Let's determine the pairs of parameters $p, q$ for which the equation

$$
\frac{2 x-2 p+q}{2 x-2 p-q}=\frac{2 q+p+x}{2 q-p-x}
$$

has exactly one solution.
Let's enclose the terms containing $x$ and $p$ in parentheses for now, and the equation can be written as:

$$
\frac{2(x-p)+q}{2(x-p)-q}=\frac{2 q+(x+p)}{2 q-(x+p)}
$$

Multiplying both sides by the product of the two denominators - assuming of course that the product is not 0 - two terms cancel out:

$$
2 q^{2}-2\left(x^{2}-p^{2}\right)=-2 q^{2}+2\left(x^{2}-p^{2}\right), \quad x^{2}=p^{2}+q^{2}
$$

We did not lose any roots by multiplying, so the equation can only be satisfied by the numbers

$$
x_{1}=\sqrt{p^{2}+q^{2}}, \quad x_{2}=-\sqrt{p^{2}+q^{2}}
$$

or the equation has no solution.

The requirement of the problem can only be met in the following two ways:
I. $x_{1}$ and $x_{2}$ are not different: $x_{1}=x_{2}=0$, and this satisfies the equation. However, the expressions (2) can only become 0 with the value pair $p=q=0$ (considering only real numbers $p, q$), but with $x=p=q=0$, both sides of the given equation are meaningless. This way does not lead to a solution.

II. One of $x_{1}$ and $x_{2}$ is a solution, the other is not, because the equation is meaningless for it.

a) The left-hand side fraction becomes meaningless if

$$
x^{*}=p+\frac{q}{2}
$$

So let's find a value pair $p, q$ for which one of $x_{1}$ and $x_{2}$ is equal to $x^{*}$:

$$
\pm \sqrt{p^{2}+q^{2}}=p+\frac{q}{2}
$$

From here, with the usual steps,

$$
q^{2}=p q+\frac{q^{2}}{4}, \quad q\left(p-\frac{3 q}{4}\right)=0, \quad \text { so }
$$

$a_{1}$) case: either $q=0$, and $p$ is any number - except the $p=0$ value already excluded in I -,

$a_{2}$) case: or $p=3 q / 4$, and again $q \neq 0$.

In the $a_{1}$) case, (2) becomes: $x_{1}=p-$ which is exactly the excluded value $x^{*}$, and $x_{2}=-p$, which should be the only root. However, with $x_{2}=-p$, the right-hand side of (1) is meaningless, so we do not get a suitable $p, q$ number pair here.

In the $a_{2}$) case, $x^{2}=25 q^{2} / 16$, but the excluded value is $x^{*}=5 q / 4$, so the only possible solution is:

$$
x=-\frac{5 q}{4}
$$

With this, the left and right sides of (1) are

$$
\frac{-\frac{5 q}{2}-\frac{3 q}{2}+q}{-\frac{5 q}{2}-\frac{3 q}{2}-q}=\frac{3}{5}, \quad \text { or } \quad \frac{2 q+\frac{3 q}{4}-\frac{5 q}{4}}{2 q-\frac{3 q}{4}+\frac{5 q}{4}}=\frac{3}{5}
$$

so (3) is indeed a solution.

b) Similarly, the right-hand side of (1) becomes meaningless if

$$
x^{* *}=2 q-p
$$

The requirement $x_{1,2}=x^{* *}$ leads to $q(3 q-4 p)=0$, or the conditions of case a), so we do not get a new suitable $p, q$ number pair.

Therefore, the requirement of the problem is met only for those and only those value pairs $p, q$ for which $p: q=3: 4$ and $q \neq 0$. For these value pairs, the only solution to the equation is $x=-5 q / 4$, or $x=-5 p / 3$.

Lehel Jenö (Budapest, Apáczai Csere J. gym. g. IV. o. t.)