H. Darmon and the reviewer [Bull. Lond. Math. Soc. 27, 513–543 (1995; Zbl 0838.11023)] showed that the title equation has only finitely many solutions in nonzero coprime integers whenever \(1/p+1/q+1/r<1\). When \(1/p+1/q+1/r=1\) one has a curve of genus one, and that theory applies. For each case when \(1/p+1/q+1/r>1\), and \(A=B=C=1\) they wrote down a parametrization that gives rise to infinitely many nonzero coprime integer solutions (when one plugs in integers for the parameters).
This raises the question as to whether their parametrization, or some other parametrization, so accounts for all nonzero coprime solutions. In this interesting paper, Beukers shows that for any nonzero integers \(A,B,C,p,q,r\) satisfying \(1/p+1/q+1/r>1\) there are a finite number of parametrized solutions from which all solutions of the title equation can be obtained by specialization of the parameters to rational integral values. Moreover every solution to the title equation gives rise to a parametrization.
In the case that \(\{p,q,r\}=\{2,2,k\}\) one can easily find a set of parametrizations yielding all integer solutions by using factoring and composition from the theory of binary quadratic forms. In Appendix A, Beukers gives such sets of parametrizations, found by Zagier, for \(\{p,q,r\}=\{2,3,3\}\) and \(\{2,3,4\}\) when \(A=B=C=1\). Beukers gives 25 different parametrizations for \(x^5+y^3=z^2\) in Appendix C, though this may not cover all coprime integer solutions! (Although his methods are effective, in principle, it seems unlikely that they can easily be made so, in practice.)
His theorem follows as a consequence of the following result: Let \(G\) be a finite subgroup of \(\text{GL}(n, \mathbb{C})\) containing no complex reflections, and let \(I_1(x),\dots, I_r(x)\) be a set of homogeneous generating elements of the ideal of \(G\)-invariant polynomials in \( \mathbb{C}[x]\) (where \(x=(x_1,x_2,\dots ,x_n)\)). Let \(V\) be the variety given by the set of relations between the \(I_j\), and suppose that \(V\) can be defined over the rationals. We let \(W\) be those integral points on \(V\) that are not equivalent modulo \(p\) to any singular point of \(V\), for any \(p\) outside of some finite set \(S\). Then there exists a finite set of parametrizations, such that every integral point of \(W\) is a specialization of one of these parametric solutions. Moreover, if there is one integral point in \(W\) there are infinitely many essentially different points (that is, different as points in the appropriate weighted projective space), and these form a Zariski dense subset of \(V\).
To apply this to our original question, Beukers finds a finite subgroup \(G\) of \(\text{GL}(2, \mathbb{C})\), as above, where \(V=\{bf^p+g^q=h^r\}\) for some integer \(b\), and \(f,g,h\) generate the \(G\)-invariant polynomials. In each case he does this by selecting a finite subgroup \(R\) of \(\text{GL}(2, \mathbb{C})\) generated by reflections, with invariants \(f(u,v)\) and \(g(u,v)\), and then lets \(G\) be the set of elements of \(R\) of determinant one (in general, this construction always leads to a variety \(V\) defined by one equation of the form \(y^r=P(x_1,\cdots , x_n)\), where \(r\) is the least common multiple of the orders of the reflections in \(R\)). In order to have arbitrary coefficients \(a,b,c\) one must then determine an appropriate number field containing \(u\) and \(v\) to obtain a prescribed solution.
Beukers’ method can be applied, in principle, to reduce the search for all solutions of equations \(Ax^7 + By^3 = Cz^2\) to the determination of rational points on a finite number of rational twists of the quartic Klein curve. One constructs four polynomials \(f,g,h,k\) which generate the invariants to Klein’s group of order 168 (contained in \(\text{GL}(3)\)), and these invariants belong to a variety \(V\) defined by a polynomial \(k^2=h^3-1728g^7+fA(f,g,h)\), where \(A\) is a polynomial with integer coefficients. Thus solutions to \(k^2=h^3-1728g^7\) are points in \(V\) with \(f=0\). From Beukers’ theorem we see that we need “only” determine the parametric solutions covering the integer points in \(V\) and then see in which cases we have \(f=0\); and so it turns out we will need to determine the rational points on a finite set of quartic curves (it is not, however, clear that this can be done, though from Faltings’ theorem, we know that there are only finitely many such points).
Zagier noted one can find all solutions to \(x^3+y^8=z^2\) by taking each parametrization for \(x^3+Y^4=z^2\) and setting \(Y=y^2\). Thus we must determine the rational points on five explicit hyperelliptic curves of genus 2. Bruin has taken a similar approach to \(x^2+y^8=z^3\).