TA’s Challenge Problem #5

**TheAbstractionist** · July 14th 2009, 05:29 AM

This will be a very easy problem indeed.

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

**NonCommAlg** · July 14th 2009, 06:44 PM

Originally Posted by TheAbstractionist

This will be a very easy problem indeed.

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

considering the condition $d \mid 2n,$ i guess you're looking for a "simple" continued fraction of $n + \sqrt{n^2+k}.$ let $2n=kd$ and $x=n+\sqrt{n^2 +k}.$ then $x=2n + \frac{1}{d + \frac{1}{x}}.$ therefore: $n+\sqrt{n^2+k}= 2n \ + \ \frac{1}{d \ + \ \frac{1}{2n \ + \ \frac{1}{d \ + \ \frac{1}{2n \ + \ \cdots}}}}.$

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