Prove that sum is an integer. Prove that: $\sqrt{2-\frac{1}{1^2+\sqrt{1^4+\frac{1}{4}}}}+\sqrt{2-\frac{1}{2^2+\sqrt{2^4+\frac{1}{4}}}}+\sqrt{2-\frac{1}{3^2+\sqrt{3^4+\frac{1}{4}}}}+\cdots+\sqrt{2-\frac{1}{119^2+\sqrt{119^4+\frac{1}{4}}}}$ is an integer. 
I noticed that $n^4+\frac{1}{4}$ is rather interesting, because it would be written as $(n^2+n+\frac{1}{2})(n^2-n+\frac{1}{2})$, but I do not know how to connect this to the rest of the problem. The square roots seem to further complicate the problem. Furthermore, I don't see what's special with 119, but upon using a program, I noticed that the only such numbers (such that the sum is an integer) below 500 are 3, 20, and 119. 

Since
$$\frac{1}{k^2+\sqrt{k^4+\frac 14}}=\frac{k^2-\sqrt{k^4+\frac 14}}{(k^2)^2-(k^4+\frac 14)}=-4k^2+2\sqrt{4k^4+1}$$
we have
$$\sum_{k=1}^{n}\sqrt{2-\frac{1}{k^2+\sqrt{k^4+\frac 14}}}=\sum_{k=1}^{n}\sqrt{2+4k^2-2\sqrt{4k^4+1}}\tag1$$
Here, since we have that
$$4k^4+1=4k^4+4k^2+1-4k^2=(2k^2+1)^2-(2k)^2=(2k^2+2k+1)(2k^2-2k+1)$$
and that
$$(2k^2+2k+1)+(2k^2-2k+1)=2+4k^2$$
we have
$$\begin{align}(1)&=\sum_{k=1}^{n}\sqrt{\left(\sqrt{2k^2+2k+1}-\sqrt{2k^2-2k+1}\right)^2}\\\\&=\sum_{k=1}^{n}\left(\sqrt{2k^2+2k+1}-\sqrt{2k^2-2k+1}\right)\\\\&=\sum_{k=1}^{n}\left(\sqrt{2k^2+2k+1}-\sqrt{2(k-1)^2+2(k-1)+1}\right)\\\\&=\sqrt{2n^2+2n+1}-\sqrt{1}\end{align}$$
This is an integer for $n=119$ since $2\cdot 119^2+2\cdot 119+1=169^2$.

Added :
$(1)$ is an integer if and only if there is an integer $m$ such that
$$2n^2+2n+1=m^2\iff (2n+1)^2-2m^2=-1$$
This is a Pell's equation, so we have 
$$n=-\frac 12+\left(\frac{1-\sqrt 2}{4}\right)\cdot (3-2\sqrt 2)^k+\left(\frac{1+\sqrt 2}{4}\right)\cdot (3+2\sqrt 2)^k$$
where $k=1,2,3,\cdots$.
These satisfy $a_n=6a_{n-1}-a_{n-2}+2$ with $a_1=3,a_2=20$ :
$$a_3=119,\quad a_4=696,\quad a_5=4059,\quad a_6=23660,\quad \cdots$$
