[Proof] The angle $22.5^{\circ}$ is a quarter of a right angle, and can be constructed as follows: on the extension of the right-angle side $CA$ of the isosceles right triangle $ABC$, from vertex $A$ outward, construct a line segment $AD = AB$, and connect points $D$ and $B$. In the isosceles triangle $DAB$, $\angle D = \frac{1}{2} \angle BAC$. This constructs a quarter of a right angle.

If the length of each leg of $\triangle ABC$ is taken as 1, then

and
$$\begin{aligned} D A & = A B \\ & = \sqrt{A C^{2} + B C^{2}} = \sqrt{2}, \\ D C & = D A + A C = \sqrt{2} + 1 \end{aligned}$$
$$D B = \sqrt{D C^{2} + B C^{2}} = \sqrt{(\sqrt{2} + 1)^{2} + 1} = \sqrt{4 + 2 \sqrt{2}} .$$

From this, we get $\square$
$$\begin{array}{l} u = \operatorname{ctg} 22.5^{\circ} = \frac{D C}{B C} = \sqrt{2} + 1 \\ v = \frac{1}{\sin 22.5^{\circ}} = \frac{D B}{B C} = \sqrt{4 + 2 \sqrt{2}} \end{array}$$

Therefore $\square$
$$\begin{array}{l} (u - 1)^{2} = 2, \\ v^{2} = 4 + 2 \sqrt{2} \text{, i.e., } \left(v^{2} - 4\right)^{2} = 8 . \end{array}$$

By expanding and simplifying (1) and (2), we obtain the equations
$$u^{2} - 2u - 1 = 0, v^{4} - 8v^{2} + 8 = 0$$