Which Diophantine equations can be solved using continued fractions? Pell equations can be solved using continued fractions. I have heard that some elliptic curves can be "solved" using continued fractions. Is this true?
Which Diophantine equations other than Pell equations can be solved for rational or integer points using continued fractions? If there are others, what are some good references?
Edit:
Professor Elkies has given an excellent response as to the role of continued fractions in solving general Diophantine equations including elliptic curves. What are some other methods to solve the Diophantine equations $$X^2 - \Delta Y^2 = 4 Z^3$$ and $$18 x y + x^2 y^2 - 4 x^3 - 4 y^3 - 27 = D z^2 ?$$
 A: In 1993, Tzanakis ( http://matwbn.icm.edu.pl/ksiazki/aa/aa64/aa6435.pdf ) showed that solving a quartic Thue equation, which correponding quartic field is the compositum of two real quadratic fields, reduces to solving a system of Pellian equations.  
Even if the system of Pellian equations cannot be solved completely, the information on solutions obtained from the theory of continued fractions and Diophantine approximations might be sufficient to show that the Thue equation (or Thue inequality) has no solutions or has only trivial solutions. 
For that purpose, very useful tool is Worley's result characterizing all rational approximations satifying $|\alpha - \frac{p}{q}| < \frac{c}{q^2}$ in terms of convergents of continued fraction of $\alpha$. You may consult the paper "Solving a family of quartic Thue inequalities using continued fractions" ( http://web.math.pmf.unizg.hr/~duje/pdf/dij.pdf ) and the references given there. 
A: [edited to insert paragraph on Cornacchia and point-counting]
Continued fractions, or (more-or-less) equivalently the Euclidean algorithm, can be used to find small integer solutions of linear Diophantine equations $ax+by=c$, and integer solutions of quadratic equations such as $x^2-Dy^2=\pm1$ ("Pell").  Continued fractions in themselves won't find rational points on elliptic curves, but there's a technique using Heegner points that calculates a close real approximation to a rational point, which is then recovered from a continued fraction — this is possible because the recovery problem amounts to finding a small integer solution of a linear Diophantine equation.  My paper

Noam D. Elkies: Heegner point computations, Lecture Notes in Computer Science 877 (proceedings of ANTS-1, 5/94; L.M. Adleman and M.-D. Huang, eds.), 122-133.

might have been the first to describe this approach.
Another application of continued fractions is Cornacchia's algorithm to solve $x^2+Dy^2=m$ for large $m>0$ coprime to $D$, given $x/y \bmod m$ which is a square root of $-D \bmod m$.  This has an application to counting points on elliptic curves $E\bmod p$ for $E$ such as $y^2 = x^3 + b$ or $y^2 = x^3 + ax$ for which the CM field ${\bf Q}(\sqrt{-D})$ is known: the count (including the point at infinity) is $p+1-t$ where $t^2+Du^2=4p$ for some integers $t$ and $u$, and this determines $t$ up to an ambiguity of at most $6$ possibilities that in practice is readily resolved.  The necessary square root mod $p$ is readily found in random polynomial time, though it is a persistent embarrassment that we cannot extract square roots modulo a large prime in deterministic polynomial time without assuming something like the extended Riemann hypothesis.  Indeed the application that Schoof gave to motivate his polynomial-time algorithm to compute $t$ for any elliptic curve mod $p$ was to recover a square root of $-D \bmod p$ for small $D$ !  (Though this would never be done in practice because the exponent in Schoof's algorithm is much larger than for the randomized algorithm.)  The reference for Schoof's paper is

René Schoof: Elliptic Curves over Finite Fields and the Computation of Square Roots $\bmod p$, Math. Comp. 44 (#170, April 1985), 483-494.

A natural generalization of the Euclidean algorithm to higher dimensions is the LLL algorithm and other techniques for lattice basis reduction (LBR), which have found various other Diophantine uses, including some other techniques for finding rational points on elliptic curves; another of my papers describes some of these Diophantine applications of LBR.
A: I was about to mention H J S Smith's algorithm for finding integer solutions to $x^2 + y^2 = p$ for $p \equiv 1$ mod 4; but this is referred to in a related thread at Applications of finite continued fractions
(Apologies if that thread is easily found from this one; but I wouldn't have noticed it without doing a Google search, and perhaps some other readers are equally inexperienced in StackOverflow ways or unobservant!)
Also, what about higher-dimensional continued fractions, expressed as matrix recurrence relations? I seem to recall that these can be used to find rational solutions of equations involving some kinds of cubic forms.
A: I should have stuck with your preferred notation, as in your $B^2 + B C - 57 C^2 = A^3$ in a comment. So the form of interest will be $x^2 + x y - 57 y^2.$The other classes with this discriminant of indefinite integral binary quadratic forms would then be given by
$ 3 x^2 \pm xy - 19 y^2.$  
Therefore, take
$$ \phi(x,y) = x^2 + x y - 57 y^2.$$ 
The  identity you need to deal with your $A= \pm 3$ is
$$  \phi(    15 x^3 - 99  x^2 y + 252 x y^2  - 181 y^3 , \; 2 x^3 - 15  x^2 y + 33 x y^2 - 28 y^3  ) \; = \; (  3 x^2 + xy - 19 y^2  )^3 $$  
This leads most directly to $\phi(15,2) = 27.$ Using $ 3 x^2 + x y - 19 y^2 = -3$ when $x=7, y=3,$ this leads directly to $ \phi(1581, -196) = -27.$ 
However, we have an  automorph of $\phi,$ 
$$ W \; = \;
 \left(  \begin{array}{rr}
  106 & 855  \\\
   15  & 121  
\end{array} 
  \right)  ,
  $$
and
$ W \cdot (1581,-196)^T = (6, -1)^T,$ so $\phi(6,-1) = -27.$ 
Finally, any principal form of odd discriminant, call it $x^2 + x y + k y^2,$ (you have $k=-57$) has the improper automorph
$$ Z \; = \;
 \left(  \begin{array}{rr}
  1 & 1  \\\
   0  & -1  
\end{array} 
  \right)  ,
  $$
while $ Z \cdot (6,-1)^T = (5, 1)^T,$ so $\phi(5,1) = -27.$ 
EDIT: a single formula cannot be visually obvious for all desired outcomes. There are an infinite number of integral solutions to $3 x^2 + x y - 19y^2 = -3.$ It is an excellent bet that one of these leads, through the identity I give, to at least one of the desired $\phi(5,1) = -27$ or $\phi(6,-1) = -27,$ but not necessarily both, largely because 
$3 x^2 + x y - 19y^2$ and $3 x^2 - x y - 19y^2$ are not properly equivalent. Worth investigating, I should think.
