Find the functions from $\mathbb{R}$ to $\mathbb{R}$ that satisfy:

$$
f(x+y)+f(x) f(y)=f(x y)+2 x y+1
$$
Let $P(x, y)$ be the proposition $f(x+y)+f(x) f(y)=f(x y)+2 x y+1$

Let:

$$
f(0)=a \quad f(1)=b \quad f(-1)=c
$$

We have:

$$
\begin{aligned}
P(x,-1) & \Longrightarrow f(x-1)+c f(x)=f(-x)-2 x+1 \quad(\star) \\
P(x-1,1) & \Longrightarrow f(x)+(b-1) f(x-1)=2 x-1 \quad(\star \star) \\
(1-b)(\star)+(\star \star) & \Longrightarrow(1+c-b c) f(x)+(b-1) f(-x)=-2 b x-b
\end{aligned}
$$

By replacing $x$ with $-x$, we obtain:

$$
(1+c-b c) f(-x)+(b-1) f(x)=2 b x-b
$$

Thus, we have the following system:

$$
\left\{\begin{array}{l}
(1+c-b c) f(x)+(b-1) f(-x)=-2 b x-b \\
(b-1) f(x)+(1+c-b c) f(-x)=2 b x-b
\end{array}\right.
$$

If the determinant is non-zero, we find affine functions, and we manually verify which ones work. We get $f(x)=-x-1$ and $f(x)=2 x-1$. If the determinant is zero, it gives: $1+c-b c= \pm(b-1)$. By substituting into the first or second equation, we get $b=0$. Therefore:

$$
\begin{aligned}
& P(0,1) \Longrightarrow b+a b=a+1 \Longrightarrow a=-1 \\
& P(-1,1) \Longrightarrow a+b c=c-1 \Longrightarrow c=0
\end{aligned}
$$

The previous system then gives $f(x)=f(-x)$. We can then take:

$$
\begin{aligned}
& P\left(\frac{x}{2}, \frac{x}{2}\right) \Longrightarrow f(x)+f\left(\frac{x}{2}\right)^{2}=f\left(\frac{x^{2}}{4}\right)+\frac{x^{2}}{2}+1 \\
& P\left(\frac{x}{2},-\frac{x}{2}\right) \Longrightarrow-1+f\left(\frac{x}{2}\right)^{2}=f\left(\frac{x^{2}}{4}\right)-\frac{x^{2}}{2}+1
\end{aligned}
$$

These two equations imply $f(x)=x^{2}-1$, which is indeed a solution.

Key ideas: Exploit symmetry, compare multiple equations.