Find all functions $f: \mathbf{R} \rightarrow \mathbf{R}$ such that for all $x, y \in \mathbf{R}$, the following holds:
$$
f(x+y)+f(x) f(y)=f(x y)+2 x y+1 .
$$
(46th IMO Shortlist)
Let $x=y=0$, we get $f^{2}(0)=1$.
If $f(0)=1$, let $y=0$, we get $f(x)=1$.
Substituting into equation (1) yields
$$
1+1 \times 1=1+2 x y+1 \Rightarrow x y=0 \text {. }
$$

This does not meet the problem's requirements.
Therefore, $f(0)=-1$.
Let $x=1, y=-1$, we get
$$
f(1) f(-1)=f(-1) \text {. }
$$
(1) If $f(1)=1$, let $y=1$, we get
$$
f(x+1)=2 x+1 \Rightarrow f(x)=2 x-1 \text {. }
$$
(2) If $f(-1)=0$, let $x=-2, y=1$, we get
$$
f(-2) f(1)=f(-2)-3 \text {; }
$$

Let $x=y=-1$, we get $f(-2)=f(1)+3$. Substituting into the right side of equation (2) yields $f(-2) f(1)=f(1)$.
(i) $f(1)=0$.

Substitute $x-1$ for $x$, let $y=1$, we get
$$
f(x)=f(x-1)+2(x-1)+1 \text {; }
$$

Substitute $-x+1$ for $x$, let $y=-1$, we get
$$
f(-x)=f(x-1)+2(x-1)+1 \text {. }
$$

Thus, $f(x)=f(-x)$.
Let $y=-x$, we get
$$
f^{2}(x)=f\left(x^{2}\right)-2 x^{2}+2 \text {; }
$$

Let $y=x$, we get
$$
f(2 x)+f^{2}(x)=f\left(x^{2}\right)+2 x^{2}+1 .
$$

Thus, $f(2 x)=4 x^{2}-1 \Rightarrow f(x)=x^{2}-1$.
(ii) $f(-2)=1$, i.e., $f(1)=-2$.

Let $y=1$, we get
$$
f(x+1)=3 f(x)+2 x+1 \text {. }
$$

Substitute $-x$ for $x$, let $y=-1$, we get
$$
\begin{array}{l}
f(-x-1)=f(x)+2 x+1 . \\
\text { Hence } f(x+1)-f(-x-1)=2 f(x) \\
\Rightarrow f(x)-f(-x)=2 f(x-1) \\
\Rightarrow f(-x)-f(x)=2 f(-x-1) \\
\Rightarrow f(x-1)=-f(-x-1) .
\end{array}
$$

Thus, $-f(x-1)=f(x)+2 x+1$.
$$
\begin{array}{l}
\text { Also } f(x)=3 f(x-1)+2(x-1)+1 \\
\Rightarrow f(x)=-3(f(x)+2 x+1)+2 x-1 \\
\Rightarrow f(x)=-x-1 .
\end{array}
$$

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

all satisfy the conditions, hence the solutions are the above three functional equations.