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1 \section{Radical support and the exact conjecture}\label{sec:foundations}
2 \subsection{The multiplicative quantity that ignores repetition}
3 Every positive integer has a unique prime factorization
4 \[
5  n=\prod_{p\mid n}p^{e_p},\qquad e_p\ge1.
6 \]
7 Its radical is
8 \[
9  \rad(n)=\prod_{p\mid n}p,\qquad \rad(1)=1.
10 \]
11 The radical retains which primes occur and discards how many times they occur. Thus $\rad(72)=\rad(2^3\cdot3^2)=6$, whereas $72/6=12$ records the repeated-power contribution. If $n$ is squarefree, then $\rad(n)=n$.
12 
13 For a positive primitive additive triple
14 \[
15  a+b=c,\qquad \gcd(a,b)=1,
16 \]
17 all three pairs are coprime. For example, a divisor of $a$ and $c$ also divides $c-a=b$, hence is one. Therefore the prime supports are disjoint and
18 \[
19  R=\rad(abc)=R_aR_bR_c,\qquad R_u=\rad(u).
20 \]
21 Put $M_u=u/R_u$. The exact decomposition is
22 \[
23  abc=R M_aM_bM_c.
24 \]
25 The conjecture compares $c$ with $R$, rather than with $abc$ itself. Since $abc$ already contains $c$, an estimate against the full product would lose the intended content.
26 
27 \figplaceholder{01_radical}{Prime support and repeated depth are different quantities. The radical keeps one copy of each active prime. The powerful part keeps the remaining copies.}
28 
29 \begin{conjecture}[Masser--Oesterl\'e $abc$]\label{conj:abc}
30 For every real $\eps>0$ there is a constant $K_\eps>0$, depending only on $\eps$, such that every positive integer triple with $a+b=c$ and $\gcd(a,b)=1$ satisfies
31 \[
32  \boxed{c\le K_\eps\,\rad(abc)^{1+\eps}.}
33 \]
34 \end{conjecture}
35 The order of the quantifiers is the statement. First choose $\eps$. Then choose a constant that works for every primitive triple. The constant may become large as $\eps$ approaches zero. It may not depend on the particular triple.
36 
37 This formulation is the one stated in Stewart and Yu's primary treatment of the conjecture~\cite{StewartYu2001}. Replacing $\le$ by $<$ does not change the all-$\eps$ conjecture after enlarging the constant. Requiring $a\le b$ also leaves it unchanged because the two inputs can be exchanged.
38 
39 \subsection{Equivalent output-sector formulations}
40 The form best adapted to this research separates the repeated depth of the output from the input prime support.
41 \begin{proposition}[Exact normalization]\label{prop:defect}
42 For every primitive positive triple,
43 \[
44  \frac cR=\frac{M_c}{R_aR_b},\qquad
45  Q:=\frac{abc}{R},\qquad \frac Q{ab}=\frac cR.
46 \]
47 Consequently Conjecture~\ref{conj:abc} is exactly equivalent to each all-$\eps$ assertion
48 \[
49  M_c\le K_\eps R_aR_bR^\eps,
50  \qquad Q\le K_\eps abR^\eps.
51 \]
52 \end{proposition}
53 \begin{proof}
54 Use $R=R_aR_bR_c$ and $c=R_cM_c$, then cancel the indicated factors. Multiplying any displayed inequality by the positive cancelled factors recovers the original conjecture, with the same constant and exponent.
55 \end{proof}
56 A local construction may measure $M_c$ naturally. A Wronskian may instead be divisible by $Q$. Both are legitimate, provided the final global height includes the matching normalization.
57 
58 \subsection{Quality measures the exponent in one example}
59 Define the quality of a primitive triple by
60 \[
61  q(a,b,c)=\frac{\log c}{\log R}.
62 \]
63 The denominator is positive because $c\ge2$ and $R\ge2$. The logarithm base does not matter. The identity $c=R^q$ is exact, although decimal qualities are approximations.
64 
65 A triple has $q>1$ exactly when $c>R$. It then has more output size than the global distinct-prime product. Such a triple is often called an $abc$ hit. A hit is an interesting example, not a counterexample to the conjecture. The constant $K_\eps$ is part of the conjecture.
66 
67 \begin{center}
68 \begin{tabular}{@{}rrrrl@{}}
69 \toprule
70 $a$&$b$&$c$&$R$&Factorization of the repeated terms\\
71 \midrule
72 $1$&$8$&$9$&$6$&$8=2^3$, $9=3^2$\\
73 $3$&$125$&$128$&$30$&$125=5^3$, $128=2^7$\\
74 $1$&$4374$&$4375$&$210$&$4374=2\cdot3^7$, $4375=5^4\cdot7$\\
75 $2$&$109\cdot3^{10}$&$23^5$&$15042$&$23^5=6436343$\\
76 \bottomrule
77 \end{tabular}
78 \end{center}
79 The last equality is $2+109\cdot3^{10}=23^5$. Every factor in the table matters. Its radical is $2\cdot3\cdot109\cdot23=15042$, giving quality about $1.629912$. The third triple has quality about $1.567887$. These are illustrative exact examples, without any claim that either is the largest known quality.
