frac34\log_2(3)\,k- \left\lfloor\frac34\log_2(3)\,k\right\rfloor <\frac{7.25}{48{,}384{,}000}.

Rational atanh-series bounds certify the logarithms. An exact integer sweep through k=737,837,167k=737{,}837{,}167finds 111 broad one-sided candidates. All 111 miss the fourteen row-specific correction limits by at least 0.0020764, against a correction error below 10−610^{-6}. Hence every non-boundary solution satisfies

k>737,837,167,v+y>26,774,635,116,096,000,N≥42,436,792,629,504,012. k>737{,}837{,}167,\qquad v+y>26{,}774{,}635{,}116{,}096{,}000,\qquad N\ge42{,}436{,}792{,}629{,}504{,}012.

This conditional height theorem remains correct, but the full valuation gate below now excludes every non-boundary lift in the cluster. The logarithmic sweep is therefore superseded as a frontier and should not be extended.

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FPRD-IC-T20 · full 2-adic gate

The residual rows are valuation-rigid or one-dimensional

PutF(v,y)=3aP(v)(2x+3y)+1F(v,y)=3^aP(v)(2^x+3^y)+1. The earlier gate determined xx, but equality with a power of two also requires

N=v2(F(v,y)). N=v_2(F(v,y)).

In eleven rows this valuation is fixed at the displayed base exponent, immediately excluding every nonboundary lift. In each of the other ten rows, solutions modulo2e2^e form a one-dimensional Hensel graph. WithLe=3⋅2e−2L_e=3\cdot2^{e-2}, there are exactly 2e−112^{e-11} states modulo (v,y)=(Le,Le)(v,y)=(L_e,L_e), and each vv residue determines a unique yy residue.

The lifting mechanism is elementary. LTE givesv2(3Le−1)=ev_2(3^{L_e}-1)=e; shiftingyy byLeL_e therefore toggles the next binary digit, so precisely one of its two lifts works for each next vv digit.

FPRD-IC-T21 · complete restricted classification

The positive-type-II even tower has only ten controls

At e=22e=22, each nonrigid row has 2,048 Hensel states moduloL22=3,145,728L_{22}=3{,}145{,}728. The prime 65,537 is aligned with both remaining exponent periods:

ord⁡65537(3)=65536,ord⁡65537(2)=32. \operatorname{ord}_{65537}(3)=65536, \qquad \operatorname{ord}_{65537}(2)=32.

None of the 20,480 states satisfies the equation modulo 65,537. The 21-row quotient therefore contains exactly ten canonical proper solutions, representing four factorizations:

3⋅5⋅17+1=28,3⋅31⋅11+1=210, 3\cdot5\cdot17+1=2^8,\qquad 3\cdot31\cdot11+1=2^{10}, 9⋅5⋅91+1=212,9⋅13⋅35+1=212. 9\cdot5\cdot91+1=2^{12},\qquad 9\cdot13\cdot35+1=2^{12}.

Their minimum displayed excess is four ones. Combined with the retained a=3a=3 and uniforma≥4a\ge4 exclusions, this classifies the complete positive-type-II tower witha≥1a\ge1 and proves that it cannot beat 2N2N.

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FPRD-IC-T22 · complementary-orientation gate

Quadratic reciprocity leaves six of twelve residue pairings

In the complementary orientation, division by the common final power of two gives

RS+3g=2K, RS+3^g=2^K,

where RR is the negative-character odd factor andSS the positive-character factor. Canonical smooth-operand sums giveR mod 24∈{5,7,13}R\bmod24\in\{5,7,13\} andS mod 24∈{1,11,17,19}S\bmod24\in\{1,11,17,19\}. Reducing the equation modulo each factor and taking Jacobi symbols yields

(2R)K=(3R)g,(2S)K=(3S)g. \left(\frac2R\right)^K=\left(\frac3R\right)^g, \qquad \left(\frac2S\right)^K=\left(\frac3S\right)^g.

These identities and the equation modulo 24 leave exactly

R mod 24S mod 24K mod 2g mod 251115110071007171013101131900. \begin{array}{c|c|c|c} R\bmod24&S\bmod24&K\bmod2&g\bmod2\\ \hline 5&1&1&1\\ 5&11&0&0\\ 7&1&0&0\\ 7&17&1&0\\ 13&1&0&1\\ 13&19&0&0 \end{array}.

