Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that

$$
f(x y+z)=f(x) f(y)+f(z)
$$

for all real numbers $x, y$, and $z$.
Answer: $f(x)=0$ or $f(x)=x$. Both of these trivially satisfy the equation.

We plug in $x=y=z=0$ to get

$$
\begin{array}{r}
f(0)=f(0)^{2}+f(0) \\
0=f(0)^{2} \\
0=f(0)
\end{array}
$$

Then we plug in $x=y=1, z=0$ to get

$$
\begin{aligned}
f(1) & =f(1)^{2}+f(0) \\
0 & =f(1)^{2}-f(1)
\end{aligned}
$$

so $f(1)=0$ or $f(1)=1$. If $f(1)=0$, we can immediately plug in $y=1$ and $z=0$ to conclude that

$$
f(x)=f(x) \cdot 0+0=0
$$

for all $x$.

If $f(1)=1$, then we can derive that $f$ has two simple properties: the addition property from plugging in $y=1$,

$$
f(x+z)=f(x)+f(z)
$$

and the multiplication property from plugging in $z=0$,

$$
f(x y)=f(x) f(y) .
$$

Plugging $z=-x$ into the addition property tells us that $f(-x)=-f(x)$, i.e. $\mathrm{f}$ is an odd function. Also, the addition property lends itself to use for an induction, starting at $f(1)=1$, that proves $f(n)=n$ for all positive integers $n$.

Suppose that $f$ is not the identity function. Then there is a number $x$ such that $f(x) \neq x$. Changing $x$ to $-x$ if necessary, we can assume that $f(x)0$ such that $f(u)0$,

$$
f(u)=f(\sqrt{u} \cdot \sqrt{u})=f(\sqrt{u}) \cdot f(\sqrt{u}) \geq 0 .
$$