Another quadratic Diophantine equation: How do I proceed? How would I find all the fundamental solutions of the Pell-like equation
$x^2-10y^2=9$
I've swapped out the original problem from this question for a couple reasons.  I already know the solution to this problem, which comes from http://mathworld.wolfram.com/PellEquation.html.  The site gives 3 fundamental solutions and how to obtain more, but does not explain how to find such fundamental solutions.  Problems such as this have plagued me for a while now.  I was hoping with a known solution, it would be possible for answers to go into more detail without spoiling anything.
In an attempt to be able to figure out such problems, I've tried websites, I've tried some of my and my brother's old textbooks as well as checking out 2 books from the library in an attempt to find an answer or to understand previous answers.
I've always considered myself to be good in math (until I found this site...).  Still, judging from what I've seen, it might not be easy trying to explain it so I can understand it.  I will be attaching a bounty to this question to at least encourage people to try.  I do intend to use a computer to solve this problem and if I have solved problems such as $x^2-61y^2=1$, which will take forever unless you know to look at the convergents of $\sqrt{61}$.
Preferably, I would like to understand what I'm doing and why, but failing that will settle for being able to duplicate the methodology.
 A: You can type it into Dario Alpern's solver and tick the "step-by-step" button to see a detailed solution. 
EDIT: I'm a little puzzled by Wolfram's three fundamental solutions, $(7,2)$, $(13,4)$, and $(57,18)$. It seems to me that there are two fundamental solutions, $(3,0)$ and $(7,2)$, and you can get everything else by combining those two with solutions $(19,6)$ of $x^2-10y^2=1$. Using mercio's formalism, $$(7-2\sqrt{10})(19+6\sqrt{10})=13+4\sqrt{10}$$ shows you how to get $(13,4)$; $$(3+0\sqrt{10})(19+6\sqrt{10})=57+18\sqrt{10}$$ shows you how to get $(57,18)$.   
A: I'm going to give you the general method to obtain the fundamental solutions of the Diofantine equation $x^2-dy^2=f^2$.
First solution:
We set $y=f-1$, $d=f^2+1$, $x=f^2-f+1$
Second solution:
$y=f+1$, $d=f^2+1$, $x=f^2+f+1$
In your case $f^2=9$ and $d=f^2+1=10$. 
So the first solution is $7^2-10(2^2)=3^2$ and $13^2-10(4^2)=3^2$.
From the two fundamental solutions we obtain infinite solutions of the equation $x^2-10y^2=3^2 $with the well known methods.
A: Since I did not receive a complete answer from you for the solution I posted and because you are interested in a simple and quick method to find solutions to Diofantine equations $x^2-dy^2=f$ for many values of $d$, I will present another method that gives solutions for any $d$. In some cases the solutions are minimal.
Let's have the Diofantine equation $x^2-dy^2=f$. We set $x=m^2\pm m+k$ and $y=m\pm1$ where $k$ any non-zero natural number and $m$ any non-zero integer. From the division $x^2/y^2$ we obtain the values of $d$ and $f$, that solve the above equation. 
Let’s have $x=m^2+m+k$ and $y=m+1$. From the division $x^2/y^2$ we obtain $d=m^2 + sk$ and $f=k^2–2km –2k$. 
If $m=2, k=3$ we have $14^2-13\times4^2=-12$ which is reduced to $7^2-13\times2^2=-3$.
Since $m$ can be any integer, for $k=2$ we obtain an infinite number of values of $d$.
Let’s have $x=m^2-m+k$ and $y=m-1$. From the division $x^2/y^2$ we obtain $d=m^2 +2k$ and $f=k^2+2km-2k$.
For $m=-5, k=3$ we obtain $33^2-31\times6^2=-27$ which is reduced to $11^2-31\times2^2=-3$.
We can continue for any value of $m$.
Besides these general methods there are other specific for each value of $k$ which means we end up with an infinite number of formulas since $k$ takes all values from 1 to infinity. From these specific solutions we can obtain other fundamental solutions; in my opinion it is better to use only the general methods. 
Lastly I will give you an example to find the solution of the Diofantine equation $x^2-61y^2=f$. The closest square to 61 is 49 and $61=49+2\times6$. From this we set $m=7, k=6$ and we obtain $62^2-61\times8^2=-60$ which reduces to $31^2-61\times4^2=-15$. If we apply the well known formulas we obtain another solution $1937^2-61\times248^2=15^2$. We can continue this process indefinitely, as you know. The general method which I present here is original mathematical work and is connected to hyperelliptic equations with global solutions.
