Let's determine the integer solutions of the equation $\frac{a+1}{b}=c \cdot \frac{a}{b+1}$.
Clearly, $b$ cannot be 0 and -1; multiplying by $b(b+1)$

$$
(a+1)(b+1)=c a b
$$

If $a=0$, then from (1) $b=-1$ follows, which we have excluded; thus $a \neq 0$. If $a=-1$, then $0=c a b=-c b$ implies $c=0$, that is, $a=-1, c=0$ with any (different from 0 and -1) $b$ is a solution. In the following, we assume that (like $b$) $a \notin\{-1 ; 0\}$.

The right side of (1) is divisible by $a$, so $a \mid (a+1)(b+1)$. Since $a$ and $a+1$ are coprime, it follows that $a \mid b+1$ and similarly $b \mid a+1$.

I. If $a, b>-1$, then $a$ and $b$ are positive, so from the divisibility conditions $a \leq b+1$ and $b \leq a+1$ follow. Then $a \leq b+1 \leq a+2$, so $b$ can only be $a-1, a$ or $a+1$.

If $b=a-1$, then from $b \mid a+1$ we get $\frac{a+1}{a-1} \geq 2$, that is, $a \leq 3$; in this case, the following solutions are obtained: $a=2, b=1, c=3$ and $a=3, b=2, c=2$. If $b=a$, then similarly from $\frac{a+1}{a} \geq 2$ we get $a \leq 1$, from which the solution $a=b=1$, $c=4$ is obtained. If $b=a+1$, then we arrive at the solutions obtained in the case $b=a-1$ by swapping the roles of $a$ and $b$: $a=1, b=2, c=3$ and $a=2, b=3, c=2$.

II. If $a-1$, then $b>0$, so from $a \mid b+1$ we get $-a \leq b+1$ and from $b \mid a+1$ we get $b \leq -a-1$; therefore $-a \leq b+1 \leq -a-1+1=-a$, that is, $b=-a-1$. Substituting back into the equation, we then get the solutions

$$
a=t, \quad b=-t-1, \quad c=1 \quad(t>0)
$$

We show that there are no more solutions apart from the ones found so far. Indeed, in the remaining case $a, b<-1$, from $a \mid b+1$ we get $a \geq b+1$, and from $b \mid a+1$ we get $b \geq a+1$, so $a \geq b+1 \geq a+2$, which is impossible.

László Siroki (Debrecen, Fazekas M. Gymn., 10th grade) György Venter (Fazekas M. Main Pract. Gymn., 12th grade)