Solving $a x^p + b y^p = c z^p$ with abc Containing an Arbitrary Number of Prime Factors

Abstract

In this paper we prove new cases of the asymptotic Fermat equation with coefficients. This is done by solving some remarkable S-unit equations and applying a method of Frey–Kraus–Mazur.

Appendices

Prime Divisors of Cyclotomic Polynomials

In this appendix, we give some lower bounds for the number of prime divisors of $\ell ^n\pm 1$ for integers $\ell \ge 3$ and $n\ge 1$.

Let $\Phi _n$ be the nth cyclotomic polynomial. A usual description of $\Phi _n$ is given by the formula

$$\begin{aligned} \Phi _n(X) =\prod _k (X-\zeta _n^k) \end{aligned}$$

where $\zeta _n = e^{2\pi i/n}$ is a primitive nth root of unity and k ranges over the units of ${\mathbb {Z}}/n{\mathbb {Z}}$. Gauss proved that $\Phi _n$ is irreducible in ${\mathbb {Z}}[X]$, hence ${\mathbb {Z}}[X]/\Phi _n \simeq {\mathbb {Z}}[\zeta _n] \subseteq {\mathbb {C}}$ is a domain. In particular

$$\begin{aligned} X^n-1 = \prod _{d\mid n} \Phi _d(X) \end{aligned}$$

(3)

is the factorization of $X^n-1$ in irreducible factors over ${\mathbb {Z}}[X]$. Similarly, write $n = 2^mn_2$ where $n_2$ is the largest odd divisor of n. Then

$$\begin{aligned} X^n+1 = \prod _{d\mid n_2} \Phi _{2^{m+1}d} (X) \end{aligned}$$

(4)

since $X^{2^{m+1}n_2}-1 = (X^n-1) (X^n+1)$.

Let k be a positive integer. The map ${\mathbb {Z}}[X]\rightarrow {\mathbb {Z}}/k{\mathbb {Z}}$, $X\mapsto \ell$ factors through ${\mathbb {Z}}[\zeta _n] \rightarrow {\mathbb {Z}}/k{\mathbb {Z}}$, $\zeta _n\mapsto \ell$ if, and only if, $k\mid \Phi _n(\ell )$.

Lemma A.1

Let $p\not \mid n$ be a prime and assume that there is a ring homomorphism $\theta :{\mathbb {Z}}[\zeta _n] \rightarrow {\mathbb {F}}_p$. Then $\theta (\zeta _n)$ has order n in ${\mathbb {F}}_p^\times$ and $n\mid p-1$.

Proof

Let $\alpha = \theta (\zeta _n)$. Then $\alpha ^n-1 = \prod _d \Phi _d(\alpha )=0$. Notice that $X^n-1$ is separable over ${\mathbb {F}}_p$ since $nX^{n-1}\ne 0$ in ${\mathbb {F}}_p[X]$. Hence $\alpha$ has order n in ${\mathbb {F}}_p^\times$ and the lemma follows. $\square$

Lemma A.2

Let p be an odd prime. There is no ring homomorphism ${\mathbb {Z}}[\zeta _p] \rightarrow {\mathbb {Z}}/p^2{\mathbb {Z}}$. There is no ring homomorphism ${\mathbb {Z}}[\zeta _4] \rightarrow {\mathbb {Z}}/4{\mathbb {Z}}$.

Proof

It is enough to prove that $\Phi _p(X) = \sum _{i=0}^{p-1} X^i$ has no roots in ${\mathbb {Z}}/p^2{\mathbb {Z}}$. The following proof is standard. Assume that there is a root a of $\Phi _p$ in ${\mathbb {Z}}/p^2{\mathbb {Z}}$. Then $a = 1 \mod p$, since $\Phi _p = (X-1)^{p-1}$ in ${\mathbb {F}}_p$. Notice that $\Phi _p(1 + pb ) = \sum _{i=0}^{p-1} 1 + i pb=p$ in ${\mathbb {Z}}/p^2{\mathbb {Z}}$ for every b. Hence $\Phi _p(a) = p$ for every $a\equiv 1\pmod p$.

