ABC Conjecture

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

Quantifiers, literature, and direction of attack.

Put $R=\operatorname{rad}(abc)$, and retain positive, pairwise coprime $a,b,c$ with $a+b=c$. The assertion asks for a constant depending only on $\varepsilon$, simultaneously for all prime supports and exponents. A high-quality example $c>R$ does not contradict it. The catalogue's disputed-claim status and sources [S2]--[S4] are retained; this notebook neither assumes a claimed proof nor purports to adjudicate those arguments. Scholze--Stix [S3] is a specific critique of an existing approach, not a proof that every possible approach must fail.

The polynomial analogue suggests a sharply testable route. In the Mason--Stothers theorem, differentiation removes one multiplicity at every root, and additivity identifies several Wronskians. Snyder [E1] gives a short proof of that established theorem. Pasten [E2] develops arithmetic derivatives tailored to one equation and proves a precise relation between small nondegenerate derivatives and $abc$. These are existing ideas. The present work first computes the exact defect for the fixed ordinary arithmetic derivative, proves the resulting restricted radical bound, and then tests how allowing adjustable prime weights changes the obstruction.

The fixed derivative and the additive defect.

For $n>0$ define \[ D(n)=\sum_{p\mid n}v_p(n)\frac np, \qquad D(1)=0. \] This integer-valued function satisfies $D(mn)=mD(n)+nD(m)$, as follows by adding the prime valuations. It is not additive. Define \[ \Delta=D(a)+D(b)-D(c),\quad W=aD(b)-bD(a),\quad A=\frac a{\operatorname{rad}(a)},\ B=\frac b{\operatorname{rad}(b)},\ C=\frac c{\operatorname{rad}(c)}. \] The letters $A,B,C$ here denote integers, not polynomials or unspecified constants. Since $a,b,c$ are pairwise coprime, so are $A,B,C$. For each $n$, the integer $n/\operatorname{rad}(n)$ divides $D(n)$: every summand $v_p(n)n/p$ contains all prime factors of that divisor. It follows that \begin{equation}\label{wsref-0-72-61-3a-31-31-3a-63-6f-6e-67-72-75-65-6e-63-65-73} AB\mid W,\qquad W=aD(c)-cD(a)+a\Delta,\qquad C\mid W-a\Delta. \end{equation} In particular, because $(a,C)=1$, \begin{equation}\label{wsref-0-72-61-3a-31-31-3a-65-78-61-63-74-2d-64-65-66-65-63-74} \gcd(C,W)=\gcd(C,\Delta),\qquad C\mid W\ \Longleftrightarrow\ C\mid\Delta. \end{equation} Thus the missing factor in the polynomial analogy is an explicit congruence. It cannot be supplied merely by applying the product rule.

Lemma 11.1 (a radical estimate with the defect retained). Suppose $W\ne0$, and put $J=C/\gcd(C,\Delta)$, with $\gcd(C,0)=C$. Then \begin{equation}\label{wsref-0-72-61-3a-31-31-3a-64-65-66-65-63-74-2d-62-6f-75-6e-64} c\le R J\,\frac{\log c}{\log2}. \end{equation}

Proof. The integer $AB\gcd(C,\Delta)$ divides $W$, by (ra:11:congruences)--(ra:11:exact-defect) and pairwise coprimality. Since $W$ is nonzero, this divisor is at most $|W|$. Also, using $p\log p\ge2\log2$ for every prime $p$, \[ \frac{D(n)}n=\sum_{p\mid n}\frac{v_p(n)}p \le\frac{\log n}{2\log2}. \] Consequently \[ \frac{abc}{RJ}=AB\gcd(C,\Delta) \le |W| \le ab\frac{\log a+\log b}{2\log2} \le ab\frac{\log c}{\log2}. \] Divide by $ab>0$ to obtain (ra:11:defect-bound). The logarithm is natural throughout.$\square$

This proves the desired near-linear estimate on the concrete subclass where $W\ne0$ and $C\mid\Delta$: there $J=1$. To check the exponent carefully, fix $\varepsilon>0$ and set $\alpha=\varepsilon/(1+\varepsilon)$. The elementary maximum of $(\log x)x^{-\alpha}$ on $x\ge1$ is $1/(e\alpha)$. Thus the lemma gives \[ c^{1-\alpha}\le\frac{R}{e\alpha\log2},\qquad c\le\left(\frac1{e\alpha\log2}\right)^{1+\varepsilon} R^{1+\varepsilon}. \] Enlarging the constant yields the strict inequality in the catalogue. This is a proved restricted result, with a uniform constant on that subclass. There is no proof that every triple belongs to it.

Exact examples and a failure of the naive argument.