This is an unbounded arithmetic gate: the six omitted canonical pairings cannot occur at any exponent height.

FPRD-IC-T23 · exact subbranch exclusion

The double-even complementary branch is empty

If f=2f=2 is split as one factor of two from each inner sum, both sums have the form1+3b1+3^b with even exponent. Their odd parts satisfy

1+3b2≡{5(mod24),b≡2(mod4),17(mod24),b≡0(mod4). \frac{1+3^b}{2}\equiv \begin{cases} 5\pmod{24},&b\equiv2\pmod4,\\ 17\pmod{24},&b\equiv0\pmod4. \end{cases}

Opposite character forces one residue of each kind, hence the pair (5,17)(5,17). That pair is forbidden by FPRD-IC-T22, proving that the entire infinite subbranch has no solution.

The same gate compresses f=1f=1to three residue rows. For the two rows whose negative factor is even, divisibility by 5 addsK≡3g(mod4)K\equiv3g\pmod4. These are the input to the valuation theorem below.

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FPRD-IC-T24 · exact complementary valuation theorem

Tied minima cannot create an exceptional f=1 lift

Let the unique even inner sum be1+3b1+3^b, withbb positive and even, and write the other odd factor asW=2h+2q3yW=2^h+2^q3^y. The reduced equation is

1+3b2W+3g=2K. \frac{1+3^b}{2}W+3^g=2^K.

After multiplying by two, removing the constant2h2^h, and applying LTE,

1+v3(K+1−h)=v3 ⁣(2h3b+2q3y+2q3b+y+2⋅3g). 1+v_3(K+1-h)=v_3\!\left( 2^h3^b+2^q3^y+2^q3^{b+y}+2\cdot3^g\right).

Ordinarily the right side has valuationmin⁡(b,y,g)\min(b,y,g). Canonical parity reduces every possible cancellation of its lowest 3-adic digit to six families. Complete exponent periods modulo 7 and 73 have empty joint survivor sets in all six; the only minimum-one exception is checked against its three possible mod-73 targets separately. Therefore every actual solution in the three f=1f=1 rows obeys

min⁡(b,y,g)=1+v3(K+1−h). \boxed{\min(b,y,g)=1+v_3(K+1-h)}.

Thus tied minima produce no exceptional lift graph: the least ternary exponent is always logarithmically small in the output exponent. This controls the branch but does not prove that everyf=1f=1 row is empty.

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FPRD-IC-T25 · exact complementary valuation theorem

The remaining f=2 split has no exceptional tied-minimum lift

In the split where one inner sum contributes both factors of two, put Vb=(1+3b)/4V_b=(1+3^b)/4, whereb≥3b\ge3 is odd, and write the other odd factor asW=2h+2q3yW=2^h+2^q3^y. The reduced equation is

VbW+3g=2K. V_bW+3^g=2^K.

Multiplying by four, removing the constant term2h2^h, and applying LTE gives

1+v3(K+2−h)=v3 ⁣(2h3b+2q3y+2q3b+y+4⋅3g). 1+v_3(K+2-h)=v_3\!\left( 2^h3^b+2^q3^y+2^q3^{b+y}+4\cdot3^g\right).

The four possible residues of VbV_bmodulo 24 leave eight placements in the quadratic gate. Canonical parity then reduces every possible cancellation at the least ternary exponent to ten families. Complete periods modulo 7 and 73 eliminate nine. The tenth leaves one class; it has 343 survivors modulo 487 and four modulo 2593, but no pair satisfies the coordinatewise CRT compatibility conditions. The sole minimum-one boundary also has no mod-73 survivor. Therefore every actual solution in this split obeys

min⁡(b,y,g)=1+v3(K+2−h). \boxed{\min(b,y,g)=1+v_3(K+2-h)}.

Equivalently,2⋅3min⁡(b,y,g)−1∣K+2−h2\cdot3^{\min(b,y,g)-1}\mid K+2-h. The least ternary exponent is logarithmically small in the output exponent, but the valuation law does not by itself make the split empty.