Notice that $\Phi _4(X)=X^2 + 1$ has no roots in ${\mathbb {Z}}/4{\mathbb {Z}}$. $\square$

Lemma A.3

Let $\ell \ge 3$, $n\ge 2$ be integers and let p be the largest prime divisor of n, then $|\Phi _n(\ell )|> p$.

Proof

The Euler’s totient function $\varphi$,

satisfies that

$$\begin{aligned} p-1\mid \varphi (n). \end{aligned}$$

Hence

$$\begin{aligned} | \Phi _n(\ell ) |= \prod _k |\ell -\zeta _n^k| \ge \prod _k 2 \ge 2^{p-1} \end{aligned}$$

and case $p\ge 3$ follows.

If $p=2$ then n is a power of 2, $n= 2^m$, and

$$\begin{aligned} \Phi _n(\ell ) = \ell ^{2^{m-1}}+1> 2. \end{aligned}$$

$\square$

The polynomial $\Phi _n$ has no real roots for $n\ge 3$, hence $|\Phi _n(\ell )| =\Phi _n(\ell )$.

Theorem A.4

Let $\ell \ge 3, n\ge 3$ be integers. There is a prime divisor p of $\Phi _n(\ell )$ not dividing 2n. Hence, $\ell$ has order n in ${\mathbb {F}}_p^\times$.

Proof

Case $n=2^m\ge 4$.

One has that $\Phi _{2^m}(X)= X^{2^{m-1}}+ 1$ and $\Phi _n(\ell ) \ge 10$. If $4\mid \Phi _n(\ell )$ then ${\mathbb {Z}}[\zeta _n]\rightarrow {\mathbb {Z}}/4{\mathbb {Z}}$, $\zeta _n\mapsto \ell$ defines a ring homomorphism that restricts to ${\mathbb {Z}}[\zeta _4]\subseteq {\mathbb {Z}}[\zeta _n]$. This contradicts Lemma A.2. Hence either $\Phi _n(\ell )$ is odd or $\Phi _n(\ell )/2\ge 5$ is odd.

Case $p\mid n$, p odd.

Notice that $\Phi _n(\ell )$ is odd. Indeed, if $2\mid \Phi _n(\ell )$ then there exists a ring homomorphism

$$\begin{aligned} {\mathbb {Z}}[\zeta _n]\rightarrow {\mathbb {F}}_2 \end{aligned}$$

which induces by restriction a map

$$\begin{aligned} {\mathbb {Z}}[\zeta _p]\rightarrow {\mathbb {F}}_2 \end{aligned}$$

hence $p\mid 2-1$ by Lemma A.1.

Let us see that either $\Phi _n(\ell )$ and n are coprime or there is a prime p such that $\Phi _n(\ell )/p, n$ are coprime. Assume that $p<q$ are prime divisors of $\Phi _n(\ell )$ and n. Then there is a ring homomorphism

$$\begin{aligned} {\mathbb {Z}}[\zeta _q] \subseteq {\mathbb {Z}}[\zeta _n] \rightarrow {\mathbb {F}}_p \end{aligned}$$

and $q\mid p-1$ by Lemma A.1 which contradicts $p<q$. Hence the greatest common divisor of $\Phi _n(\ell )$ and n is a possibly trivial power of an odd prime p. If $p\mid n$ then $p^2\not \mid \Phi _n(\ell )$ by Lemma A.2. Hence either $\Phi _n(\ell ), 2n$ are coprime or there is an odd prime divisor p of $\Phi _n(\ell )$ such that $\Phi _n(\ell )/p$ and 2n are coprime. In the second case $\Phi _n(\ell )/p$ is an odd integer $>1$ by Lemma A.3 and the first part of the theorem follows. The order of $\ell$ is computed in Lemma A.1. $\square$

Corollary A.5

Let $\ell \ge 3$. Assume that $n_1, \dots , n_r$ are pairwise different integers $\ge 3$. Then

$$\begin{aligned} \prod _i \Phi _{n_i}(\ell ) \end{aligned}$$

has at least r odd prime divisors.

