## Task A-4.2.

Determine all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that for all real numbers $x$ and $y$,

$$
f(x) f(y)=f(x+y)+x y
$$
## Solution.

By substituting $y=0$, we get $f(x) f(0)=f(x)$ for all real numbers $x$.

If $f(x)=0$ for all $x$, then the initial equation would read $0=0+xy$ for all $x, y$, which is impossible. Therefore, it must be that $f(0)=1$.

By substituting $x=1, y=-1$, we get $f(1) f(-1)=f(0)-1$, or $f(1) f(-1)=0$.

If $f(1)=0$, then by substituting $y=1$ into the initial equation, we get $0=f(x+1)+x$ for all $x$, and by substituting $z=x+1$, we get $f(z)=1-z$ for all $z$.

Similarly, if $f(-1)=0$, by substituting $y=-1$, we get $0=f(x-1)-x$ for all $x$, and by substituting $z=x-1$, we get $f(z)=1+z$ for all $z$.

By substituting, we see that the functions $f(x)=1-x$ and $f(x)=1+x$ indeed satisfy the initial equation, and these are the solutions we are looking for.