Counting Type I and Type II solutions of the Erdos-Straus equation: concentration, local obstructions, and a pointwise comparison

RESUMO

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).