For $(a,b,c)=(1,8,9)$, one has $D(8)=12$, $D(9)=6$, $\Delta=6$, $C=3$, and $W=12$. Here $R=6<c=9$, and $J=1$. The triple is therefore a nontrivial test of the restricted result. For $(1,80,81)$, however, \[ D(80)=176,\quad D(81)=108,\quad \Delta=68, \quad (A,B,C)=(1,8,27),\quad W=176. \] The required factor $C=27$ does not divide $W$; indeed $ABC=216>176$. Hence the unqualified claim $abc/R\le|W|$, obtained by copying the polynomial proof, is false for this explicit coprime additive triple. The corrected lemma retains $J=27$ and survives this test. This is a counterexample to a proposed intermediate argument, not to the $abc$ conjecture.

Degeneracy is also an arithmetic issue.

The condition $W\ne0$ cannot simply be deleted. There is an exact description of its failure. Write $a=\prod_{p\mid a}p^{e_p}$ and $b=\prod_{q\mid b}q^{f_q}$. Then \begin{equation}\label{wsref-0-72-61-3a-31-31-3a-7a-65-72-6f-2d-63-6c-61-73-73} W=0\quad\Longleftrightarrow\quad \begin{cases} p\mid e_p\text{ for every }p\mid a,\\ q\mid f_q\text{ for every }q\mid b,\\ \displaystyle\sum_{p\mid a}e_p/p=\sum_{q\mid b}f_q/q. \end{cases} \end{equation} To prove the forward implication, multiply $D(a)/a=D(b)/b$ by the product $P$ of all primes dividing $ab$. Reduction modulo a prime $p\mid a$ leaves only the term $e_pP/p$, so $p\mid e_p$. The argument on the other side is identical. The remaining equality is already the displayed derivative identity. Conversely, the three conditions immediately give $W=0$. Empty prime supports are allowed; thus $(1,1,2)$ is included.

For example $a=4$, $b=27$, $c=31$ gives $D(a)/a=D(b)/b=1$ and $W=0$. More generally $(a,b)=(2^{2t},3^{3t})$ has $W=0$ for every positive integer $t$. These are valid coprime additive triples after setting $c=a+b$, but no radical estimate for their sums follows from a zero Wronskian. Their existence blocks a universal proof using this one fixed derivative; it does not imply that their radical ratios are large.

Allowing prime weights: a known framework, an explicit test.

Let $z_p\in\mathbb Z$ be zero outside the primes dividing $abc$, and define \[ D_z(n)=\sum_{p\mid n}v_p(n)\frac np z_p, \qquad H(z)=\max_{p\mid abc}|z_p|. \] These are Pasten's arithmetic derivatives [E2], expressed in coordinates. Impose $D_z(a)+D_z(b)=D_z(c)$ and require $W_z=aD_z(b)-bD_z(a)\ne0$. Theorem 3.3 of [E2] gives \begin{equation}\label{wsref-0-72-61-3a-31-31-3a-70-61-73-74-65-6e} c\le R H(z)\frac{\log c}{\log2}. \end{equation} The divisibility calculation above is exactly the local mechanism, now with zero additive defect. This bound and the general derivative strategy are established work; they are not being claimed as new progress on $abc$.

For the failed example $(1,80,81)$, the additivity equation is \[ 160z_2+16z_5=108z_3, \quad\hbox{equivalently}\quad40z_2+4z_5=27z_3. \] The weights $(z_2,z_3,z_5)=(3,4,-3)$ give $D_z(80)=D_z(81)=432$, $W_z=432$, and $H(z)=4$. The required divisor $216$ now divides $W_z$. Moreover height four is minimal: reducing the equation modulo four forces $4\mid z_3$, while nondegeneracy here requires $z_3\ne0$ because $W_z=D_z(80)=108z_3$. A height below four is therefore impossible. This is an exact optimization for one triple, rather than a numerical lattice guess or an asymptotic statement.

Existence alone is inexpensive. Except at $(a,b)=(1,1)$, interchange $a,b$ if necessary so $b>1$, choose $p\mid b$ and $r\mid c$, and set \[ U=bv_p(b)/p,\quad V=cv_r(c)/r,\quad g=\gcd(U,V), \qquad z_p=V/g,\quad z_r=U/g, \] with all other weights zero. Their supports are disjoint by coprimality. Then $D_z(a)=0$, $D_z(b)=D_z(c)=UV/g>0$, so $W_z\ne0$. But this construction only gives $H(z)\le c\log c/(2\log2)$. Substitution in (ra:11:pasten) cancels the useful factor $c$ and does not prove the conjecture. The obstruction has moved from existence to size.

There is a separate short-vector trap. In the model lattice $x_1+x_2+Mx_3=0$, the nonzero vector $(1,-1,0)$ has height one. If the required nondegeneracy is $x_3\ne0$, every acceptable vector has height at least $M/2$, since $|x_1|+|x_2|\ge M|x_3|$. This is not asserted to be the lattice of an actual triple; it disproves the abstract inference that finding a short nonzero vector automatically finds a short vector outside a forbidden hyperplane. The Wronskian condition must remain in any lattice estimate.

