Solve the functional equation $f(x+a+f(y))=f(f(x))+a+y$

Question

So let $a$ be a real number. Find all functions $f:\mathbb{R}\rightarrow \mathbb{R}$ so that $f(x+a+f(y))=f(f(x))+a+y$, for all real $x,y$.

Answer 1

Let $P(x,y)$ be the property that $f(x+a+f(y)) = f(f(x))+a+y$ [1].

Then, $P(-a,x)$ gives us $f(f(x)) = f(f(-a))+a+x$ [2].

Substituting [2] into [1] yields $f(x+a+f(y)) = x+y+2a+f(f(-a))$, call this $Q(x,y)$.

Then, $Q(x-a-f(f(-a)),f(-a))$ yields $f(x) = x+a+f(-a) = x+C$ where $C = a+f(-a)$.

It is easy to verify that $f(x) = x+C$ is indeed a solution for any $C \in \mathbb{R}$.

Answer 2

$f(f(y))=f(f(0))+y=c+y$

Let $x=-c,y=f(0)$: we get $f(a)=a+f(0)+f(f(-c))=a+f(0)$

Then, for $f(x)=x+f(0)$ (substituting into orginial equation), get $x+a+y+2f(0)=x+a+y+2f(0)$ so $f(0)$ arbitrary and we have found the most general solution.