When $\sqrt{1+12n^2} = k \in \mathbb{Z}_+^*$ we have $k^2 - 3(2n)^2 = 1$. The Pell equation $k^2 - 3\ell^2 = 1$ has all solutions given by $(k_m,\ell_{m})$, $m\geq 1$, where $k_m \pm \ell_m\sqrt{3} = (2\pm \sqrt{3})^m$ (by the classical theory behind it). But $\ell_m$ is even (thus qualifies for being $2n$) if and only if $m = 2t$ is even (easily seen from the recurrence relations for $k_m,\ell_m$). Then $2 + 2k_{2t} = 2 + (2 + \sqrt{3})^{2t} + (2 - \sqrt{3})^{2t} = \left ((2 + \sqrt{3})^{t} + (2 - \sqrt{3})^{t}\right )^2 = (2k_t)^2$, thus a perfect square.

$\sqrt{1+12n^2}$ is an integer$\Longrightarrow $ roots of $x^2-x-3n^2=0$ are integers. Call them $a, b$ where $a<b$. Indeed, $\frac{1-\sqrt{1+12n^2}}{2}=a<b=\frac{1+\sqrt{1+12n^2}}{2}$. Now, $ab=-3n^2, a+b=1\Longrightarrow (a, b)=1$. Also, $a<b$ and $ab<0\Longrightarrow a<0<b$. So, $(a, b)=(-3p^2, q^2), (-p^2, 3q^2)$ from $ab=-3n^2$ for some natural $p, q$. If it is the former case, then we are done as $2+2\sqrt{1+12n^2}=4b=(2q)^2$ would be a perfect square. Suppose $(a, b)=(-p^2, 3q^2)$, then, since $a+b=1, 3q^2-p^2=1\Longrightarrow 3|(p^2+1)$ which is not possible as any perfect square never leaves remainder $2$ when divided by $3$. So, we are done.

This is problem 6 in BMO-1 2006/7. See here or here for a similar problem. PS. Can this be solved: [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1860280]slagscat[/url] wrote: I wonder if we can find a set of naturals K, such that for all k in K, $2+2\sqrt{1+km^2}$ is an integer iff it is a perfect square. ?

barcelona wrote: Let n is a natural number,for which $\sqrt{1+12n^2}$ is a whole number.Prove that $2+2\sqrt{1+12n^2}$ is perfect square. Let $\sqrt{1+12n^2}=m \Leftrightarrow 12n^2=(m-1)(m+1)$. Because $m$ is odd, $\text{gcd}(m-1,m+1)=2$, so $m-1=2a^2$ and $m+1=6b^2$ or $m-1=6u^2$ and $m+1=2v^2$. In first case, if we subtract the expressions we get that $3b^2=a^2+1\Rightarrow a^2\equiv 2$ $($mod $3)$, witch is contradiction, so $m+1=2v^2$ and $m=2v^2-1$. On the other hand $2+2\sqrt{1+12n^2}=2(m+1)=2(2v^2-1+1)=(2v)^2$ witch is perfect square.