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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!

Hugh Entwistle
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1 Answers1

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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.

Ross Millikan
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  • Thank you so much! I haven't done much stuff with the floor function, so I didn't really make the connection that I could simplify it so easily! Thanks again – Hugh Entwistle Nov 11 '16 at 15:20
  • Floor functions do not play nice with algebra, so you generally need to get rid of them. Once you know that $a$ is odd, you can find that half rounded down is half the next lower number and get the expression that I used. – Ross Millikan Nov 11 '16 at 15:58