gpt 6 astra pro attempt on Math-35

ABC Conjecture

Definitions and mathematical statement

For a positive integer $n$, its radical is $\operatorname{rad}(n)=\prod_{p\mid n}p$, with $\operatorname{rad}(1)=1$. A triple $a,b,c>0$ with $a+b=c$ and $\gcd(a,b)=1$ is automatically pairwise coprime. The conjecture is \[ \forall\varepsilon>0\ \exists C_\varepsilon>0\ \forall a,b,c\in{\mathbb{Z}}_{>0},\qquad \begin{cases}a+b=c,\\\gcd(a,b)=1\end{cases} \Longrightarrow c<C_\varepsilon\operatorname{rad}(abc)^{1+\varepsilon}. \] The constant depends on $\varepsilon$ only. A counterexample must defeat every constant for some fixed positive $\varepsilon$; one unusually large triple does not suffice.

Formulation and concept references: [S1].

Short English statement

Can the size of a coprime additive triple always be controlled, up to any small exponent loss, by its distinct prime factors?

Research attempt

Current frontier.

Research check: 27 September 2026. The selected assertion has exponent $1+\varepsilon$ for every $\varepsilon>0$, not one fixed large exponent. Pasten's arithmetic-derivative framework relates carefully quantified small-derivative statements to $abc$-type bounds; it motivates the lattice attack below. [E1], Sections 2--4. His September 2026 manuscript reports improved discriminant--conductor bounds toward Szpiro, not the target power bound here. [E2]. Letendre's July 2026 paper proposes another conjecture and conditional applications. [S4]. The inherited proof claims are not adopted as established results: Scholze--Stix explain a specific objection to the IUT argument and record that its proponents disagree. [S3]. This notebook does not purport to audit the complete IUT literature or Joshi's later claim.

Attack route.

The polynomial $abc$ analogy suggests replacing differentiation by an integer-valued map satisfying Leibniz and the single additive relation $a+b=c$. The missing ingredient is then a small nondegenerate integer solution of a linear constraint. Rather than assume that any short lattice vector works, we compute exactly the image of the Wronskian functional on that lattice and test its transverse direction. This is an investigation within an established framework, not a claim to have invented arithmetic derivatives.

Attempt: the derivative lattice and its arithmetic image.

Fix coprime positive $a+b=c$ with $(a,b)\ne(1,1)$, let $P=\{p:p\mid abc\}$ and $R=\prod_{p\in P}p$. For integer weights $w=(w_p)_{p\in P}$, extended by zero outside $P$, define \[ D_w(n)=n\sum_{p\mid n}\frac{v_p(n)}p w_p\quad(n\geq1). \] This is an integer and obeys $D_w(uv)=uD_w(v)+vD_w(u)$. Define two integer row vectors indexed by $P$: \[ A_p=\begin{cases}a v_p(a)/p,&p\mid a,\\ b v_p(b)/p,&p\mid b,\\-c v_p(c)/p,&p\mid c, \end{cases}\qquad B_p=\begin{cases}-ab v_p(a)/p,&p\mid a,\\ ab v_p(b)/p,&p\mid b,\\0,&p\mid c. \end{cases} \tag{11.1} \] Pairwise coprimality makes these cases disjoint. The allowed weights form $K=\ker(A:{\mathbb{Z}}^P\to{\mathbb{Z}})$; for them $D_w(a)+D_w(b)=D_w(c)$ and the Wronskian is $W(w)=aD_w(b)-bD_w(a)=Bw$. The rows $A,B$ are independent over ${\mathbb{Q}}$: $A$ has a nonzero coordinate at a prime dividing $c$, where $B$ vanishes, and $B$ is nonzero at a prime dividing $ab$. Thus $B(K)\ne\{0\}$.

Here is an exact lattice lemma, proved rather than inferred from a norm heuristic. Put $g=\gcd_p|A_p|$ and \[ d=\frac{\gcd_{p<q}|A_pB_q-A_qB_p|}{g}>0. \qquad B(K)=d{\mathbb{Z}}. \tag{11.2} \] To prove this, the primitive row $A/g$ can be transformed by a unimodular integer column matrix $U$ to $(1,0,\ldots,0)$, using Bezout's identity and successive Euclidean column operations. Write $BU=(b_1,\ldots,b_s)$. In these coordinates $K$ is generated by the last $s-1$ standard basis vectors, so its image is $\gcd(b_2,\ldots,b_s){\mathbb{Z}}$. The ideal generated by all two-by-two minors is unchanged by unimodular column operations: each new minor is an integer combination of old ones, and applying $U^{-1}$ gives the reverse inclusion. For the transformed rows $A/g,B$, that ideal is exactly $(b_2,\ldots,b_s)$. Multiplying the first row by $g$ proves $11.2$. In particular, the smallest nonzero Wronskian magnitude is explicitly computable from the valuations; this says nothing yet about the height of weights achieving it.

The nonzero minors in $11.2$ occur only between different prime blocks. For example, if $p\mid a$ and $q\mid b$, the minor equals $abc\,v_p(a)v_q(b)/(pq)$. Between the $a,c$ or $b,c$ blocks the same formula holds up to sign with the corresponding valuations. This gives a direct finite computation of $d$, without enumerating all $w$.

Attempt: divisibility versus height.

