The integer factorization used in the cubic abc proof

P699 research · The cubic abc proof

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This note collects the unconditional algebra needed in Sections 1 and 4 of the cubic abc proof. Its input is the endpoint reduction. These are necessary conditions for badness, not a converse: the higher Lucas digits are not imposed by the factorization.

1. The factorization for all hypothetical bad triples

For a bad triple write

$$n=2^hu,\quad c=n-j,\quad d=c-j>0,\quad t=v_2(j),\quad g=\gcd(u,3),\quad v=u/g,$$

where $u$ is odd, $h\ge2$, $4\le j<n/2$, and $3u^3<2^h$. The endpoint conditions give

$$u(n-1)\mid3jc,\qquad u(n-1)(n-2)\mid j(j-1)(j-2),$$

and the corresponding condition for $c$. At each active odd component $p^a\Vert n-i$, they give $j\bmod p^a\in\{0,\ldots,i\}$. Only a component $3^1$ can be inactive. We also use the proved alternating-coefficient inequality $d^2>3n-2$.

The normalization gives $m:=h-t\ge2$, $v_2(c)=t$, and $v_2(d)=t+1$. Here is a short justification. Put $W=u(n-1)(n-2)/6$. The four coefficients $\binom j{3-i}\binom ci/W$, $0\le i\le3$, are positive integers and sum to $2^h$. If $t\ge h$, all four are divisible by $2^h$, which is impossible. For $t<h$ their exact common two-part is $2^t$: each is divisible by $2^t$, and their derivative sum is $3c/u$, of valuation $t$. Dividing by $2^t$ proves $m\ge2$. The assertions about $c,d$ follow. These coefficient divisibilities also follow directly from the residue labels at each active prime.

Set

$$K=\frac{jc}{n-1},\qquad M=\frac{n-2}{2},\qquad \lambda=\operatorname{den}(K)\in\{1,3\},\qquad k=\lambda K.$$

The denominator is in lowest terms. If $\lambda=3$, then $3\mid n-1$ and $3\nmid k u M$; hence in all cases $\gcd(k,k-\lambda)=\gcd(\lambda,M)=1$. Partition the full prime-power factors of $M$ by their labels $j\equiv0,1,2$, and call their products $U,B,V$, respectively. Thus $U=C_0$, $B=C_1$, and $V=C_2$ in residue-group notation. The component $3^1$, if present, also has one of these three labels. Thus $M=UBV$, with $U,B,V$ pairwise coprime. Put $A=UV$. The labels and

$$K-1=\frac{(j-1)(c-1)}{n-1}$$

give $A\mid k$ and $B^2\mid k-\lambda$. Define positive integers $R,s$ by

$$AR=k,\qquad B^2s=k-\lambda.\tag{1}$$

In particular $\gcd(AR,Bs)=1$ and $\gcd(\lambda,s)=1$. The label-zero conditions at the active components of $u$ give $v\mid j,c,d/2$; removing $g$ accounts for its possibly inactive single factor three. Since $\gcd(v,A(n-1))=1$, we have $v^2\mid R$. Also $v_2(R)=2t$. Define

$$w=2^{2t}v^2,\qquad R=wr,\qquad L=2^tvB.$$

Then $r$ is a positive odd integer. To define the remaining parameter, put $e=d/2$. The middle label gives $B\mid j-1$, so the identity

$$e=M-(j-1)=AB-(j-1)$$

proves $B\mid e$ without assuming any auxiliary quotient integral. As $\gcd(B,v)=1$, the preceding divisibilities and $v_2(e)=t$ give the positive odd integer $f=e/L$.

We have obtained

$$\boxed{\begin{aligned} AB+1&=2^{h-1}u,\\ wAr-B^2s&=\lambda,\\ \lambda w f^2&=\lambda A^2-s(2AB+1),\\ A^3r-\lambda f^2&=2^{2m-2}g^2s. \end{aligned}}\tag{2}$$

For the third identity use $e^2=(M+1)^2-K(2M+1)$, then substitute (1) and $e=Lf$. For the fourth, subtract the third from $A^2(wAr-B^2s)=\lambda A^2$ and use $(AB+1)^2/w=2^{2m-2}g^2$.

