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?
 A: 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
A: $m^2 = 12 n^2 + 1$ implies $m^2 - 12 n^2 = 1$ so 
$$(m + n\sqrt{12} )= (7 + 2\sqrt{12} )^N$$
for some $N\ge 0$  (see also http://en.wikipedia.org/wiki/Pell%27s_equation, just mind you, their $n$ is our $12$).
We also have the conjugate equality:
$$(m - n\sqrt{12} )= (7 - 2\sqrt{12} )^N$$
and , therefore, for $\sqrt{12n^2 + 1} = m$ we get, as @Thomas Andrews indicated
$$\sqrt{12n^2 + 1} = m = \frac{1}{2}(\, (7 + 2\sqrt{12} )^N + (7 - 2\sqrt{12} )^N)  $$
and so 
$$2 + 2\sqrt{12n^2 + 1} =  (7 + 2\sqrt{12} )^N + (7 - 2\sqrt{12} )^N) + 2$$
Note however that 
$$7 \pm \sqrt{12} = (2 \pm \sqrt{3})^2 $$ and so we get 
$$2 + 2\sqrt{12n^2 + 1} = (2 + \sqrt{3} )^{2N} + (2 - \sqrt{3} )^{2N}) + 2\cdot  (2 + \sqrt{3} )^N \cdot (2 - \sqrt{3} )^N = \\ = (\,( 2 + \sqrt{3})^N + (2-\sqrt{3})^N)^2$$
and inside the bracket we have a natural number. 
