Erdos-Straus 方程第 I 类与第 II 类解计数：集中性、局部障碍与逐点比较

中文摘要 AI 辅助

对于素数 p ≥ 5，令 f_I(p) 和 f_II(p) 分别按 Elsholtz 和 Tao 的意义计数 Erdos-Straus 方程 4/p = 1/x + 1/y + 1/z 的第 I 类与第 II 类解。我们不讨论 Erdos-Straus 猜想本身；我们的主题是解的数量。我们将 Bradford 的一个双射转化为精确计数恒等式：f_I(p) 和 f_II(p) 是对与 p 互素的整数 n 求和，每一项是 n^2 的因子中落在模 e_n = 4n - p 的单个剩余类里的因子个数。一个对偶恒等式将两个计数置于同一个整数和同一个模数上，仅由剩余类 -p 和 -1 区分。我们的主要无条件结果对所有分子 m 成立：方程 m/p = 1/x + 1/(pY) + 1/(pZ) 且 Y ≤ K 的第 II 类解的数量至多为 sum_{n ≤ K} tau(n) = K log K + O(K)，该界对 p 和 m 一致；在 p ≤ N 上平均时，该数量仅为 O_m((log K)^3 log log K)。因此第 II 类解集中在 x/p → 1/m 处，而第 I 类解不能如此。分子以两种独立方式携带算术信息。模 e_n 的实特征 chi 在整除 n 的素数上平凡，当 chi(-1) = -1 时消去第 II 类解，当 chi(-1)chi(m_0) = -1 时消去第 I 类解，其中 m_0 是 m 的无平方因子核；这两者恰好当 m 为平方数时重合，因此对于分子 4，没有任何模平方数的判据能区分这两类。另外，当 4 整除 m 时，对每个奇完全平方数 n 有 f_I(n) = f_II(n) = 0，这推广了 Elsholtz 和 Tao 的一个定理。最后我们证明当 e_n 属于 {1,3,5} 时的锐利局部不等式 2B(n) ≥ A(n)，并构造例子表明该不等式对每个满足 7 ≤ e ≤ 2*10^4 的奇数 e 失效；我们猜想对所有 p > 5 且 p 不 ≡ 1 mod 8，有 f_I(p) > f_II(p)，对 p ≤ 10^6 验证了该猜想，并将其归约为有限窗口。我们还精确定位了 Elsholtz-Tao 关于 f_I(p) 之和的上界中的因子 log log N。

英文摘要

For a prime p >= 5 let f_I(p) and f_II(p) count the Type I and Type II solutions of the Erdos-Straus equation 4/p = 1/x + 1/y + 1/z, in the sense of Elsholtz and Tao. We do not address the Erdos-Straus conjecture itself; our subject is the number of solutions. We turn a bijection of Bradford into an exact counting identity: f_I(p) and f_II(p) are sums, over the denominators n coprime to p, of the number of divisors of n^2 in a single residue class modulo e_n = 4n - p. A dual identity puts both counts on the same integer and the same modulus, distinguished only by the classes -p and -1. Our main unconditional result holds for every numerator m: the number of Type II solutions of m/p = 1/x + 1/(pY) + 1/(pZ) with Y <= K is at most sum_{n <= K} tau(n) = K log K + O(K), uniformly in p and m; on average over p <= N it is only O_m((log K)^3 log log K). Hence Type II solutions concentrate at x/p -> 1/m, while Type I solutions cannot. The numerator carries arithmetic information in two independent ways. A real character chi modulo e_n that is trivial on the primes dividing n annihilates Type II when chi(-1) = -1, and Type I when chi(-1)chi(m_0) = -1, where m_0 is the squarefree kernel of m; these coincide exactly when m is a square, so for the numerator 4 no criterion modulo squares can separate the two types. Separately, f_I(n) = f_II(n) = 0 for every odd perfect square n whenever 4 divides m, generalising a theorem of Elsholtz and Tao. Finally we prove the sharp local inequality 2B(n) >= A(n) for e_n in {1,3,5}, with a construction showing it fails for every odd e with 7 <= e <= 2*10^4; we conjecture f_I(p) > f_II(p) for all p > 5 with p not congruent to 1 mod 8, verify this for p <= 10^6, and reduce it to a finite window. We also locate exactly the factor log log N in the Elsholtz-Tao upper bound for the sum of f_I(p).

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