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! 
 A: The fact that if $x_1=2x_0+3y_0$ then $x_1\gt x_0$ is immediate: you cannot have both $x_0$ and $y_0$ zero; likewise with $y_1$.
That $x_1^2+3y_1^2$ is also equal to $n$ if you assume that $x_0^2 - 3y_0^2=n$ should follow by simply plugging in the definitions of $x_1$ and $y_1$ (in terms of $x_0$ and $y_0$), and chugging.
Finally, what you have just done is show that if you have one solution, you can come up with another solution. Do you see how this implies the final thing you "need to show"?
A: HINT $\: $ Put $\rm\: z = x+\sqrt{3}\ y\:,\:$ norm $\rm\:N(z)\: = z\:z' = x^2 - 3\ y^2\:.\:$ Then $\rm u = 2 + \sqrt{3}\ \Rightarrow\ N(u) = u\:u' = 1\:$ so $\rm\ N(u\:z)\ =\ (u\:z)\:(u\:z)' =\ u\:u'\:z\:z'\ =\ z\:z'\:,\:$ where $\rm\ u\:z\ =\ 2\:x+3\:y + (x+2\:y)\ \sqrt{3}\:.\:$ Therefore the composition law (symmetry) $\rm\ z\to u\:z\ $ on the solution space $\rm\:\{z\ :\ N(z) = n\}$ arises simply by multiplying by an element of $\rm\:u\:$ of norm $1\:,\:$ using the multiplicativity of the norm: $$\rm\ N(u) = 1\ \ \Rightarrow\ \ N(u\:z)\ =\ N(u)\:N(z)\ =\ N(z) = n$$
A: 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) $
