34.12. Find all continuous solutions of the functional equation

$$
f(x+y)=f(x)+f(y)+f(x) f(y)
$$
34.12. It is clear that

$$
f(2 x)=2 f(x)+(f(x))^{2}=(1+f(x))^{2}-1
$$

We will prove by induction on $n$ that for any natural $n$ the equality holds

$$
f(n x)=(1+f(x))^{n}-1
$$

Indeed, from this equality it follows that

$$
\begin{aligned}
& f((n+1) x)=f(n x+x)=f(n x)+f(x)+f(x) f(n x)= \\
& \quad=(1+f(x))^{n}-1+f(x)+f(x)(1+f(x))^{n}-f(x)=(1+f(x))^{n+1}-1 .
\end{aligned}
$$

Let $f(1)=c$. Write equality (2) for $x=1$:

$$
f(n)=(1+c)^{n}-1
$$

Write this equality also for $x=m / n$:

$$
f\left(n \frac{m}{n}\right)=\left(1+f\left(\frac{m}{n}\right)\right)^{n}-1
$$

But $f\left(n \frac{m}{n}\right)=f(m)=(1+c)^{m}-1$, therefore

$$
\left(1+f\left(\frac{m}{n}\right)\right)^{n}=(1+c)^{m}, \quad \text { i.e. } \quad f\left(\frac{m}{n}\right)=(1+c)^{m / n}-1
$$

Here, we always need to take the positive value of the root, because if we replace $x$ with $x / 2$ in equality (1), we get $f(x)+1=(1+f(x / 2))^{2} \geqslant 0$.

In the end, by continuity considerations, we obtain that if $x>0$, then $f(x)=k^{x}-1$, where $k=c+1 \geqslant 0$.

Substitute $x=0$ and then $y=-x$ in the original functional equation:

$$
\begin{aligned}
& f(y)=f(0)+f(y)+f(0) f(y) \\
& f(0)=f(x)+f(-x)+f(x) f(-x)
\end{aligned}
$$

If $f(y) \not \equiv-1$, then $f(0)=0$. In this case, for $x>0$ we get

$$
0=k^{x}-1+f(-x)+\left(k^{x}-1\right) f(-x)=k^{x}-1+k^{x} f(-x)
$$

from which $f(-x)=k^{-x}-1$.

Thus, we have obtained that either $f(x) \equiv-1$, or $f(x) \equiv 0$, or $f(x)=k^{x}-1$, where $k>0, k \neq 1$. A simple check shows that all these functions satisfy the given functional equation.