# The Fermat-Catalan conjecture with signature $(2,n,4)$, $n\ge4$

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.

• This is a pretty good question. I suggest that you contact Bennett, and share his answer here. Aug 24, 2020 at 18:45

[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

• Well done! Thank you. Sep 4, 2020 at 13:50