模24余1的Erdős-Straus猜想的筛维与搜索深度
Sieve dimension and search depth for the Erdős-Straus conjecture, $n \equiv 1 \pmod{24}$
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中文总结 AI 辅助
本文针对模24余1的素数，研究Erdős-Straus猜想的无因式分解搜索深度，证明了相关集合的上界，展示了无条件障碍，还提出了可判定区间内素数猜想的无因式分解方法，该方法存在I/II型遍历的代价不对称性。
中文摘要 AI 辅助
对于满足n≡1 mod24的素数n，我们研究在不进行因式分解的情况下，将4/n分解为三个单位分数的显式搜索需要达到的深度。记E₂(N;J)为所有不存在深度至多为J的见证的此类素数n≤N构成的集合，这里的见证是指满足互素参数u,a≤J的定理3.9中的双参数准则。我们证明，对于每个固定的J，|E₂(N;J)|≪_J N/(log N)^(1+𝔄(J)/2)，其中指数是证明所基于的覆盖的精确维数。该证明用无不动点的对合取代了由素因子生成的子群（当(ℤ/4m)^×的指数大于2时，该方法会失效），且对每个J都是无条件的。其次，我们展示了一个无条件障碍：在偏移量c=7处，存在≍N(log N)^(-3/2)个素数n≤N、n≡1 mod24，使得除数准则的两个分支均失效。K₇=(n+7)/4由判别式为-7的主型表示必须是本原的，因此相关输入来自Fuchs、Hsu、Rickards、Schindler和Stange[25]的本原表示定理，而非Iwaniec的经典结果；因此，固定偏移量无法留下有限的剩余集合。第三，一种不进行因式分解的方法可判定区间[N,2N]内所有素数的猜想：其II型遍历的代价为𝒪(N(log N)³)，而I型遍历的代价为Θ(N²)，这将整个二次代价定位于4u²d+1的除数提取中，并展现出I型/II型二分法两部分之间的不对称性。该猜想本身仍未解决。
英文摘要
For primes $n\equiv1\pmod{24}$ we study how deep an explicit, factorization-free search for a decomposition of $4/n$ into three unit fractions has to go. Write $E_2(N;J)$ for the set of such primes $n\le N$ at which no witness of depth at most $J$ exists, in the sense of the two-parameter criterion of Theorem 3.9 with coprime parameters $u,a\le J$. We prove that for every fixed $J$ $$|E_2(N;J)| \ll_J \frac{N}{(\log N)^{1+\mathfrak{A}(J)/2}},$$ the exponent being the exact dimension of the covering on which the proof rests. The proof replaces the subgroup generated by the prime factors, an approach that breaks down as soon as $(\mathbb{Z}/4m)^{\times}$ has exponent greater than $2$, by a fixed-point-free involution, and is unconditional at every $J$. Second, we exhibit an unconditional obstruction. At the shift $c=7$ there are $\asymp N(\log N)^{-3/2}$ primes $n\le N$, $n\equiv1\pmod{24}$, at which both branches of the divisor criterion fail. The representation of $K_7=(n+7)/4$ by the principal form of discriminant $-7$ has to be primitive, so that the relevant input is the primitive-representation theorem of Fuchs, Hsu, Rickards, Schindler and Stange [25] rather than the classical results of Iwaniec; a fixed shift therefore cannot leave a finite residual set. Third, a factorization-free procedure decides the conjecture for all primes of an interval $[N,2N]$. Its Type II pass costs $\mathcal{O}(N(\log N)^{3})$ while its Type I pass costs $Θ(N^{2})$, which locates the whole quadratic cost in the extraction of the divisors of $4u^{2}d+1$ and exhibits an asymmetry between the two halves of the Type I/Type II dichotomy. The conjecture itself remains open.
