Solve the quadratic equation

x^{2} = 7 x + 1

. This equation is equivalent to

x = 7 + \frac{1}{x}

which has solutions given by the infinite continued fraction

$x = 7 + \frac{1}{7 + \frac{1}{7 + \frac{1}{7 + \dots}}} .$

This is because, if we think of this last equation as being $x = 7 + \frac{1}{y}$ , then clearly $y = x$ . Show that the sequence of numbers

$7 + \frac{1}{7}, 7 + \frac{1}{7 + \frac{1}{7}}, 7 + \frac{1}{7 + \frac{1}{7 + \frac{1}{7}}}, \dots$

gives better and better approximations to one of the solutions of the original quadratic equation. [Note that to find these approximations you can simply repeat the steps: 'take reciprocal, add 7', over and over again].

Find integers $a$ and $b$ , with $b$ less than 400, such that $\frac{a}{b}$ , is equal to $\sqrt{5} 3$ correct to six significant figures.

Now consider $x^{2} = 5 x + 1$ , ...

[See the articles Continued Fractions I and Continued Fractions II. ]