Exercise 10. Find all functions $\mathrm{f}$ from $\mathbb{R}$ to $\mathbb{R}$ such that for all real numbers $x, y$,

$$
f(x) f(y)=f(x y)+x+y
$$
Solution to Exercise 10 By evaluating at $y=0$ we obtain $f(x) f(0)=f(0)+x$ for all real $x$. For $x=1$, we get that $f(0) \neq 0$, otherwise we would have $1=0$. By evaluating at $x=0$ we get $f(0)^{2}=f(0)$, so $f(0)=1$ and thus $f(x)=1+x$ for all real $x$.

Conversely, let $f$ be the function such that $f(x)=1+x$ for all real $x$ and let $(x, y) \in \mathbb{R}^{2}$, then $\mathrm{f}(\mathrm{x}) \mathrm{f}(\mathrm{y})=(1+\mathrm{x})(1+\mathrm{y})=1+\mathrm{xy}+\mathrm{x}+\mathrm{y}=\mathrm{f}(\mathrm{xy})+\mathrm{x}+\mathrm{y}$, so $\mathrm{f}$ satisfies the equation. The unique solution to the equation is therefore the function $f$ such that $f(x)=1+x$ for all real $x$.

Grader's Comment The exercise is generally well done. Be careful when dividing by a number to ensure that the number is non-zero. Also, do not forget to check that the solutions found do indeed satisfy the functional equation.