The coprimalities needed below are

$$\gcd(Ar,s)=\gcd(B,r)=\gcd(A,f)=\gcd(s,f)=1.\tag{3}$$

The first two follow from (1). At a prime of $A$, the outer label makes $e\equiv1$ or $-1$ modulo that prime, proving the third. The third identity of (2), together with $\gcd(\lambda A,s)=1$, proves the fourth. The last identity of (2) also shows that a common prime of $r,f$ can only be three. Finally $d^2>3n-2$ gives

$$K<M/2,\qquad R<\lambda B/2.\tag{4}$$

For the first abc application, the three terms $A^3r$, $\lambda f^2$, and $2^{2m-2}g^2s$ have common gcd at most nine. Indeed, (3) forces a common prime to divide $g$; if it is three, then $3\nmid s$ and the last term has valuation two.

2. The conic and parameter sizes

Put

$$T=n/2=AB+1,\qquad q=\lambda B,\qquad F=Lf=d/2.$$

The notation separates the full coefficient $R=wr$ from its odd factor $r$; $q$ is the middle-group coefficient, while $L$ contains the column's two-part, $v$, and the middle group $B$. Since $K/M=R/q$ and $F^2=T^2-K(2T-1)$, we obtain

$$\boxed{qF^2=(q-2R)T^2+3RT-R,\qquad 0<R<q/2.}\tag{5}$$

All four parameters are positive integers. Also $\gcd(q,R)=1$, because $\gcd(B,R)=1$ and $\lambda=3$ implies $3\nmid R$. The elementary size and support relations used in the abc proof are

$$Bv\le L,\qquad q\le3L,\qquad u\le3L,\qquad \operatorname{rad}(T)\le2u.$$

3. Paired factors at n minus one and n minus two

Choose either orientation $(J,C)=(j,c)$ or $(c,j)$. Reassign $U,V$ to the zero and two labels of $J$; reflection exchanges them and leaves $A,B,R,q$ unchanged. Partition

$$\frac{n-1}{\lambda}=D_0D_1,\qquad D_0\mid J,\quad D_1\mid C.$$

These factors are coprime and contain the full components with labels zero and one, respectively. When $\lambda=1$, integrality of $K$ also supplies the label at a possibly inactive component three. When $\lambda=3$, that component is removed; an active component at three would already force $K$ integral. All factors of $D_0D_1$ are coprime to $UBV$. Define

$$\alpha=\frac{J}{D_0U},\quad \beta=\frac{C}{D_1V},\quad \kappa=\frac{n+J-2}{D_1U},\quad \ell=\frac{n+C-2}{D_0V}.\tag{6}$$

The labels prove that these are positive integers, for even as well as odd columns. Their products and linear relations are

$$\boxed{\begin{gathered} \alpha\beta=R,\qquad \kappa\ell=R+4q,\\ \kappa U+\beta V=2\lambda D_0,\qquad \alpha U+\ell V=2\lambda D_1,\\ \ell D_0-\beta D_1=2BU,\qquad \kappa D_1-\alpha D_0=2BV. \end{gathered}}\tag{7}$$

These follow by multiplying or adding (6); for the second product use $(n+J-2)(n+C-2)=JC+2(n-1)(n-2)$. Moreover $v_2(\alpha)=v_2(\beta)=t$, and $\gcd(\lambda,R(R+4q))=1$. Reflection interchanges $\alpha\leftrightarrow\beta$ and $\kappa\leftrightarrow\ell$.

4. The paired abc identity

The same labels give positive integers

$$X_1=\frac{J-1}{BD_1},\qquad Z_1=\frac{J-2}{V}.$$

They satisfy the identity used in the second abc application:

$$\boxed{\lambda^2\alpha D_0^3Z_1+\beta\kappa =4B^3T X_1^2.}\tag{8}$$

To prove it, multiply by $D_1^2UV$ and substitute (6), $\lambda D_0D_1=n-1$, and $2BUV=n-2$. The result is

$$(n-1)^2J(J-2)+C(n+J-2)=n(n-2)(J-1)^2,$$

which follows from $(n-1)^2-(J-1)^2=C(n+J-2)$. Every term in (8) is positive. The abc proof controls the common divisor and radical before applying abc.

The identities here are repository deductions, extracted from the earlier factorization calculations. No elliptic-curve or number-field assumption is used in this note.