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FPRD-IC-T26 · exact modular lift classification

The first two complementary minimum layers lock to forty periodic graphs

The two valuation theorems have the uniform form

m:=min⁡(b,y,g)=1+v3(K+f−h),f∈{1,2}. m:=\min(b,y,g)=1+v_3(K+f-h),\qquad f\in\{1,2\}.

At m=1m=1 andm=2m=2, this fixesK+f−hK+f-h modulo 18. Keeping the exact coordinates attaining the minimum produces 32 canonical strata. A complete sequence of CRT-compatible order-modulus quotients proves that nine are empty and that all states in the other 23 lie on exactly 40 lift graphs. Their common coordinate period is

P=20,652,025,680=2435⋅5⋅11⋅13⋅17⋅19⋅23. P=20{,}652{,}025{,}680 =2^4 3^5\cdot5\cdot11\cdot13\cdot17\cdot19\cdot23.

Centering the graph coordinates moduloPP gives seventeen distinct proper presentations on 29 graphs and nine distinct improper rational boundary identities on eleven graphs. The proper roots are harmless: their displayed costs exceed2N2N by at least six ones. Any other proper solution must move at least one exponent by a full period; the defining equation then gives

N≥20,652,025,680. \boxed{N\ge20{,}652{,}025{,}680}.

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FPRD-IC-T27 · complete complementary-layer classification

The forty lift graphs contain exactly seventeen proper presentations

The next ternary digit closes the gap left by the lift lock. Because the order of two modulo36=7293^6=729 is 486 and divides the common graph period, each ternary coordinate is either its small centered value or contributes zero modulo 729. Exhausting these low/high patterns eliminates all eleven graphs rooted at improper rational identities. It also fixes every ternary coordinate on 26 of the 29 control-rooted graphs; a direct 2-adic divisibility check then fixes their binary coordinate.

The three residual graphs reduce to only two equations:

2K+1=11⋅3b+173,2K=3g+295. 2^{K+1}=11\cdot3^b+173, \qquad 2^K=3^g+295.

Here bb orgg is congruent to six modulo 20,652,025,68020{,}652{,}025{,}680. Matveev's explicit lower bound for rational linear forms in logarithms givesb,g<1015b,g<10^{15}. Exact rational bounds on log⁡23\log_2 3 leave 117,363 possible lift-index pairs for each equation. Successive exact reductions modulo 29, 43, 59, 83, and 101 leave none.

Exactly seventeen proper presentations occur, all with cost at least 2N+6. \boxed{\text{Exactly seventeen proper presentations occur, all with cost at least }2N+6.}

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FPRD-IC-T28 · classical S-unit corollary

The normalized complementary family is finite—but the theorem gives no height cutoff

After multiplying away the fixed factor2f2^f, every complementary presentation has the form

(2r+2s3v)(2u+2w3y)+2f3g=2K+f. (2^r+2^s3^v)(2^u+2^w3^y)+2^f3^g=2^{K+f}.

Expansion and division by the right side makes five positive{2,3}\{2,3\}-units sum to one. Their multiplicative group has rank at most six, and positivity makes every solution nondegenerate. The Evertse–Schlickewei–Schmidt theoremtherefore bounds each fixed factor-shape pattern byexp⁡(7⋅3015)\exp(7\cdot30^{15})solutions. Across the eight patterns under study,

# normalized complementary presentations≤8exp⁡(7⋅3015). \boxed{\#\text{ normalized complementary presentations} \le 8\exp(7\cdot30^{15}).}

This established theorem bounds a number of solutions, not their exponent heights. It therefore proves finiteness but does not supply the uniform Matveev cutoff needed to compare directly with the growing 3-adic period. That distinction redirects the attack to exact finite quotients in the unopenedf=0f=0 branch.

FPRD-IC-T29 · exact complementary f=0 classification

The two triple-minimum cancellation strata contain one harmless control

For RS+3g=2KRS+3^g=2^K, with each factor of type 1+2x3v1+2^x3^v or2x+3v2^x+3^v, the quadratic gate leaves forty ordered canonical placements. The first ternary digit permits twenty tied-minimum cancellations. Only two can have v=y=g=mv=y=g=m.

