$x^{4}-2 a x^{2}+b^{2}=0$ equation, express its roots as the sum of two simple square roots!
The solutions for $x^{2}$ of the quadratic equation $x^{4}-2 a x^{2}+b^{2}=0$ are:

$$
x^{2}=a \pm \sqrt{a^{2}-b^{2}}, \quad \text { or } \quad x= \pm \sqrt{a \pm \sqrt{a^{2}-b^{2}}}
$$

Here, we have obtained four values for $x$; each is a complex square root. It is sufficient to express just one of them as the sum of two simple square roots. Let

$$
\sqrt{a+\sqrt{a^{2}-b^{2}}}=\sqrt{u}+\sqrt{v} \text { and thus } a+\sqrt{a^{2}-b^{2}}=u+v+2 \sqrt{u v}
$$

The latter equation can only hold if

$$
u+v=a \quad \text { and } \quad 2 \sqrt{u v}=\sqrt{a^{2}-b^{2}} \quad \text { or } \quad u v=\frac{a^{2}-b^{2}}{4}
$$

Accordingly,

$$
u \text { and } v \text { are the roots of the equation } z^{2}-a z+\frac{a^{2}-b^{2}}{4}=0:
$$

$$
\begin{aligned}
& u=z_{1}=\frac{1}{2}\left(a+\sqrt{a^{2}-4 \frac{a^{2}-b^{2}}{4}}\right)=\frac{1}{2}(a+b) \\
& v=z_{2}=\frac{1}{2}\left(a-\sqrt{a^{2}-\frac{a^{2}-b^{2}}{4}}\right)=\frac{1}{2}(a-b)
\end{aligned}
$$

Or perhaps $u=z_{2}$ and $v=z_{1}$. The exchange of $u$ and $v$ does not yield different solutions, and thus

$$
\sqrt{a+\sqrt{a^{2}-b^{2}}}=\sqrt{\frac{a+b}{2}}+\sqrt{\frac{a-b}{2}}
$$

Similarly:

$$
\sqrt{a+\sqrt{a^{2}-b^{2}}}=\sqrt{\frac{a+b}{2}}-\sqrt{\frac{a-b}{2}}
$$

The roots of the equation are:

$$
x= \pm\left(\sqrt{\frac{a+b}{2}} \pm \sqrt{\frac{a-b}{2}}\right)
$$

Judit Komlós (Szent-Erzsébet Girls' High School, Pécs)