Proof

Let $p_i$ be a prime divisor of $\Phi _{n_i}(\ell )$ coprime to $2n_i$ as in Theorem A.4. Then $\ell$ has order $n_i$ in ${\mathbb {F}}_{p_i}^\times$, thus $p_i\ne p_j$ for different i, j. $\square$

For an integer k let $\omega (k)$ denote the number of prime divisors of k and let $\sigma (k)$ denote the number of divisors of k.

Corollary A.6

Let $\ell \ge 3$, $n\ge 1$ be integers. If $(\ell , n)\ne (3,even)$ then

$$\begin{aligned} \omega (\ell ^n-1)\ge \sigma (n). \end{aligned}$$

Otherwise

$$\begin{aligned} \omega (3^{2t}-1)\ge \sigma (2t)-1. \end{aligned}$$

Proof

Let $i\in \{1,2\}$ such that $n \equiv i \pmod 2$. Then

$$\begin{aligned} A:=\prod _{\begin{array}{c} {d\mid n}\\ {d\ge 3} \end{array}}\Phi _d(\ell )=\dfrac{\ell ^n-1}{\ell ^i-1} \end{aligned}$$

has at least $\sigma (n)-i$ odd prime divisors $S=\{p_d\}_{d\mid n, d\ge 3}$ as in Theorem A.4. Notice that $p_d\not \mid \ell ^i-1$ for every $p_d\in S$. Indeed, if an odd prime p divides $\ell ^i-1$ then $\ell$ has order $\le i$ in ${\mathbb {F}}_{p}^\times$ by Lemma A.1. Thus

$$\begin{aligned} \omega (\ell ^n-1) \ge \sigma (n) - i + \omega (\ell ^i-1). \end{aligned}$$

It is enough to prove that $\omega (\ell ^i-1) \ge i$ if and only if $(\ell ,i) \ne (3,2)$. If $i=1$ then $\ell -1\ge 2$ and $\omega (\ell -1) \ge 1$. If $i=2$ then $gcd(\ell -1,\ell +1)\le 2$. Assume $\omega (\ell ^2-1) < 2$ then $\ell -1, \ell +1$ are powers of two. Hence $\ell =3$. $\square$

Corollary A.7

Let $\ell \ge 3$, $n\ge 1$ be integers and let $n_2$ be the largest odd divisor of n. Then

$$\begin{aligned} \omega (\ell ^n+1)\ge \sigma (n_2). \end{aligned}$$

Proof

Let $n=2^m n_2$ then $\ell ^n+1 = \prod _{d\mid n_2} \Phi _{2^{m+1} d}(\ell )$ by the polynomial factorization of (4). For every d such that $2^{m+1} d\ne 2$ consider a prime $p_d\mid \Phi _{2^{m+1} d}(\ell )$ as in Theorem A.4. If $m=0$ let $p_1$ be an arbitrary prime divisor of $\Phi _{2}(\ell )=\ell +1$. Then $\prod _{d\mid n_2} p_d$ is a squarefree divisor of $\ell ^n+1$. $\square$

1.1 Catalan Conjecture

One deduces a case of Catalan’s Conjecture.

Theorem A.8

(Partial Catalan’s Conjecture) Let $\ell \ge 3$ be an integer and assume that

$$\begin{aligned} 2^m - \ell ^n \in \{\pm 1\} \end{aligned}$$

for some integers $m, n\ge 2$. Then $m=\ell =3$, $n=2$.

Proof

Assume that $2^m = \ell ^n+1$, $n\ge 2$ and let $n_2$ be the largest odd divisor of n. Then $\ell$ is odd and $\ell ^n+1 \ge 4$. By Corollary A.7 we have that $1=\omega (\ell ^n +1)\ge \sigma (n_2)$, hence $n_2=1$ and $n=2^r$ for some positive r. Since $2^m=\ell ^{2^r}+1 \equiv 2 \pmod 4$ one has that $m=1$ and $2= \ell ^{2^r} + 1$.

