Number theory - rational number Are there any $x, y$ that fit in below 
$\sqrt{4y^2-3x^2}$ such that an rational number is yielded.
Appreciate if explanation is given.
 A: You want $4y^2-3x^2=n^2 \iff (2y-n)(2y+n) = 3x^2$
So for e.g. we can set $2y-n = 3, 2y+n = x^2 \implies y= \dfrac{x^2+3}4, n = \dfrac{x^2-3}2$ and both will be integer if $x \equiv 1 \pmod 2$.  Thus $x = 2k+1, y = k^2+k+1$ should work for any $k \in \mathbb{Z}$, and we find
$$4y^2-3x^2 = 4(k^2+k+1)^2-3(2k+1)^2 = (2k^2+2k-1)^2$$
giving you at least one infinite family of solutions. You could try other ways of factoring $3x^2$ if you need to find other solutions.
A: We can find infinitely many solutions by considering the Pell equation $y^2-3s^2=1$. 
We look for perfect squares $4y^2-3x^2$, where $x$ is even, say $2s$. Then we want $4y^2-12s^2$ to be a perfect square, say $(2t)^2$. So we want to solve the Diophantine equation $y^2-3s^2=t^2$.
There are already infinitely many solutions with $t=1$. For $(2+\sqrt{3})(2-\sqrt{3})=1$.  Thus $(2+\sqrt{3})^n(2-\sqrt{3})^n=1$. 
But $(2+\sqrt{3})^n=y_n+s_n\sqrt{3}$ for integers $y_n$, $s_n$. It follows that $y_n^2-3s_n^2=1$. 
A: The easiest solution is to try different integers. Some quick ones we could try are $x = 0, y = 1$, and $x = y = 1$.
For $x = 0, y = 1: \sqrt{4-0} = \sqrt{4} = 2$ and for $x = y = 1: \sqrt{4 - 3} = \sqrt{1} = 1$.
A: You can view $4x^2-3y^2=1=(2x)^2-3y^2$ as a form of Pell's equation.  All solutions of this will do what you want, though there will be more solutions to your problem.  As others have commented $x_1=1,y_1=1$ is a solution.  Given $(x_n,y_n)$ is a solution,  $x_{n+1}=7x_n+6y_n,y_{n+1}=8x_n+7y_n$ will be another solution.  The next few are $(13,15),(181,209),(2521,2911)$  You can find solutions to equations of this sort (with explanations if you want) at Dario Alpern's site
