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.
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Notes
We use the terminology non-trivial local obstructions to distinguish from the ones introduced in Proposition 1.2.
Consider the degree p morphism \(\phi : F_p^{a, b, c} \rightarrow {\mathbb {P}}_1\), \([x:y:z]\mapsto [x:y]\). It is ramified at p points with constant ramification index p.
The set \(F_2^{a, b, c}({\mathbb {Q}})\) is infinite if and only if it is not empty. If \(\mathcal O\in F_3^{a, b, c}({\mathbb {Q}})\) then \((F_3^{a, b, c},\mathcal O)\) is an elliptic curve over \({\mathbb {Q}}\) and \(F_3^{a, b, c}({\mathbb {Q}})\) is a finitely generated group.
Indeed, if \(q\mid x,y\) then \(q^p\mid c\) and \(p\le v_q(c)\le v_q(abc)\).
For example one can take B even and \(A\equiv -1\pmod 4\).
More precisely, \({\bar{\rho }}\) is finite at \(\ell \) if there is a finite flat \({\mathbb {F}}_p\)-vector space scheme H over \({\mathbb {Z}}_\ell \) such that \(H(\bar{{\mathbb {Q}}}_\ell )\) is isomorphic to \({\bar{\rho }}\vert _{G_\ell }\) as \({\mathbb {F}}_p[G_\ell ]\)-modules.
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Acknowledgements
We would like to thank Samuele Anni, Henri Cohen, Nuno Freitas, Roberto Gualdi, Xavier Guitart, Mariagiulia De Maria, Artur Travesa, Carlos de Vera and Gabor Wiese for helpful conversations and comments. The second author is very grateful to Marc Masdeu and Alberto Soto for their help on computational aspects. We would like to thank the anonymous referees for a thorough reading of our paper, and for the numerous helpful suggestions they made to improve the exposition.
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Luis Dieulefait is partially supported by MICINN grant MTM2015-66716-P. Eduardo Soto is partially supported by MICINN grant MTM2016-78623-P.
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
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
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
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
Hence
and case \(p\ge 3\) follows.
If \(p=2\) then n is a power of 2, \(n= 2^m\), and
\(\square \)
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
which induces by restriction a map
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
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
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
Otherwise
Proof
Let \(i\in \{1,2\}\) such that \(n \equiv i \pmod 2\). Then
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
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
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
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
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
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
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,
since \(C^{-1} \equiv C \pmod 8\) and
since \(B,C\in \{\pm 1\} \mod 3\). Thus
-
(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.
\(\square \)
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Dieulefait, L., Soto, E. Solving \(a x^p + b y^p = c z^p\) with abc Containing an Arbitrary Number of Prime Factors. Mediterr. J. Math. 18, 48 (2021). https://doi.org/10.1007/s00009-020-01678-1
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DOI: https://doi.org/10.1007/s00009-020-01678-1