(M) We are given $a, b \in \mathbb{Q}$, find the $f: \mathbb{Q} \rightarrow \mathbb{Q}$ such that

$$
f(x+a+f(y))=f(x+b)+y
$$
We fix $x$, and obtain the bijectivity of $f$. We set $x=-b$ and have $f(a-b+f(y))=f(0)+y$ and we set $y=f(0)$ to obtain $f(x+a+f(f(0)))=f(x+b)+f(0)=f(a-b+f(f(x+b))$ and by injectivity we have

\[
f(f(x))=f(f(0))+x
\]

we then set $x=f(w)-b$ and obtain $f(f(w)+f(y)+a-b)=w+y+f(f(0))=f(f(w+y))$ and by injectivity

\[
f(w)+f(y)+a-b=f(w+y)
\]

Thus, $g=f-\frac{a-b}{2}$ satisfies the Cauchy equation whose solutions in $\mathbb{Q}$ are linear. Hence, $f(x)=A x+\frac{a-b}{2}$. Properly synthesize this.