20.19. (Belgium, 77). Find all infinitely differentiable functions $f: \mathbf{R} \rightarrow \mathbf{R}$, satisfying the identity

$$
f(x+y) \equiv f(x)+f(y)+2xy, \quad x, y \in \mathbf{R}
$$
20.19. Substituting $x=y=0$ into the original identity, we get $f(0)=$ $=2 f(0)$, i.e., $f(0)=0$. For any $x \in \mathbb{R}$ from the identity

$$
f(x+y)-f(x)=f(y)+2 x y \quad(y \in \mathbf{R})
$$

## we have

$$
\begin{aligned}
f^{\prime}(x)=\lim _{y \rightarrow 0} \frac{f(x+y)-f(x)}{y}= & \lim _{y \rightarrow 0} \frac{f(y)+2 x y}{y}= \\
& =2 x+\lim _{y \rightarrow 0} \frac{f(y)-f(0)}{y}=2 x+f^{\prime}(0)
\end{aligned}
$$

therefore, the sought functions must satisfy the condition

$$
f(x)=f(x)-f(0)=\int_{0}^{x} f^{\prime}(y) d y=x^{2}-f f^{\prime}(0) x_{1}
$$

Note that for any value $a \in \mathbb{R}$, the function $f(x)=x^{2}+a x$ satisfies the required identity in the problem.