Solve the following equation:

$$
\frac{1-a x}{1+a x} \sqrt{\frac{1+b x}{1-b x}}=1 .
$$

For which values of $a$ and $b$ will the roots of the equation be real?
Solve the following equation:

$$
\frac{1-a x}{1+a x} \sqrt{\frac{1+b x}{1-b x}}=1
$$

For which values of $a$ and $b$ will the roots of the equation be real?

If we square both sides of the equation and remove the denominators, we get the following equation:

$$
(1-a x)^{2}(1+b x)=(1+a x)^{2}(1-b x)
$$

or, after rearranging,

$$
a^{2} b x^{3}-(2 a-b) x=0
$$

The roots of this equation are

$$
\begin{aligned}
& 1^{\circ} \quad x=0 \\
& 2^{\circ} \quad x= \pm \frac{1}{a} \sqrt{\frac{2 a-b}{b}}
\end{aligned}
$$

For these to be real, it is necessary that

$$
\frac{a}{b} \geq \frac{1}{2}
$$

# Editor's note: 

The roots of the equation under 2) satisfy the equation under 1), but in some cases only the positive value of the square root, and in others the negative value. For the roots of 2) to satisfy 1) with the positive value of the square root, it is necessary that the expression

$$
\frac{1-a x}{1+a x}
$$

and with it the expression

$$
(1-a x)(1+a x)=\left(1-a^{2} x^{2}\right)
$$

be positive when the roots of 2) are substituted for $x$, i.e., if

$$
1-\frac{2 a-b}{a}>0
$$

or if

$$
\frac{a}{b} > 1
$$

If $\frac{a}{b} < 1$, the roots of 2) satisfy the equation

$$
\frac{1-a x}{1+a x} \sqrt{\frac{1+b x}{1-b x}}=-1
$$

where the square root is again taken to be positive.

If $\frac{a}{b}=1$, only the root $x=0$ of 2) satisfies the equation under 1), while $x=+\frac{1}{a}=+\frac{1}{b}$ is not a root of 1) at all.

The problem was solved by: VII. class of the Budapesti ág. hitv. ev. fögymn.; Grossmann Gusztáv, VIII. class, Budapest; Grünhut Béla, VI. class, Pécs; Imre János, VIII. class, Nyíregyháza; Reif Jenő, VI. class, Pécs; Weisz Lipót, VI. class, Pécs.

10 more solutions were received, some of which were partially incorrect, and others not as complete as the above 6 young colleagues'.