Assume that $2^m = \ell ^n-1$, $n\ge 2$. If $(\ell ,n)=(3, 2t)$ with t an integer then $1=\omega (3^{2t}-1) \ge \sigma (2t)-1$ by Corollary A.6. Hence $t=1$.

If $(\ell , n)\ne (3, even)$, by Corollary A.6 one has that $1=\omega (\ell ^n-1) \ge \sigma (n)$. Hence $n=1$. $\square$

This partial result is well known to experts, see [21, B3.3]. See ibid for a complete treatment of Catalan’s conjecture written before Preda Mihăilescu’s proof [19]. See also Bilu–Bugeaud–Mignotte’s book [2] for a minimalistic approach of the proof or Schoof’s book [24] based on two sets of lecture notes by Yuri Bilu.

The Conductor of E[p]

The j-invariant of a Frey curve is given by the formula

$$\begin{aligned} j_E=\dfrac{2^8 (C^2 - AB)^3}{A^2 B^2 C^2}. \end{aligned}$$

Thus one has for the case $(A,B,C) = (ax^p , by^p ,cz^p)$ being pairwise coprime that $C^2-AB$ and ABC are coprime. Let $\ell$ be a prime divisor of ABC. Then

$$\begin{aligned} v_\ell (j_E) = 8 v_\ell (2) - 2v_\ell (ABC) \equiv 8v_\ell (2) - 2v_\ell (abc)\pmod p. \end{aligned}$$

Thus $p\mid v_\ell (j_E)$ if and only if

• $\ell$ is odd and $p\mid v_\ell (abc)$, or

• $\ell =2$ and $v_2(abc) \equiv 4 \pmod p$.

Proposition B.1

Let $E=E_{A,B,C}$ be the Frey curve as in Theorem 3.4. Let f be a newform in $S_2(M)$ for some divisor M of $2^s{\mathrm{rad}}'(abc)$ and let $\mathfrak p$ be a prime ideal such that

$$\begin{aligned} E[p] \simeq \bar{\rho }_{f,\mathfrak p} \end{aligned}$$

as ${\mathbb {F}}_p[G_{{\mathbb {Q}}}]$-modules. Then $M= 2^s {\mathrm{rad}}'(abc)$.

Proof

Let R be the largest (square-free) divisor of $2^s{\mathrm{rad}}'(abc)$ coprime to 2p. By Tate’s uniformization E[p] is ramified at every prime divisor $\ell$ of R and so is $\bar{\rho }_{f,\mathfrak p}$. Thus, $R\mid M$.

Let $\ell =2$. If $s \in \{ 3,5\}$ then Carayol [3] predicts that the lifting $\rho _{f,\mathfrak p}$ of $\bar{\rho }_{f,\mathfrak p}$ has conductor exponent s. Thus $2^s\mid M$. If $s=0$ then M is odd and so is R. If $s=1$ then E[p] is ramified at 2 and so is ${\bar{\rho }}_{f,\mathfrak p}$. Hence $2\mid M$.

One could just avoid case $p\mid M$ since we will consider big primes p with respect to ${\mathrm{rad}}(abc)$. Still, if $p\mid {\mathrm{rad}}'(abc)$ then E[p] is not finite at p. That is, $E[p]\vert _{G_p}$ is reducible and not peu ramifié by [9, Proposition 8.2]. If $p\not \mid M$ then $\bar{\rho }_{f,\mathfrak p}\vert _{G_p}$ is either irreducible or reducible and peu ramifié. Thus $E[p]\vert _{G_p}\not \simeq \bar{\rho }_{f,\mathfrak p}\vert _{G_p}$. This completes the proof. $\square$

Mod 24 Exercises

Proof of Lemma 2.13

Let (A, B, C) be a primitive S-unit point of height $\ge 3$. Assume $A=2^r$, $r\ge 3$. Then $B+C\equiv 0\pmod 8$ and $B+C\not \equiv 0\pmod 3$. Hence,

$$\begin{aligned} BC\equiv -1 \pmod 8 \end{aligned}$$

since $C^{-1} \equiv C \pmod 8$ and

$$\begin{aligned} BC\equiv 1 \pmod 3 \end{aligned}$$

since $B,C\in \{\pm 1\} \mod 3$. Thus

$$\begin{aligned} \pm q^s \ell ^t = BC \equiv 7 \pmod {24}. \end{aligned}$$

• (1) By hypothesis $(q,\ell ) \equiv (-5, 5)$ or $(11,-11)\pmod {24}$. Notice that

  $$\begin{aligned} q^s \ell ^t\equiv \pm q^{s+t}\not \equiv \pm 7 \pmod {24}, \end{aligned}$$

  hence A is not a power of two.

