10.4. Find all values of the parameter $a$ for which the equation $x^{4}-a x^{2}+1=0$ has 4 roots forming an arithmetic progression.
Answer: $\frac{10}{3}$. Solution. Let $t=x^{2}$. Then the equation $t^{2}-a t+1=0$ must have two positive roots $t_{1}, t_{2}\left(t_{1}<t_{2}\right)$, and the roots of the original equation will be $-\sqrt{t_{2}},-\sqrt{t_{1}}, \sqrt{t_{1}}, \sqrt{t_{2}}$. The condition of the problem about the arithmetic progression gives the relation $\sqrt{t_{2}}-\sqrt{t_{1}}=2 \sqrt{t_{1}}$, i.e., $\sqrt{t_{2}}=3 \sqrt{t_{1}} \Rightarrow t_{2}=9 t_{1}$. From Vieta's theorem, we have $t_{1} t_{2}=1$ and $t_{1}+t_{2}=a$. Hence: $9 t_{1}^{2}=1 \Leftrightarrow t_{1}=\frac{1}{3}$ (considering the positivity of $t_{1}$) and $a=t_{1}+9 t_{1}=10 t_{1}=\frac{10}{3}$.