Finite certificates and their scope

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These are author-recorded reductions, not constructed solutions. The arguments below need no numerical search or executable evidence.

A sufficient certificate for Problem 203

Problem 203 asks for an integer $m$ coprime to $6$ such that $m2^k3^\ell+1$ is composite for all nonnegative $k,\ell$. Choose a finite set $Q$ of distinct primes greater than $3$ and nonzero residues $c_q\pmod q$. Put

$$T_2=\operatorname{lcm}_{q\in Q}\operatorname{ord}_q(2),\qquad T_3=\operatorname{lcm}_{q\in Q}\operatorname{ord}_q(3).$$

The exact finite test is

$$\text{for every }(u,v)\in\mathbb Z/T_2\mathbb Z\times\mathbb Z/T_3\mathbb Z, \quad c_q2^u3^v\equiv-1\pmod q\text{ for some }q\in Q.\tag{1}$$

If (1) holds, the Chinese remainder theorem gives an arithmetic progression of $m$ satisfying $m\equiv c_q\pmod q$ for all $q\in Q$ and $m\equiv1\pmod6$. Choose a positive member larger than every $q\in Q$. Reduction of $(k,\ell)$ modulo $(T_2,T_3)$ supplies a divisor $q$ of $m2^k3^\ell+1$. Since $m2^k3^\ell+1>m>q$, this is a proper divisor, including when $k=\ell=0$. This proves the required compositeness and coprimality, conditional only on the finite certificate (1).

Conversely, any certificate using this fixed set $Q$ to divide every term has exactly this periodic form, with $c_q=m\bmod q$ after removing primes dividing $m$, which cannot divide any term. Thus the reduction is exact for finite prime covers of the stated form. It does not show that every solution must admit one.

No $Q$ or residues satisfying (1) are supplied here. The hard step is covering the entire product of exponent cycles with one compatible value of $m$ modulo each prime. A bounded test must specify $Q$ and both periods, then either retain all covering residues or identify an uncovered pair. Repeated failure for bounded prime sets is no nonexistence theorem; stop that finite test at its declared set of primes, without extrapolation to all $Q$.

Finite covers and non-finite covers ask opposite questions

For Problem 7, a candidate is a finite family of classes with distinct odd moduli greater than $1$ covering every residue modulo their least common multiple. Squarefree restrictions or restrictions on the number of prime factors leave unrestricted odd moduli outside their scope. For Problem 273, the moduli must specifically be distinct numbers $p-1$ with $p\ge5$ prime. A finite-period checker must preserve that extra condition. Neither question is resolved by a long covered prefix shorter than the period.

A finite set of allowed moduli can instead be excluded by a sieve certificate covering every residue choice at once. The [[research/leads/odd_covering_distortion/e273_calibration|digit-sieve calibration]], whose exact computation is checked by its own evidence program, does this for the $149$ moduli $p-1$ with $5\le p\le877$: a distortion sieve along base-$p$ digits bounds the loss of the $49$ moduli with prime factors at most $7$ by $115/126$, and every later digit completes fewer moduli than it has values, so it loses nothing. That count bound is specific to this domain; it does not hold uniformly over all finite families of distinct odd moduli. The limits of the same sieve at the small primes are recorded on the [[research/leads/odd_covering_distortion/cube_free_obstruction|cube-free obstruction page]].

In contrast, the no-finite-prime-cover requirement in Problem 276 demands escape from every fixed finite set of primes. For a fixed positive integer sequence $a_n>1$, no finite prime set divides at least one factor of every term if and only if its least prime factors are unbounded. Indeed, primes at most $B$ cover every term exactly when every least prime factor is at most $B$. The all-composite requirement is additional. Finding another finite cover therefore cannot establish the requested escape. The construction on Problem 1113 likewise needs its own no-finite-cover argument if that extra property is claimed; algebraic compositeness alone does not provide it.

The restricted assertion that every non-perfect-power Sierpinski parameter has a finite covering set is Conjecture 3 in the held Filaseta--Finch--Kozek account. Proving it would leave perfect-power parameters unresolved. A counterexample to that assertion would instead answer Problem 1113 positively. The source starts exponents at one; any such parameter $m$ exceeds one, so the extra term at exponent zero is the composite even number $m+1$. Adding one term does not change whether the least prime factors are unbounded. This is a logical implication between the stated questions, not a counterexample or an audit of an external formal variant.

For an exact partition of a finite group into cosets in Problem 274, subgroup indices must remain distinct in the claimed conclusion. Passing through a normal subgroup $N$ gives $[G:H]=[G:HN][N:H\cap N]$. Distinct products need not have distinct first or second factors. Results for simple and solvable groups do not compose through this identity without an extra argument controlling repeated indices. The discriminating test for an extension argument is precisely that index control; stop before asserting closure under extensions if it is absent.

Hall certificates and cover decompositions

Fix a block $n+1,\ldots,n+k$ and write $P_i$ for the set of prime divisors of $n+i$. The distinct-prime selection in Problem 375 is exactly a system of distinct representatives for these finite sets. Hall's theorem therefore gives a finite obstruction: a subset $I\subseteq\{1,\ldots,k\}$ with $|\bigcup_{i\in I}P_i|<|I|$, accompanied by complete factorizations of the affected integers, with every member of the whole block proved composite. Checking a fixed block answers only that block. Having a prime factor greater than $k$ in every integer is sufficient for distinctness, since such a prime cannot divide two numbers of the block, but is not necessary. The existence question in Problem 962 has a different quantifier over blocks.

For Problem 467, fix $x$ and a choice of one residue class for every prime $p\le x$. Make these primes the vertices and, for each integer $1\le n<x$, make an edge consisting of the primes whose chosen class contains $n$. This choice splits into two covers of $[1,x)$ exactly when the vertices have a two-coloring giving both colors to every edge. The required classes and coloring may be chosen afresh for each sufficiently large $x$; the task is not to decompose every given double cover. Nor is coverage of a whole common period required.

Double coverage says only that every edge has size at least two. That property alone does not imply the required coloring: the three two-element edges on three vertices are a counterexample for abstract hypergraphs. This is a logical obstruction to an argument using double coverage alone, not an arithmetic counterexample to Problem 467. A successful proof must construct the classes together with a coloring for every sufficiently large $x$. A bounded test checks every integer of $[1,x)$ for its declared $x$ and residue choices. Failure of one choice, or an abstract non-colorable hypergraph, does not refute that existential construction problem.