# Solution

Express $\sqrt{17}$ in continued fraction form.

First, we separate the integer part and fractional part of $\sqrt{17}$. Since $4^2 \lt 17 \lt 5^2$,

$$\sqrt{17} = 4 + (\sqrt{17} - 4)$$

Then,

$$\begin{array}{rcl} \sqrt{17} &=& 4 + (\sqrt{17} - 4)\\ &=& 4 + \frac{1}{\frac{1}{\sqrt{17}-4}}\\ &=& 4 + \frac{1}{\frac{1}{\sqrt{17}-4} \cdot \frac{\sqrt{17} + 4}{\sqrt{17}+4}}\\ &=& 4 + \frac{1}{\frac{\sqrt{17}+4}{1}}\\ &=& 4 + \frac{1}{8 + (\sqrt{17} - 4)} \end{array}$$

Seeing the duplication of $(\sqrt{17}-4)$, we immediately know that the continued fraction form we seek is $[4;\overline{8}]$.