Periodic Continued Fraction Formula for $\sqrt{a^2+4}$ for any non-zero integer $a$ If $$x = \sqrt{ a^2 + 4 } $$ then I have conjectured that:
$$ x = [ a ; \overline{ \frac{a}{2} , 2a }] \ \ \ \text{ if a is even}$$
and $$ x = [ a ; \overline{ \frac{a}{2}, 1, 1, \frac{a}{2}, 2a } ]\ \ \  \text{            if a is odd} $$ where we take the floor of $\frac{a}{2} $
The even version is easy enough to prove. But for the odd version, we note that $$\frac{a}{2} = [\frac{a}{2}] + \frac{1}{2} = [\frac{a}{2}] + \frac{1}{1+\frac{1}{1}}$$
and I suspect that this is why the pair of ones emerges, but still a bit stuck! 

Follow the usual development of a continued fraction.  It is important to express the first term as $\frac {a-1}2$ as that can be used in algebra instead of $\frac a2$ rounded down.
$$\frac 1{\sqrt{a^2+4}-a}=\frac {a-1}2+\frac {\sqrt{a^2+4}-a+2}4\\
\frac 4{\sqrt{a^2+4}-a+2}=1+\frac {\sqrt{a^2+4}-2}a\\
\frac a{\sqrt{a^2+4}-2}=1+\frac{\sqrt{a^2+4}+2-a}a\\\frac a{\sqrt{a^2+4}-(a-2)}=\frac{a-1}2+\frac {\sqrt{a^2+4}-a}4\\
\frac a{\sqrt{a^2+4}-a}=2a+(\sqrt{a^2+4}-a)$$
and you can read off the desired repeat.
