35. Find the function $f: \mathbf{R} \rightarrow \mathbf{R}$, such that for $\forall x \in \mathbf{R}, \forall y \in$ $\{f(x) \mid x \in \mathbf{R}\}$. We have $f(x-y)=f(x)+x y+f(y)$, where $\mathbf{R}$ is the set of real numbers.
(15th Korean Mathematical Olympiad)
Solution: First, assume $f(x)=c$ is a constant, then
$$
c=f(x-c)=f(x)+c x+f(c)=c x+2 c, x \in \mathbf{R} \text {. }
$$

Therefore, $c=0$, hence $f(x)=0$.
Second, assume $f(x)$ is not a constant. Then there exist $a, b \in \mathbf{R}$, such that $f(a)=b \neq 0$. Thus for all $x \in \mathbf{R}$, we have
$$
\begin{array}{l}
f(x-b)=f(x)+b x+f(b) \\
\Rightarrow f(x-b)-f(x)=b x+f(b) . \\
\text { Therefore, }\{f(x-b)-f(x) \mid x \in \mathbf{R}\} \\
=\{b x+f(b) \mid x \in \mathbf{R}\}=\mathbf{R} .
\end{array}
$$

Observing, if $b \in D=\{f(x) \mid x \in \mathbf{R}\}$, then
$$
\begin{array}{l}
f(b-b)=f(b)+b^{2}+f(b) \\
\Rightarrow f(b)=\frac{-b^{2}+f(0)}{2} .
\end{array}
$$

From equation (1), for any given $x \in \mathbf{R}$, there exist $y_{1}, y_{2} \in$ $D$, such that $x=y_{1}-y_{2}$. From equation (2) we have
$$
\begin{array}{l}
f(x)=f\left(y_{1}-y_{2}\right)=f\left(y_{1}\right)+y_{1} y_{2}+f\left(y_{2}\right) \\
=\frac{-y_{1}^{2}+f(0)}{2}+\frac{-y_{2}^{2}+f(0)}{2}+y_{1} y_{2} \\
=-\frac{\left(y_{1}-y_{2}\right)^{2}}{2}+f(0)=-\frac{x^{2}}{2}+f(0) .
\end{array}
$$

Therefore, $f(b)=-\frac{b^{2}}{2}+f(0)=-\frac{b^{2}}{2}+\frac{f(0)}{2}$, which implies $f(0)=0$.
Hence $f(x)=0$ or $f(x)=-\frac{x^{2}}{2}$.