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.

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.
