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.

  • 2
    $\begingroup$ This is a pretty good question. I suggest that you contact Bennett, and share his answer here. $\endgroup$
    – GH from MO
    Aug 24, 2020 at 18:45

1 Answer 1


[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

  • $\begingroup$ Well done! Thank you. $\endgroup$
    – GH from MO
    Sep 4, 2020 at 13:50

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