<s>[INST] 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 $$18 x y + x^2 y^2 - 4 x^3 - 4 y^3 - 27 = D z^2 ?$$
 [/INST] 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.
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