A Systematic Recursive Construction for the ABC Conjecture via Pythagorean Structures

Abstract We study a recursively generated family of primitive ABC triples obtained from the classical Pythagorean identity. The construction has a deliberately single-input form: one fixes a single odd integer a0 > 1, and the index n = 0,1,2,... is allowed to vary naturally. No later values of A,B,C,b, or c are independently selected. Define an = a2n 0 , bn = An−1 An =a2 n, 2 , cn=An+1 2 , Bn =b2 n, Then a2 n + b2 n = c2 n, Cn =c2 n. An +Bn =Cn, and the recursive relations cn = bn +1, an+1 = a2 n, An+1 = A2 n, bn+1 = 2bncn, Bn+1 = 4BnCn follow identically. The principal arithmetic object for ABC quality is not Bn itself but rad(Bn). We prove the exact radical accumulation formula n−1 ∏ rad(Bn) = rad(b0) k=0 rad(ck), where the factors are separated bycoprimality; repeatedpowersarecountedonlyoncebytheradical.Consequently n ∑ log rad(AnBnCn) = lograd(a0) + lograd(b0) + We then define qn = logCn log rad(AnBnCn) , k=0 log rad(ck). εn = qn −1. The key unconditional recursive statement is qn > 1 =⇒qn+1 >1, which follows from the exact identity logCn = 2logcn, the growth inequality logcn+1 > 2logcn, and the estimate rad(cn+1) ≤ cn+1. Thus a single verified base index produces a recursively persis tent positive quality state for all later indices. The manuscript distinguishes this infinite recursive statement from the stronger fixed-ε requirement in the standard formulation of the ABC conjecture. The latter requires a fixed ε0 > 0 for infinitely many triples and is stated separately so that no finite computation is confused with the universal recursive argument. 中文摘要。本文研究一族由經典畢氏恆等式所轉換而來的原始ABC三元組。構造採用明 確的單一輸入形式：先固定一個奇數a0>1，之後只讓n=0,1,2,...自然變化；後續的 A,B,C,b,c 均不是另外選取，而是由同一公式自動生成。定義 an = a2n 0 , bn = An−1 An =a2 n, 2 , cn=An+1 2 , Bn =b2 n, 則有 a2 n + b2 n = c2 n, Cn =c2 n. An +Bn =Cn, 以及 cn = bn +1, an+1 = a2 n, An+1 = A2 n, bn+1 = 2bncn, Bn+1 = 4BnCn. 本文特別強調，ABC質量的關鍵不是Bn本身，而是rad(Bn)。我們證明其精確的radical 累積式 n−1 ∏ rad(Bn) = rad(b0) k=0 rad(ck), 2 並利用互質性將因子分開；任何重複冪次在radical中只計一次。因此 n ∑ log rad(AnBnCn) = lograd(a0) + lograd(b0) + 再定義 qn = logCn log rad(AnBnCn) , k=0 log rad(ck). εn = qn −1. 本文核心的無條件遞迴命題是 qn > 1 =⇒qn+1 >1, 其依據為恆等式logCn =2logcn、成長不等式logcn+1 >2logcn，以及rad(cn+1) ≤ cn+1。 因此只要有一個已驗證的起始階，後續所有階均可由同一個一般n的證明遞迴得到。本文 特別把此無窮遞迴命題與ABC標準形式所要求的固定ε0>0無窮多次條件分開陳述，避 免把有限數值驗證誤認為普遍證明。