To find all integral solutions of $3x^2 - 4y^2 = 11$ I have to find all integral solutions of $3x^2 - 4y^2 = 11$. 
I looked at the equation $\text{mod}\ 3$, and $\text{mod}\ 4$ to see that $x^2 \equiv 1\ \text{mod}\ 4$ and $y^2 \equiv 1\ \text{mod}\ 3$
How do I proceed from here? Or is there a better way to approach this? 
 A: Two good books. I have given more detailed answers for many of this type of problem.
http://www.maths.ed.ac.uk/~aar/papers/conwaysens.pdf
https://www.math.cornell.edu/~hatcher/TN/TNbook.pdf
Here is a good one http://math.stackexchange.com/questions/739752/how-to-solve-binary-form-ax2bxycy2-m-for-integer-and-rational-x-y/739765#739765 
As $11$ is prime, there are two orbits; one sequence of $x$ values is
$$ 3, \; \; 37, \; \; 515, \; \; 7173, \ldots  $$
with $$ x_{n+2} = 14 x_{n+1} - x_n,  $$ with matching $y$ values
$$  2, \; \; 32, \; \;  446, \; \; 6212, \ldots  $$
$$ y_{n+2} = 14 y_{n+1} - y_n.  $$
Here is the second orbit:
sequence of $x$ values is
$$ 5, \; \; 67, \; \; 933, \; \; 12995, \ldots  $$
with $$ x_{n+2} = 14 x_{n+1} - x_n,  $$ with matching $y$ values
$$  4, \; \; 58, \; \;  808, \; \; 11254, \ldots  $$
$$ y_{n+2} = 14 y_{n+1} - y_n.  $$
The linear recurrences above come from Cayley-Hamilton for the matrix
$$
\left(
\begin{array}{cc}
7 & 8 \\
6 & 7 \\
\end{array}
\right)
$$