The first is

(1+4⋅3m)(1+2x3m)+3m=2K, (1+4\cdot3^m)(1+2^x3^m)+3^m=2^K,

with x,Kx,K even andmm odd. Its complete solution tables modulo 7 and 73 have no CRT-compatible pair. The second family has odd x,m,Kx,m,K:

(2+3m)(1+2x3m)+3m=2K. (2+3^m)(1+2^x3^m)+3^m=2^K.

Reduction modulo 2x2^x andv2(1+3m)=2v_2(1+3^m)=2 forcex=3x=3. Withz=3mz=3^m, completing the square gives

(8z+9−2(K+3)/2)(8z+9+2(K+3)/2)=65. (8z+9-2^{(K+3)/2})(8z+9+2^{(K+3)/2})=65.

The two positive factor pairs of 65 leave exactly(m,x,K)=(1,3,7)(m,x,K)=(1,3,7):

(2+3)(1+8⋅3)+3=27,cost=18=2K+4. \boxed{(2+3)(1+8\cdot3)+3=2^7,\qquad \text{cost}=18=2K+4.}

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FPRD-IC-T30 · complete complementary f=0 cancellation closure

Every exceptional lowest-digit cancellation is now classified

The remaining eighteen strata have exactly two ofv,y,gv,y,g equal to the least ternary exponent mm. The squarefree quotient

Q=7⋅73⋅5⋅13=33215,ord⁡Q(2)=36,ord⁡Q(3)=12 Q=7\cdot73\cdot5\cdot13=33215, \qquad \operatorname{ord}_Q(2)=36, \quad \operatorname{ord}_Q(3)=12

makes seven families empty and reduces the other eleven to 23 complete states modulo(36,12,12,36)(36,12,12,36). Modulo 27, an exponent congruent to one or two modulo 12 can be its exact low value or a high value whose power of three vanishes; every other positive representative is necessarily high. Exhausting the feasible low/high patterns leaves one state:

(4+3m)(1+2x3m)+3g=2K,(x,m,g,K)≡(3,1,4,8)(mod(36,12,12,36)), (4+3^m)(1+2^x3^m)+3^g=2^K, \qquad(x,m,g,K)\equiv(3,1,4,8) \pmod{(36,12,12,36)},

with m=1m=1 exactly. Reduction modulo 2x2^x gives2x∣7+3g2^x\mid7+3^g. Since4∣g4\mid g, the latter number has exact 2-adic valuation three, so x=3x=3. The equation becomes

(2K/2−3g/2)(2K/2+3g/2)=175. (2^{K/2}-3^{g/2})(2^{K/2}+3^{g/2})=175.

Its factor pairs (1,175),(5,35),(7,25)(1,175),(5,35),(7,25)leave only (K,g)=(8,4)(K,g)=(8,4). Hence the entire pair-minimum cancellation locus contains the single control

(4+3)(1+8⋅3)+34=28,cost=29=2K+13. \boxed{(4+3)(1+8\cdot3)+3^4=2^8, \qquad\text{cost}=29=2K+13.}

Together with FPRD-IC-T29, this classifies all twenty possible first-digit cancellations. If2r2^r and2u2^u are the constant binary terms of the two factors, every otherf=0f=0 solution therefore obeys

min⁡(v,y,g)=1+v3(K−r−u). \boxed{\min(v,y,g)=1+v_3(K-r-u).}

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Evidence and reproducibility

