Solve the following equation:

$$
\frac{a-x}{a+x}+\frac{b-x}{b+x}=\frac{a+x}{a-x}+\frac{b+x}{b-x} .
$$
By moving the terms on the right side of our equation to the left side, we obtain:

$$
\left(\frac{a-x}{a+x}-\frac{a+x}{a-x}\right)+\left(\frac{b-x}{b+x}-\frac{b+x}{b-x}\right)=0
$$

or

$$
\frac{(a-x)^{2}-(a+x)^{2}}{a^{2}-x^{2}}+\frac{(b-x)^{2}-(b+x)^{2}}{b^{2}-x^{2}}=0
$$

After performing the indicated operations in the numerators,

$$
\frac{4 a x}{a^{2}-x^{2}}+\frac{4 b x}{b^{2}-x^{2}}=0
$$

or

$$
4 x\left[a\left(b^{2}-x^{2}\right)+b\left(a^{2}-x^{2}\right)\right]=0
$$

One of the roots of this equation is $x=0$, which also satisfies the original equation. The other root of the equation is provided by

$$
a\left(b^{2}-x^{2}\right)+b\left(a^{2}-x^{2}\right)=0 \quad \text { or } \quad(a+b) x^{2}=a b(a+b)
$$

If $a+b=0$ now, then

$$
x^{2}=a b \quad \text { and } \quad x= \pm \sqrt{a b}
$$

The roots of the given equation are: $x=0$ and $x= \pm \sqrt{a b}$.

If $a+b=0$, that is, $b=-a$, then our equation can be written in the form

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

Clearly, we have an identity, which is satisfied by any value of $x$.

Freud Géza (Berzsenyi g. VI. o. Bp. V.)