2
$\begingroup$

The Fermat-Catalan conjecture is that for coprime $x,y,z$ and positive integers $a,b,c$ with $1/a+1/b+1/c<1$, the generalized Fermat equation $x^a + y^b = z^c$ has only finitely many solutions. I'm considering signatures $(a,b,c)$ which are solved.

Table 1 of [BCDY] surveys known results and states that $(2,n,4)$, $n\ge4$ has been solved completely and that this is 'Immediate from Bennett–Skinner [BS], Bruin [Br3]'. [Br3] covers the case $n=5$. Fermat dealt with $n=4$.

This leaves $n=6, 9$ and prime $n\ge7$, but I can't see how [BS] is relevant to that. Can someone explain and/or point me to the relevant part of [BS].

[BCDY] 'Generalized Fermat equations: A miscellany', Bennett, Chen, Dahmen, Yazdani, International Journal of Number Theory, Vol. 11, No. 1 (2015)

[BS] 'Ternary Diophantine Equations via Galois Representations and Modular Forms', Bennett, Skinner, Canad. J. Math. Vol. 56(1), 2004 p23-54.

[Br3] 'Chabauty methods using elliptic curves', Bruin, J.reine angew. Math. 562 (2003), 27-49.

Note: This question was originally posted in MSE on 2020-07-03. It's had some upvotes, but no answers as of 2020-08-24.

$\endgroup$
  • 1
    $\begingroup$ This is a pretty good question. I suggest that you contact Bennett, and share his answer here. $\endgroup$ – GH from MO Aug 24 at 18:45
2
$\begingroup$

[Br2] Theorem 1 covers the case $n=6$. So this leaves $n=9$ and prime $n\ge7$.

As suggested in a comment, I contacted Michael Bennett directly and he kindly explained the rest to me:

We have $x^2+y^n=z^4$ with $x,y,z$ coprime integers.

So $(z^2 - x)(z^2 + x) = y^n$. The gcd of $(z^2 - x)$ and $(z^2 + x)$ is $1$ or $2$.

For a gcd of $1$ we have $x$ and $z$ of opposite parity, and can write

$z^2-x = u^n$ and $z^2+x = v^n$

so that $u^n+v^n = 2z^2$.

This is solved for coprime integer $u,v,z$ for $n\ge4$ by [BS] Theorem 1.1.

For a gcd of $2$ we have $x$ and $z$ both odd, and one of

$z^2-x = 2 u^n$ and $z^2+x = 2^{n-1}v^n$, or

$z^2+x = 2 u^n$ and $z^2-x = 2^{n-1}v^n$.

In either case, $u^n + 2^{n-2} v^n = z^2$.

This is solved for coprime integer $u,v,z$ for prime $n\ge7$ by [BS] Theorem 1.2.

This leaves the case $n=9$ with $y$ even.

Going back to the original equation, we have $x^2+y^9=z^4$. [Co, Section 14.4.1] gives complete parametrizations of $x^2+w^3=z^4$ in terms of $s$ and $t$. In our case, $w$ is an even cube and from this and the parity constraints on $s$ and $t$ given in [Co] it follows that there exist coprime integers $s$ and $t$ with $s t (s^3 - 16 t^3) (s^3 + 2 t^3)$ a cube.

The factors on the left hand side are pairwise coprime. ($s^3-16t^3$ and $s^3+2t^3$ could possibly have a common factor of $3$, but if they do, the whole expression is divisible by $9$ but not by $27$, and hence is not a cube.)

Since the factors are pairwise coprime, $s^3+2t^3$ is a cube.

This corresponds to a rational point on the curve $A^3+2B^3$=1 which is isomorphic to the elliptic curve $Y^2=X^3-1728$ via standard transformations. The latter curve has rank $0$ (and only the rational points corresponding to the point at infinity and $(X,Y)=(12,0)$). Tracing these back to $A^3+2B^3=1$, we find that $(A,B)=(1,0)$ or $(A,B)=(-1,1)$.

These points lead to either $t=0$ (which gives $z=0$ in $x^2+y^9=z^4$) or to $st = -1$ (which does not make $s t (s^3 - 16 t^3) (s^3 + 2 t^3)$ equal to a cube).

[Br2] 'The Diophantine Equations $x^2 \pm y^4 = \pm z^6$ and $x^2 + y^8 = z^3$', Bruin, Compositio Mathematica 118: 305-321, 1999.

[Co] 'Number Theory Volume II: Analytic and Modern Tools', Henri Cohen

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Well done! Thank you. $\endgroup$ – GH from MO Sep 4 at 13:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.