Elementary number theory , when is $12n^2 + 1$ a square Prove that if
$$k = 2 + 2\sqrt{12n^2 + 1}$$
is an integer then it is a square.
Can anyone help me with this? All I know is that k is an integer if and only if ${12n^2 + 1}$ is a square. What do I do next?

let
$$\sqrt{12n^2+1}=m\Longrightarrow 12n^2+1=m^2\Longrightarrow \dfrac{m-1}{2}\cdot\dfrac{m+1}{2}=3n^2$$
because $m$ is odd numbers,so $\dfrac{m-1}{2},\dfrac{m+1}{2}\in N^{+}$
since
$$\gcd\left(\dfrac{m+1}{2},\dfrac{m-1}{2}\right)=1$$
case 1:
$$\dfrac{m-1}{2}=3u^2,\dfrac{m+1}{2}=v^2,uv=n$$
$$\Longrightarrow 2+2\sqrt{12n^2+1}=2m+2=4v^2$$
case 2:
$$\dfrac{m-1}{2}=u^2,\dfrac{m+1}{2}=3v^2\Longrightarrow 3v^2=u^2+1$$
since $3v^2\equiv 0,3\pmod 4$,and 
$u^2+1\equiv 1,2\pmod 4$
so that's impossible
