This will be a very easy problem indeed.

Find the continued-fraction expansion of $\displaystyle n+\sqrt{n^2+k}$ where $\displaystyle n,\,k$ are positive integers and $\displaystyle k$ divides $\displaystyle 2n.$

Printable View

- Jul 14th 2009, 05:29 AMTheAbstractionistTA’s Challenge Problem #5
This will be a very easy problem indeed.

Find the continued-fraction expansion of $\displaystyle n+\sqrt{n^2+k}$ where $\displaystyle n,\,k$ are positive integers and $\displaystyle k$ divides $\displaystyle 2n.$ - Jul 14th 2009, 06:44 PMNonCommAlg
considering the condition $\displaystyle d \mid 2n,$ i guess you're looking for a "simple" continued fraction of $\displaystyle n + \sqrt{n^2+k}.$ let $\displaystyle 2n=kd$ and $\displaystyle x=n+\sqrt{n^2 +k}.$ then $\displaystyle x=2n + \frac{1}{d + \frac{1}{x}}.$ therefore: $\displaystyle n+\sqrt{n^2+k}= 2n \ + \ \frac{1}{d \ + \ \frac{1}{2n \ + \ \frac{1}{d \ + \ \frac{1}{2n \ + \ \cdots}}}}.$