How to prove infinitely many integer values for a square root equation? I have the equation $y = \sqrt{3x^2 + 1}$, and I need to prove that there will be infinitely many integer solutions. I saw possible solutions with things like Pell's equation, but I did not fully understand these methods. It would be awesome if someone could help me out with simpler methods.
Thanks,
John
 A: $y^2 - 3 x^2 = 1$ is a Pell's equation, so you can't ask for something "simpler" than that. 
The basic idea is this.  Consider a Pell's equation $y^2 - d x^2 = 1$ (where $d$ is a positive integer that is squarefree (i.e. not divisible by the square
of an integer $> 1$).
First find the smallest solution in positive integers: in your case $(x_0, y_0) = (1,2)$. Then for any $x$ and $y$, 
$$(y_0 y + d x_0 x)^2 - d (x_0 y + y_0 x)^2 = y^2 - d x^2$$
(expand out the squares to see why this is the case).
As a result, for any solution $(x,y)$ you can get a 
bigger solution $(x_0 y + y_0 x, y_0 y + d x_0 x)$ (in your case $(y + 2 x, 2 y + 3 x)$.
This can be repeated, obtaining an infinite family of solutions:
in your case $(1,2)$, $(2 + 2\times 1, 2 \times 2 + 3\times 1) = (4,7)$, $(7 + 2\times 4, 2 \times 7 + 3\times 4) = (15,26)$, ...
A: If, $y^2-3x^2=1$ write it as $(y-\sqrt3x)(y+\sqrt3x)=1$ and cube both sides to obtain
$(y^3-3\sqrt3x^3-3\sqrt3xy(y-\sqrt3x))(y^3+3\sqrt3x^3+3\sqrt3xy(y+\sqrt3x))=1$
multiply and simplify to get $(y^3+3x^2y)^2-3(3y^2x+3x^3)^2=1$
So, once you find a solution $(x_0,y_0)$ you can iteratively generate infinite solutions (x_n,y_n), given by $x_{n+1}=3y_n^2x_n+3x_n^3$ and $y_{n+1}=y_n^3+3x_n^2y_n$.
