Setting $x=a-2005$ and $y=2004$ gives us $f(a) = a-2005 + f(2005)$. Setting $a=2004$ yields $2005=(-1) + f(2005)$ or, equivalently, $f(2005)=2006$. Thus, $f(a) = a+1$. . Since $f(a)=a+1$ is clearly a solution, we're done.

let x=1 and f(y)=k than we have :$f(1+k)=1+f(k)$ since f is defined on R we can have $f(1+k)=1+f(k),\Rightarrow f(1)=1+f(0),f(2)=1+f(1)=2+f(0)...f(2004)=2004+f(0)$ now since $f(2004)=2005\wedge f(2004)=2004+f(0) \Rightarrow f(0)=1$ the functoin i $f(x)=1+x$ maybe im wrong but if someone can tell me i;ll be thankfull

@rahman: You need to prove that $f$ is surjective before you can assume $k = f(y)$. (Otherwise, there are some values of $k$ that you cannot find a $y$ for it, which means you cannot plug in that value of $k$.)

But how to prove that a function is surjective

Prove that for any $x$, there exists $y$ such that $f(y )=x$. In this case, it is actually easy to prove: Fix $y$ and let $x = x-f(f(y))$. We have $f( \text{something} ) = x$. Since $x$ can take all values, we have proven that $f$ is surjective. In other functional equations, this might be harder or even impossible (if $f$ is indeed not surjective, obviously you cannot "prove" that it is surjective).

JSGandora wrote: Find all functions $f:\mathbb{R}\rightarrow \mathbb{R}$ satisfying \[f(x+f(y))=x+f(f(y))\]for all real numbers $x$ and $y$, with the additional constraint $f(2004)=2005$. Here is my solution We claim that without the constraint $f(x)=x+c$ is the only solution. Let $P(x,y)$ denotes the assertion. $$P(-f(y),y) \implies f(0)=-f(y)+f(f(y)) \implies f(f(y)) = f(y) +c$$where $c=f(0)$. Substituting this back to original equation, we get that $f(x+f(y)) = x+f(y)+c$. Now then $$P(x-f(y),y) \implies \boxed{f(x)=x+c}$$. With Constraint : Using the given constraint, we get $a=1$. So $f(x)=x+1$ is the only solution atlast. $\blacksquare$

it's very easy We have clarity $f$ is surjective hence : $P(f(x),y)-P(f(y),x)\implies f(x)=x+c$ put $x=2004\implies c=1\implies f(x)=x+1$

$P(x-2005,2004)\Rightarrow f(x)=x+c$, testing gives $\boxed{f(x)=x+1}$.

JSGandora wrote: Find all functions $f:\mathbb{R}\rightarrow \mathbb{R}$ satisfying \[f(x+f(y))=x+f(f(y))\]for all real numbers $x$ and $y$, with the additional constraint $f(2004)=2005$. Let $P(x,y)$ the assertion of the given FE: $P(x-f(f(y)),y)$ $$f(x+f(y)-f(f(y)))=x \implies f \; \text{surjective}$$Since $f$ is surjective set $f(y)=t$ and $P(-t,y)$ $$f(t)=t+f(0)$$So now set $t=2004$ $$f(2004)=2004+f(0) \implies f(0)=1$$Thus the only solution is: $\boxed{f(x)=x+1 \; \forall x \in \mathbb R}$ Thus we are done

As usual, let $P(x,y)$ denote the given assertion. $P(x-f(f(y)),y): f(f(x-f(f(y))+f(y))=x$. From here, we know $f$ is surjective. Let $k$ be a real number such that $f(k)=2004$. $P(-2004,k): f(0)=f(2004)-2004=1$. Let $f(y)=-x$. Now, we have $1=x+f(-x)\implies f(-x)=-x+1$, so $\boxed{f(x)=x+1}$ is the only solution, and it clearly works.

Setting $x=x-f(0)$ and $y=0$ gives $f(x)=x+k.$ But only $f(x)=x+1$ works.