general solution of equation and relation I am interested in learning the below question in some elementary way. Please discuss this problem and help me to get mind free state.
How to get solutions for $x^2 - 10y^2$ = $1$? I would like to learn some easy method. Also, I have seen in some text book $x= 19$ and $y = 6$ is one of the solution and general solution can be prepared by using $x = 19$ and  $y = 6$ in terms of $n$. where $n$ is some positive integer. Please explain this general solution and relation with $x = 19$ and $y =6$. 
Advance thanks to all.
 A: This is what is known as Pell's equation.
This is how to find all solutions in an elementary way.
$$x^2-10y^2=1$$ can be factored to give $$(x+\sqrt{10}y)(x-\sqrt{10}y)=1$$
You can find more solutions by taking $(19+\sqrt{10}6)$ to some integer power, collecting like terms and putting $y$ equal to the coefficient of $\sqrt{10}$ and $x$ to the integer part of the result. These will always be solutions because $1^k=1$ and because the binomial theorem confirms that the absolute value of the negative term in the second bracket will always be equal to the value of $\sqrt{10}y$. 
Note we have not shown that this is a complete set of solutions. That is much harder to do, and not as elementary as you would probably want. If you do want an explanation however, just leave a comment below.
Why $$x=\frac{(x+\sqrt{10}y)^n+(x-\sqrt{10}y)^n}{2}$$ is always a solution:
We know from the above that $x$ is integer part of $(19+\sqrt{10} \cdot 6)^n$.
Note that in the expansion of $(x+\sqrt{10}y)^n+(x-\sqrt{10}y)^n$ all the irrational terms cancel out (by the binomial theorem http://en.wikipedia.org/wiki/Binomial_theorem). What is left is twice the required coefficient so we divide by 2 to get $x$.
A: We add a few comments. First we should mention the Brahmagupta Identity, which for this case is 
$$(s^2-10t^2)(u^2-10v^2)=(st+10uv)^2-10(sv+tu)^2.\tag{$1$}$$
This identity can be verified by expanding both sides. 
From this identity, if we have a representation of $m$ in the form $x^2-10y^2$, and also a representation of $n$ of that form, then the identity gives us a representation of $mn$.
$(1.)$ In particular, let $s=u=3$ and $t=v=1$. Then $s^2-10t^2=-1$. Thus by the Brahmagupta Identity, we have
$$(s^2-10t^2)(s^2-10t^2)=(9+10)^2-10(3+3)=(-1)(-1)=1.$$
Thus the solution $(19,6)$ can be thought of as coming from the simpler "almost" solution $(3,1)$.
$(2.)$ Let $a_1=19$, $b_1=6$. For $n\ge 1$, define $a_{n+1}, b_{n+1}$ by
$$a_{n+1}=19a_n+6b_n,\qquad\text{and}\qquad b_{n+1} =6a_n+ 19b_n.\tag{$2$}$$
The recurrences $(2)$ give an easy way to produce more solutions of our equation $x^2-10y^2=1$. For example, $(a_2,b_2)=(721,228)$.
It turns out that all positive solutions come from the recurrence $(2)$.
$(3.)$ You asked how one obtains $(19,6)$, the fundamental solution. There is a mechanical and relatively quick way of doing it, using the continued fraction expansion of $\sqrt{10}$. Computations can be arranged so that they are pure integer computations, meaning we do not need to know anything about the decimal expansion of $\sqrt{10}$.
There is a very nice book on the Pell equation by Ed Barbeau. Many number theory texts discuss the Pell equation in detail.  
