17. Find the number of real solutions of the equation
$$
x^{2}+\frac{1}{x^{2}}=2006+\frac{1}{2006} \text {. }
$$
17. Ans: 4

Consider the equation $y+1 / y=a$, where $a>0$. It can be changed into
$$
y^{2}-a y+1=0 .
$$

Observe that it has two positive real solutions:
$$
y=\frac{a \pm \sqrt{a^{2}-4}}{2}>0 .
$$

Thus the equation
$$
x^{2}+\frac{1}{x^{2}}=2006+\frac{1}{2006} .
$$
has four real solutions (i.e., $\pm \sqrt{200} \overline{6}, \pm 1 / \sqrt{2006}$ ).