For every $w\in K$, one has \[ \frac{abc}{R}\mid W(w). \tag{11.3} \] Indeed, if $p^e\Vert n$, the defining formula for $D_w(n)$ is divisible by $p^{e-1}$: the $p$-term has this factor and every other term has $p^e$. For $p\mid a$ or $p\mid b$, apply this to $W=aD_w(b)-bD_w(a)$. For $p\mid c$, use the equal expression $W=aD_w(c)-cD_w(a)$. Multiplying the resulting prime-power divisibilities proves $11.3$. This recovers the core divisibility in Pasten's Theorem 3.3. [E1].

If $W(w)\ne0$ and $H=\|w\|_\infty$, then \[ \frac{abc}{R}\leq |W(w)| \leq abH\sum_{p\mid ab}\frac{v_p(ab)}p \leq \frac{abH\log(ab)}{2\log2} \leq \frac{abH\log c}{\log2}. \tag{11.4} \] The middle inequality follows termwise from $p\log p\geq2\log2$. Thus $c\leq RH\log c/\log2$. The exact obstruction is now a transverse minimum \[ \kappa(a,b)=\min\{\|w\|_\infty:w\in K,\ Bw\ne0\}, \tag{11.5} \] not the length of the shortest nonzero vector of $K$. For example, $a=1,b=8,c=9$ gives $A=(12,-6)$, $B=(12,0)$, $g=6$, $d=12$, and $w=(1,2)$ has $W=12=abc/R$.

A sufficient, deliberately stronger hypothesis would be: for every $\eta>0$ there is $C_\eta$ such that $\kappa(a,b)\leq C_\eta c^\eta$ for the nonexceptional triples described below. Given a target $\varepsilon>0$, choose $\eta>0$ and $\delta>0$ with $\eta+\delta<\varepsilon/(1+\varepsilon)$. Using $\log c\leq C_\delta c^\delta$ in $11.4$ yields $c^{1-\eta-\delta}\ll R$, hence $c\ll R^{1/(1-\eta-\delta)}\leq R^{1+\varepsilon}$. Increasing the constant makes the final inequality strict. This supplies the exact exponent bookkeeping, but the hypothesized transverse bound is not proved or asserted plausible in this full strength.

Stress test: the short direction can be useless.

For the abstract integer rows $A=(H,1,0)$ and $B=(1,0,0)$, $K$ has the length-one vector $(0,0,1)$, yet every $w\in K$ with $Bw\ne0$ has $\|w\|_\infty\geq H$. Moreover $11.2$ gives $d=1$. So even the conjunction of a very short kernel vector and a small Wronskian image generator does not yield a short transverse vector. This is a countermodel to the lattice inference, not an arithmetic counterexample to $abc$.

The arithmetic example $1+108=109$ gives $A=(108,108,-1)$ and $B=(108,108,0)$ on primes $2,3,109$. The vector $(1,-1,0)$ is short but has $W=0$; any nonzero $W$ forces $|w_{109}|\geq108$. This illustrates the exceptional prime configurations also discussed in [E1], Example 3.8. They should not be used to claim an unconditional lower bound along an infinite sequence that has not been constructed. Such triples are easy for $abc$ itself: if one entry is $1$ and another is prime $q$, then $c\leq q+1\leq2R$. They can therefore be excluded from the sufficient hypothesis after handling them separately. The case $(1,1,2)$ is also immediate.

The remaining gap is not solved by computing a Smith form: that finds integer generators and the exact image, but may produce very large weights. Nor does a bound $\kappa\ll c^\eta$ for one fixed $0<\eta<1$ alone give every exponent $1+\varepsilon$ through $11.4$. It gives only a fixed larger power after absorbing the logarithm. No stronger implication from the literature is silently substituted for this calculation.

Outcome.

Outcome: Exploratory approach with unresolved gap. The investigation gives an exact minors formula for all achievable Wronskians, separates image size from the transverse minimum, and exhibits both abstract and arithmetic failures of a shortest-vector shortcut. The $abc$ exponent would follow from the stated much stronger transverse estimate, but that estimate remains unproved. These are not a proof of $abc$, a counterexample to it, or a claim of historical novelty for the underlying derivative framework or lattice algebra.

Sources

• S1 Distance Between Cubics and Rationals, Results in Mathematics $2026$, statement of the abc conjecture. https://link.springer.com/article/10.1007/s00025-026-02616-5. Catalogue formulation source.
• S2 Construction of Arithmetic Teichmuller Spaces IV: Proof of the abc-conjecture — Kirti Joshi. https://arxiv.org/abs/2403.10430. Catalogue research/status reference; not a proof endorsement.
• S3 Why abc is still a conjecture — Peter Scholze and Jakob Stix, July 16, 2018. https://www.math.uni-bonn.de/people/scholze/WhyABCisStillaConjecture.pdf. Catalogue research/status reference; not a proof endorsement.
• S4 The abc Conjecture Revisited — Patrick Letendre, July 2026. https://arxiv.org/abs/2607.07641. Catalogue research/status reference; not a proof endorsement.
• E1 Hector Pasten, Arithmetic derivatives through geometry of numbers, arXiv:2106.16165v2 (20 July 2021), Sections 2--4, Theorem 3.3 and Example 3.8. https://arxiv.org/html/2106.16165v2. Added research source. Arithmetic derivatives, Wronskians and the obstruction from small but degenerate derivatives. Consulted 27 September 2026.
• E2 Hector Pasten, Improved bounds for Szpiro's conjecture, arXiv:2609.17390v1 (15 September 2026), abstract and main bounds. https://arxiv.org/abs/2609.17390. Added research source. Recent manuscript reporting discriminant--conductor estimates; not a proof audit or an abc resolution. Consulted 27 September 2026.