A precise sufficient next statement.

One possible stronger-than-necessary target is: for every $\delta>0$ there is $K_\delta$ such that every coprime triple with $c>R$ admits additive weights with $W_z\ne0$ and $H(z)\le K_\delta c^\delta$. Together with (ra:11:pasten), this would imply the conjecture. Indeed use $\log c\ll_\delta c^\delta$ and choose $2\delta=\varepsilon/(1+\varepsilon)$; then $c^{1-2\delta}\ll_\delta R$, giving exactly exponent $1+\varepsilon$. Triples with $c\le R$ already satisfy the target after enlarging the constant. No equivalence between this particular formulation and $abc$ is claimed. Pasten's more refined equivalence has its own exponents and exceptional configurations [E2, Section 4]. The elementary construction above falls well short of even this sufficient target.

What a disproof search would have to establish.

For a fixed $\varepsilon>0$, define the normalized ratio $Q_\varepsilon(a,b,c)=c/R^{1+\varepsilon}$. A disproof requires an unbounded sequence of these ratios with the same $\varepsilon$. Maximizing the quality $\log c/\log R$ over one finite box does not provide that sequence, nor does allowing $\varepsilon$ to shrink separately for every record triple. The defect estimate gives a more directed necessary condition on a hypothetical counterexample sequence with $W\ne0$: \[ J\ge\frac{c\log2}{R\log c} =Q_\varepsilon\,\frac{R^\varepsilon\log2}{\log c}. \] Thus a candidate family must escape the class with small defect factor, or repeatedly fall into the exactly described zero-Wronskian class. This inequality is a filter, not a construction: a large $J$ is compatible with a perfectly ordinary radical ratio. The same caution applies to the minimum weighted height; failure to find small weights in a bounded search does not prove they do not exist outside that box. The height-four example avoided this problem by a modular lower bound matching an explicit vector. A useful computational extension would retain such certificates of minimality and organize difficult triples by their divisibility obstruction, rather than reporting large quality values alone. No family meeting the required unbounded-ratio condition was obtained here.

Finite verification.

The local standard-library script exhausts all $608294$ coprime triples with $1\le a\le b$ and $c\le2000$. It verifies (ra:11:congruences), (ra:11:exact-defect), and the full equivalence (ra:11:zero-class) by exact integer arithmetic. There are three zero-Wronskian triples in that range. Among the other triples, $467698$ satisfy $C\mid\Delta$; most are easy because $C=1$. Exactly seven of these also have $c>R$, including $(1,8,9)$, $(1,48,49)$, and $(1,512,513)$. Exhaustion of the bounded weight boxes independently confirms the height-four minimum for $(1,80,81)$; the modular argument above explains it. No finite count is used to infer a density or a universal bound. Reproduction files are checks.py and results.json under _Local/top_problems/continuation_2026_09_27/analytic/.

First gap and outcome.

The fixed-derivative calculation proves the radical bound with its exact defect factor, and hence the full $\varepsilon$-family of inequalities on the specified subclass. It also classifies zero Wronskians and gives an explicit counterexample to the uncorrected polynomial analogy. The first universal gap is controlling the defect or finding sufficiently small additive weights while preserving $W_z\ne0$. The example and the lattice stress test show that neither requirement follows from formal differentiation or from a nonzero lattice vector alone.

Outcome: UNRESOLVED. This is a checked restricted estimate and an obstruction analysis within an established derivative approach. It does not prove $abc$, disprove it, or certify a disputed proof. Direct derivation, primary-source comparison, and exact finite tests were used; no formal proof assistant or independent expert review was involved.

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. Formulation source.
• S2 Construction of Arithmetic Teichmuller Spaces IV: Proof of the abc-conjecture — Kirti Joshi. https://arxiv.org/abs/2403.10430. Further 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. Further reference; not a proof endorsement.
• S4 The abc Conjecture Revisited — Patrick Letendre, July 2026. https://arxiv.org/abs/2607.07641. Further reference; not a proof endorsement.

• E1 Noah Snyder, An Alternate Proof of Mason's Theorem, Elemente der Mathematik 55 $2000$, no. 3, 93--94, DOI: 10.1007/S000170050074. https://ems.press/journals/em/articles/609. https://ems.press/content/serial-article-files/45146. Primary source for the polynomial analogy; it does not prove the integer conjecture. Consulted 27 September 2026.
• E2 Hector Pasten, Arithmetic derivatives through geometry of numbers, Canadian Mathematical Bulletin 65 $2022$, no. 4, 906--923, DOI: 10.4153/S0008439521000990; arXiv:2106.16165v3 (11 December 2021), Sections 2--4, especially Theorem 3.3. https://arxiv.org/pdf/2106.16165v3. Primary source for weighted arithmetic derivatives, the known Wronskian radical estimate, and the small-derivative obstruction. Consulted 27 September 2026.

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.