f(x + f(y)) = f(x) + y on the rationals

Let $c=f(0)$.

Injectivity. $x=0$: $f(f(y))=c+y$. If $f(y_1)=f(y_2)$ then $f(f(y_1))=f(f(y_2))$, so $y_1=y_2$.

$c=0$. $y=0$: $f(x+c)=f(x)$. By injectivity $x+c=x$, so $c=0$. Hence $f(f(y))=y$ ($f$ is an involution).

Additivity. Replace $y$ by $f(y)$ in the equation: $f(x+f(f(y)))=f(x)+f(y)$, i.e.
$$f(x+y)=f(x)+f(y).$$

Solving on $\mathbb{Q}$. As in the Cauchy problem, $f(q)=qa$ for every rational $q$, where $a=f(1)\in\mathbb{Q}$. Now $f(f(1))=1$ gives $a\cdot a=1$, so $a=\pm1$.

Answer. $f(x)=x$ and $f(x)=-x$. Check: for $f(x)=-x$, $f(x+f(y))=-(x-y)=y-x=f(x)+y$ ✓.

Remark. Over $\mathbb{R}$ without regularity, the same argument shows that $f$ is an *additive involution*, and there are exotic ones, so the problem on $\mathbb{R}$ needs an extra hypothesis (continuity, for example).

Lesson. Injectivity from $f(f(y))=\ldots$ is the first thing to look for; then show an involution and replace $y$ by $f(y)$ to simplify.