24. The numbers $a, b, c$ satisfy the equations

$$
a b+a+b=c, \quad b c+b+c=a, \quad c a+c+a=b
$$

Find all possible values of $a$.
24. Answer: $0,-1,-2$.

Add 1 to both sides of all three equations and factorize the left sides. Let $x=a+1, y=b+1, z=c+1$, we get the system of equations

$$
x y=z, \quad y z=x, \quad x z=y .
$$

We can proceed similarly to the first solution, but we can also do it completely differently. Multiply all three equations: $(x y z)^{2}=x y z$. Therefore, $x y z=0$ or $x y z=1$. Recalling that $y z=x$, we get $x^{2}=0$ or $x^{2}=1$, hence $x=0,1,-1$.

It is easy to verify that all three cases are realized: $x=y=z=0 ; x=y=z=1 ; x=y=-1, z=1$.

It remains to make the reverse substitution $a=x-1$ and so on, and obtain the final answer.