Is there a rational pair of numbers $a, b$ for which the following equality holds?

$$
\frac{a+b}{a}+\frac{a}{a+b}=b
$$
I. solution. If there exists a suitable pair of numbers $a, b$, then the quotient $a / b=r$ is also a rational number. (It cannot be $b=0$, because then (1) would not be satisfied, as its left side could only be 2.) Substituting $a=b r$ into (1), we get:

$$
\frac{r+1}{r}+\frac{r}{r+1}=b
$$

From this, we immediately see that every rational number, except for $r=0$ and $r=-1$, gives a suitable pair of numbers $b, a$:

$$
b=\frac{r+1}{r}+\frac{r}{r+1}, \quad a=r+1+\frac{r^{2}}{r+1}
$$

Pintz János (Budapest, Fazekas M. High School)

Remark. The following is even simpler than the above solution. The two terms on the left side are rational numbers and reciprocals of each other (so neither can be 0). Denoting the first by $r^{\prime}$,

$$
\begin{aligned}
& b=r^{\prime}+\frac{1}{r^{\prime}}, \quad \text { where } \frac{a+b}{a}=r^{\prime}, \text { so } \\
& a=\frac{1}{r^{\prime}-1} \cdot b=\frac{1}{r^{\prime}-1}\left(r^{\prime}+\frac{1}{r^{\prime}}\right), \text { if } r^{\prime} \neq 1
\end{aligned}
$$

If $r^{\prime}$ is a rational number different from 0 and 1, then (2) and (3) give rational $b$ and $a$. (It can also be seen that we have obtained all rational solutions.)

II. solution. Removing the fractions and rearranging (1) in terms of $a$, we get:

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

For $b$, only values can be used for which the discriminant $r_{1}$ has a rational square root:

$$
D=b^{2}(2-b)^{2}-4(2-b) b^{2}=b^{2}\left(b^{2}-4\right)=r_{1}^{2}
$$

that is,

$$
\begin{aligned}
& b^{2}-4=r_{2}^{2} \\
& b^{2}=r_{2}^{2}+4
\end{aligned}
$$

where $r_{2}$ is also a rational number.

Looking for solutions in integers, only $r_{2}=0, b^{2}=4, b= \pm 2$ are possible. However, $b=2$ leads to a contradiction with (4); for $b=-2$, $a=1$ is the only root, so the pair of numbers $a, b=1,-2$ is a solution, and we can already state that the answer to the question is affirmative.

Takács László (Sopron, Széchenyi I. High School)

Remarks. 1. Generally, looking for rational $b$ and $r_{2}$ in the form $B / N$, and $R / N$ respectively, where $B, R, N$ are integers, (5) transforms into:

$$
R^{2}+(2 N)^{2}=B^{2}
$$

According to this, every Pythagorean triple provides a rational solution:

$$
b=\frac{u^{2}+v^{2}}{u v},\left(r_{2}=\frac{u^{2}-v^{2}}{u v}\right), a=\ldots=\frac{u^{2}+v^{2}}{v(v-u)}
$$

where $u, v$ are integers different from each other and from 0.

Takács László

2. However, a single pair of numbers $a, b$ could have been more easily obtained from the rearrangement of (4) in terms of $b$:

$$
(1-a) b^{2}+a(2-a) b+2 a^{2}=0
$$

indeed, this equation simplifies to a first-degree equation for $a=1$, and thus its solution is rational: $b=-2$.