Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ satisfying for all $x, y \in \mathbb{R}$,

$$
f(x+y)=f(x-y)
$$
Here, it is judicious to choose $y=x$ in the equation, in order to simplify the right-hand side. We obtain:

$$
\forall x \in \mathbb{R}, \quad f(2 x)=f(0)
$$

By replacing $x$ with $\frac{x}{2}$, we get that $f$ is a constant function: for all $y \in \mathbb{R}, f(y)=C$, with $C \in \mathbb{R}$. Conversely, all constant functions are clearly solutions.

## Remark 3.

If $a$ and $b$ are real numbers, we could also set $x=\frac{a+b}{2}$ and $y=\frac{a-b}{2}$ to obtain $f(a)=f(b)$, and we find that $f$ must be constant.

## Remark 4.

In the previous exercise, we can notice that in the left-hand side, the variables $x$ and $y$ play a symmetric role, which is not the case in the right-hand side. To exploit such symmetry/asymmetry properties, it is often useful to swap $x$ and $y$.