Question

Show that
${(a+\frac{1}{2a+}\frac{1}{2a+}\cdots )}^2+{(a-\frac{1}{2a-}\frac{1}{2a-}\cdots )}^2=2a^2$,
and $(a+\frac{1}{2a+}\frac{1}{2a+}\cdots )(a-\frac{1}{2a-}\frac{1}{2a-}\cdots )=a^2-\frac{1}{2a^2-}\frac{1}{2a^2-}\cdots$.

Answer 1

$\Rightarrow$Let x & y be the continued fraction

$x-a=\frac{1}{2a+x-a}=\frac{1}{x+a}$

$x^2-a^2=1$

$x=\sqrt{a^2+1}$

$y-a=-\frac{1}{2a+y-a}=\frac{-1}{y+a}$

$y^2-a^2=-1$

$y=\sqrt{a^2-1}$

$\therefore x^2+y^2=a^2+1+a^2-1=2a^3$

$x.y=\sqrt{a^4-1}$

$\sqrt{a^4-1}=a^2-(a^2-\sqrt{a^4-1})=a^2-\frac{1}{a^2+\sqrt{a^4-1}}$

$a^2+\sqrt{a^4-1}=2a^2-(a^2-\sqrt{a^4-1})=a^2-\frac{1}{a^2+\sqrt{a^4-1}}$

and so on

$\therefore xy=\sqrt{a^4-1}=a^2-\frac{1}{2a^2-}\frac{1}{2a^2-}.....$