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The deductions and obstructions here are author-recorded by Plasma AI. Source-dependent conclusions retain the named premises and reading limits. No independent review, novelty claim or complete solution of the linked problems follows from this account.

The relevant questions are E0445, E0479, E0676, E1107.

E0445: interval packing does not improve this theorem's exponent

The target is every translate (n,n+p^c), for all sufficiently large primes p, for each c>1/2; a,b need not differ. Browning--Haynes Theorem 1 guarantees an inverse pair in one of J interval pairs of lengths H,K, with disjoint first intervals, when JH^2K^2>=C p^3 log^4 p.

To force the pair inside one prescribed nonwrapping interval I of length L, put all first and second intervals inside I. Then JH<=L and K<=L, so

$$JH^2K^2\le L^4/J\le L^4.$$

The criterion can therefore only be invoked if L^4>=Cp^3 log^4 p. Its best packing is J=1; subdividing loses. For L=p^c this is still c>3/4, with the logarithm excluding the endpoint. Intervals outside I prove existence elsewhere. Splitting a wrapping interval modulo p only changes constant factors. This obstructs this particular internal-packing use, not the target or all Kloosterman methods. Next test: uniform control for one specified interval or new use of the same-interval condition. Stop if the gain is only choosing one successful interval among many.

E0479: fixed-prime-multiplier lifting fails for generic k

The target is infinitely many positive n for every fixed integer k!=1, including negative k and zero. Suppose n=mp with m fixed and p prime. Fermat gives 2^(mp)=(2^m)^p congruent to 2^m modulo p, also for p=2. Thus a solution forces p|(2^m-k). If 2^m!=k, only finitely many p can work. This holds even when m was already a solution and p is confined to a progression.

For k=2^m the obstruction vanishes. The held Tang theorem and the page's order/Dirichlet argument choose p in a reduced class making m divide the difference too, excluding primes dividing m. This is the known power-of-two family, not a lifting mechanism for k=3. The known k=0,-1,-2 mechanisms are not refuted by the fixed-m statement.

Next test: a growing-seed mechanism producing a new prime with all needed order conditions. Stop at multiplication of one fixed seed by arbitrary primes.

E0676: finite prime sieves leave an exact moving-cutoff gap

For a given n the condition is exactly that some prime p<=sqrt(n) satisfies n mod p^2<p. The cutoff follows from a>=1. A complete finite test of one n must therefore check all those primes; an eventual assertion has no supplied finite reduction.

For a finite prime set S, let M=product_(p in S)p^2. All sufficiently large n congruent to -1 modulo M avoid S, since p^2-1>=p. Thus no fixed finite set of primes covers all large integers. These n can still be represented by other primes: this is not a counterexample family to E0676.

Conversely the density avoiding all primes p<=y is exactly product_(p<=y)(1-1/p), by CRT: precisely p of each p^2 residue classes succeed. The finitely many n<p^2 do not affect density. The full exceptional set is contained, up to finitely many terms, in each such avoidance set. Euler's divergence of sum_p 1/p makes the products tend to zero. This proves density zero for exceptions, not finiteness; no stronger sieve bound is used.

Next test: control y<p<=sqrt(n) along an explicit growing CRT family, or obtain an eventual coverage bound. Stop if y stays fixed while n grows, or only density/finite examples are supplied.

E1107: no fixed congruence obstruction exists

The target is all sufficiently large integers as sums of at most r+1 positive r-powerful numbers for every fixed r>=2; 1 is allowed and no coprimality is required. The Problem 939 transfer was already excluded because it makes the total powerful too; that obstruction is not new here.

For every modulus M>=1, every residue is a sum of at most three residues containing positive r-powerful integers. For M=1 this is immediate. If M is odd, each residue is a sum of two units: at each prime divisor choose a nonzero first residue different from the target, and choose the second as the difference modulo the full prime power; combine by CRT. If M is even, an even target is a sum of two units by the same choices at odd primes and 1,target-1 at the power of 2. An odd target is a sum of three units: use 1,1,target-2 at the power of 2, and 1 plus a two-unit decomposition of target-1 at odd prime powers. CRT combines these choices.

Choose e>=r with e congruent to 1 modulo phi(M). For a positive unit representative u, Euler gives u^e congruent to u modulo M. Every u^e is r-powerful, since nonzero valuations are multiples of e. The sum therefore has the desired residue, and 3<=r+1. No fixed-modulus test can disprove the target. There is no size control relative to the integer being represented, so this is not a local-to-global theorem.

No novelty is claimed for this local observation. The missing step is a global argument with the exact summand count and size control. Congruence coverage or a finite represented suffix does not supply that step.

Source interfaces

These source locators identify the uses in this account. Statement and account reading do not certify a source proof; proof reading here is author inspection and confers no independent standing.

• [[../library/diophantine_problems/browning_2013_incomplete_kloosterman_sums/_index|Browning and Haynes (2013): Incomplete Kloosterman sums]]: Many-interval corollary and constant/packing qualification (account reading). Theorem 1 and J=1 criterion; proof pointer only (statement reading).
• [[../library/number_theory/tang_2025_erdos_479_congruences/_index|Tang (2025): A note on Erdős Problem 479]]: Theorem 3.1 and order/Dirichlet proof pointer, pp.4-5 (account reading).