  Assume that

  $$\begin{aligned} 0\equiv 2^r q^{s} = \ell ^t + \varepsilon \equiv (- 3)^t +\varepsilon \pmod 8 \end{aligned}$$

  for some $\varepsilon \in \{\pm 1\}$. Then $\varepsilon =-1$ and t is even. Proposition 2.6 implies

  $$\begin{aligned} (q,\ell ) \in \{(3,5),(5,3),(3,7),(3,17)\}. \end{aligned}$$

  Condition $q\equiv -\ell \pmod {24}$ leads to a contradiction. Similarly, $2^r \ell ^t = q^r+\varepsilon$ has no solution.

• (2) Assume that $(2^r,- q^s \ell ^t, \varepsilon )$ is an S-point for some unit $\varepsilon$. Then $- \varepsilon q^s \ell ^t\equiv 7\pmod {24}$. Thus s, t are odd and $\varepsilon = -1$. That is

  $$\begin{aligned} 2^r = q^s \ell ^t + 1\equiv -1\pmod 3, \end{aligned}$$

  hence r is odd, $r=2f+1$. Thus, 2 is a square in ${\mathbb {F}}_q$, i.e. $q\equiv \pm 1\pmod 8$. Indeed

  $$\begin{aligned} \genfrac(){}0{1}{q}=\genfrac(){}0{2}{q}^r = \genfrac(){}0{2}{q}. \end{aligned}$$

  Assume that $2^r +(-1)^{a}q^{s} +(-1)^{b}\ell ^{t}=0$. Then

  $$\begin{aligned} (-1)^{a+b} q^s \ell ^t \equiv 7\pmod {24}. \end{aligned}$$

  Hence a, b have same parity and s, t are odd. Thus

  $$\begin{aligned} 2^r = q^s + \ell ^t\equiv 1\pmod 3 \end{aligned}$$

  and r is even. Thus q is a square in ${\mathbb {F}}_\ell$.

  Assume that $(2^r q^s , -\ell ^t , \varepsilon )$ is an S-point. Then $\ell ^t \equiv \varepsilon \pmod 8$ and hence t is even and $\varepsilon =1$.

  Assume that $(2^r \ell ^t , -q^s , \varepsilon )$ is an S-point. Then $\varepsilon =1$ and s is even. By Proposition 2.6$q\in \{ 3,5,7,17\}$, hence

  $$\begin{aligned} q\not \equiv 11 \pmod {24}. \end{aligned}$$

• (3) By hypothesis

  $$\begin{aligned} \ell \equiv -1 \pmod {24} \end{aligned}$$

  and $q\equiv \pm 5$ or $\pm 11 \pmod {24}$ since $q \ge 5$. Thus $q^s \ell ^t\not \equiv \pm 7 \pmod {24}$ and A is not a power of two.

  Assume that $2^r q^{s} = \ell ^{t} +1$. Then t is either 1 or an odd prime by Lemma 2.8. Case $t=1$ implies $\ell \equiv -1 \pmod q$. Case t odd prime implies $\ell$ Mersenne hence

  $$\begin{aligned} \ell \equiv 0, 1\pmod 3. \end{aligned}$$

  Assume that $2^r q^{s}=\ell ^{t} -1$. Hence t is even and Proposition 2.6 implies $\ell \in \{3,5,7, 17\}$, then $\ell \not \equiv -1\pmod {24}$. Similarly, case $2^r \ell ^s = q^t\pm 1$ is not allowed by Lemma 2.6.

$\square$