Show that $x^2 - 3y^2 = n$ either has no solutions or infinitely many solutions I have a question that I have problem with in number theory about Diophantine,and Pell's equations. Any help is appreciated!
We suppose $n$ is a fixed non-zero integer, and suppose that $x^2_0 - 3 y^2_0 = n$, where $x_0$ and $y_0$ are bigger than or equal to zero. Let $x_1 = 2 x_0 + 3 y_0$ and $y_1 = x_0 + 2 y_0$. We need to show that we have $x^2_1 - 3 y^2_1 = n$, with $x_1>x_0$, and $y_1>y_0$. Also, we need to show then that given $n$, the equation $x^2 - 3 y^2 = n$ has either no solutions or infinitely many solutions. 
Thank you very much! 

There are an infinite number of solutions to $x^2-3y^2=1$, which you can generate from the base solution $2^2-3\cdot 1^2 = 1$ as follows:
Given $(x,y) \in \mathbb{Z}^2$ such that $x^2-3y^2 = 1$:
\begin{align*}
x^2-3y^2 & =(x-\sqrt{3}y)(x+\sqrt{3}y)\\
&= 1 \\
(x-\sqrt{3}y)^n(x+\sqrt{3}y)^n &= (X - \sqrt{3}Y)(X+\sqrt{3}Y)\\
& = 1^n = 1
\end{align*}
for $(X,Y) \in \mathbb{Z}^2$.
Numbers of the form $a+\sqrt{3}b$ are closed under multiplication, so if we find  one solution to $x^2-3y^2=(x-\sqrt{3}y)(x+\sqrt{3}y) = n$ and infinitely many pairs $(a,b)$ with $a^2-3b^2=1$, we can then generate more solutions with:
$(x -\sqrt{3}y)(a - \sqrt{3}b)(x+\sqrt{3}y)(a+\sqrt{3}b) = (n)(1) $