ResultProof evidenceBoundary
Final-addend restrictionDisplayed residue proof plus independent reconstructionRestricted normalized grammar
Cyclotomic ladderTwo independent quotient evaluators and complete periodsOne y=0 branch
Odd-N closureExact congruence tables, primitive-divisor boundary, bounded controlsOne g=0 positive-type-I branch
Even-layer reductionIndependent residue cover, valuation controls, and 140-step Hensel liftIntermediate input; the lift is closed below
Exceptional x=4 liftFirst Hensel digit and an independently reconstructed nine-residue certificateComplete inside the inherited branch
Large-x frontierExact 2-adic lifts, cost thresholds, and a complete modulus-19441 quotientIntermediate reduction superseded by the complete type-II2 closure
Complete type-II2 closureExact congruence proof and independently reconstructed finite quotientsComplete for a=1 type-II2; superseded frontier below
Complete a=1 layerDisplayed modulus-7 table, modulus-13 separation, and independent reconstructionComplete for the active canonical branch at a=1
a=2 method boundaryAlgebraic formal-boundary proof on the coarse classesExplains why valuation must precede coprime order moduli
First a=2 family closureExact 3-adic lift and complete modulus-271 residue separationComplete for type-I2 at a=2; the last family closes below
Complete a=2 layerDisplayed modulus-73 quotient, modulus-577 logarithm table, and independent residue reconstructionComplete for the active canonical branch at a=2
a=3 entry theoremExact second-gate quotients, complete finite covers through x=20, and independent reconstructionIntermediate reduction; the remote tail closes below
Positive-type-II a=3 closureComplete 524,288-state quotient and independent reconstructionComplete for the complementary third layer
Uniform high-x obstructionCoupled aligned-order quotient and 162 terminal liftsForces x at most ten throughout a=1,2,3
Residual lift quotientExact 21-row quotient and bounded positive controlsFinite input to the completed second-gate classification
Shared a=2, x=6 boundary lockTwelve aligned prime quotients and independent reconstructionExact intermediate theorem; its lifts are closed below
Non-boundary height barrierRational logarithm enclosures and an exact 737,837,167-step one-sided sweepCorrect conditional bound, superseded by the exclusion below
Full 2-adic gateValuation rigidity and one-dimensional Hensel lifting with independent reconstructionClassifies eleven rows and reduces the other ten to 20,480 finite states
Positive-type-II tower classificationComplete 2-adic state set followed by the aligned prime 65,537Exactly ten controls for a≥1; all cost at least 2N+4
Complementary quadratic gateJacobi-symbol proof and independent complete residue tablesNecessary six-row quotient for a=0<g
Double-even f=2 exclusionExact odd-part cycle modulo 24 and the forbidden (5,17) rowComplete for the (1,1) valuation split only
Complete a=3 layerModulus-262657 quotient, terminal moduli 7 and 13, and independent discrete-log reconstructionComplete for positive type I at a=3
Uniform high-layer obstructionExact zero-class tables modulo 73 and 262657 for both positive canonical typesComplete for every a at least four in the f=g=0 proper fork
Complete positive-type-I branchExplicit synthesis of the cyclotomic, odd-N, and all even-layer theoremsOne branch of the normalized three-addition grammar
Complementary f=1 valuation lawLTE reduction plus complete residue intersections modulo 7 and 73Controls, but does not empty, the three surviving f=1 rows
Complementary f=2 valuation lawLTE reduction; complete periods modulo 7, 73, 487, and 2593; terminal CRT incompatibilityControls, but does not empty, the split-(2,0) branch
Complementary minimum-layer lift lockComplete 26-prime CRT lift quotient, centered-root verification, and independent Python/C++ reconstructionNine empty strata and 40 periodic graphs at m=1,2; a height barrier, not full closure
Complete complementary m=1,2 classificationModulo 729, exact 2-adic gates, Matveev's explicit bound, and finite modular descent with independent reconstructionExactly seventeen proper controls, all costing at least 2N+6
Uniform complementary finitenessFive-term positive S-unit embedding and the Evertse–Schlickewei–Schmidt solution-count theoremUniformly finite, but without an effective exponent-height cutoff
Complementary f=0 triple minimumExact 40/20/2 enumeration, complete mod-7/mod-73 tables, 2-adic gate, and independent reconstructionOne empty family and one cost-18 control; completed by the pair-minimum theorem below
Complete f=0 cancellation locusFour-prime quotient, complete mod-27 low/high lift, terminal factorization of 175, and independent Python/C++ reconstructionExactly two harmless exceptional controls; all other f=0 solutions obey the universal valuation law

Normalization and thin-branch archive · Cyclotomic closure archive · Odd-branch archive · Even-layer archive · Exceptional-lift closure and large-x archive · Complete type-II2 closure archive · Complete first-layer closure archive · First-layer and first a=2 closure archive · Complete second-layer closure archive · Third-layer entry-table archive · Positive-type-I and high-layer closure archive · a=2, x=6 boundary-lock archive

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