80 
81 \figplaceholder{03_high_quality}{Exact examples compare the output with the radical. Decimal quality is a diagnostic derived from the exact integer data. A finite collection of high qualities does not decide the all-triple conjecture.}
82 
83 \begin{proposition}[Finiteness formulation]
84 Conjecture~\ref{conj:abc} is equivalent to the assertion that, for every $\eps>0$, there are only finitely many primitive positive triples with $q(a,b,c)>1+\eps$.
85 \end{proposition}
86 \begin{proof}
87 Assume $abc$, using its bound with $\eps/2$. A triple of quality above $1+\eps$ satisfies
88 \[
89  R^{1+\eps}<c\le K_{\eps/2}R^{1+\eps/2}.
90 \]
91 Hence $R^{\eps/2}<K_{\eps/2}$. The radical is bounded. The same conjectural inequality then bounds $c$, leaving finitely many positive triples.
92 
93 Conversely, fix $\eps>0$ and assume the finiteness assertion. Outside its finite exceptional set, $c\le R^{1+\eps}$. Choose $K_\eps$ to be at least one and at least every ratio $c/R^{1+\eps}$ in the exceptional set. This single constant works for all triples.
94 \end{proof}
95 This equivalence explains why large searches can be valuable and still be logically limited. A computation can list hits in a finite range. It cannot certify the finiteness of the unsearched set.
96 
97 \subsection{Why the epsilon cannot be removed}
98 The uniform statement $c\le KR$ for one absolute constant $K$ is false. There is an elementary family that proves this without using $abc$.
99 \begin{theorem}[Failure of the zero-epsilon estimate]\label{thm:epsilon-zero}
100 The ratios $c/\rad(abc)$ are unbounded among primitive positive additive triples.
101 \end{theorem}
102 \begin{proof}
103 For $n\ge0$, take
104 \[
105  a_n=1,\quad b_n=4^{3^n}-1,\quad c_n=4^{3^n}.
106 \]
107 These are primitive, $b_n$ is odd, and $\rad(c_n)=2$. We claim
108 \[
109  v_3(b_n)=n+1.
110 \]
111 At $n=0$ this is $v_3(3)=1$. If $x=1+3^ku$ with $k\ge1$ and $3\nmid u$, then
112 \[
113  x^3-1=3^{k+1}u\left(1+3^ku+3^{2k-1}u^2\right).
114 \]
115 The bracket is a unit modulo three, so cubing increases the valuation by exactly one. This proves the claim by induction.
116 
117 Because $b_n$ contains $3^{n+1}$, its powerful part contains $3^n$. Thus
118 \[
119  \rad(b_n)\le \frac{b_n}{3^n},\qquad
120  R_n=2\rad(b_n)\le\frac{2b_n}{3^n}.
121 \]
122 Therefore
123 \[
124  \frac{c_n}{R_n}\ge\frac{3^n c_n}{2b_n}>\frac{3^n}{2}\longrightarrow\infty.
125 \]
126 \end{proof}
127 The displayed lower bound does not contradict the positive-$\eps$ conjecture. It does not exhibit infinitely many triples with a fixed positive quality gap. The output grows doubly exponentially with $n$, while the proved defect lower bound grows exponentially. A large defect lower bound and a fixed positive exponent gap are different assertions.
128 
129 \subsection{A known unconditional bound}
130 Stewart and Yu proved an effective bound of a different scale~\cite[Theorem 1]{StewartYu2001}: there is an effectively computable absolute $C>0$ such that every primitive positive triple satisfies
131 \[
132  c<\exp\!\left(CR^{1/3}(\log R)^3\right).
133 \]
134 It controls the height in terms of the radical unconditionally. Its exponential dependence is much larger than the power $R^{1+\eps}$ required by $abc$. It therefore provides a benchmark, without closing the gap this research addresses.
135 
136 \subsection{A published claim and a continuing verification dispute}
137 Mochizuki's four IUT papers were published in 2021. The fourth paper states an application to $abc$~\cite{MochizukiIUTIV}. Publication and establishment of the claimed proof are distinct questions. Scholze and Stix documented an objection to the comparison underlying IUT III, Corollary 3.12, and Mochizuki published a response disputing their simplifications~\cite{ScholzeStix2018,MochizukiComments2018}.
138 
139 A 2026 Project LANA verification report withholds a verdict on whether IUT proves $abc$. It identifies a needed compatibility between the relevant analytic and anabelian comparison maps~\cite{LANA2026}. That report also distinguishes its objection from accepting every identification in the earlier criticism. This edition follows that precise status: a published proof claim and a documented, unresolved verification dispute. None of the project proofs below depends on treating IUT as an established proof of $abc$, and none claims to repair its disputed comparison.

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