Let $ P(x,y) $ denote assertion of given functional equation. 1) $ P(0, y )$ gives us that $ f(f(y))=y+f(0) $, so we conclude that $f$ is bijective. 2) $ P(0,0 )$ gives us $f(f(0))=f(0)$, since $f$ is bijective, then $f(0)=0$ and $f(f(y))=y$ 3) $P(x,0)$ gives us that $f(2x)=x+f(x)$ 4)$ P(x, f(y))$ gives us that $f(2x+y)=f(2x)+f(y)$. It means that $f$ is additive. As a result $ f(2x)=2f(x)=x+f(x) $ , which implies that $ f(x)=x$

$ My $ $ solution $ $ P(0,y)\implies f(f(y))=y+f(0) $ $ f(2x+f(y))=x+f(x)+y\implies f-surjective\implies f(t)=0 $ $ P(0,t)\implies f(0)=0+f(0)+t\implies t=0\implies f(0)=0 $ $ f(f(y))=y+f(0)=y\implies f-bijective $ $ P(x,0)\implies f(2x)=x+f(x) $ $$ P(f(x),0)\implies f(2f(x))=f(f(x))+f(x)=x+f(x)=f(2x)\implies f(2f(x))=f(2x) $$ $ f(2f(x))=f(2x)\implies 2f(x)=2x\implies f(x)=x $

Same as my solution

We can clearly see that $f$ is injective. Setting $x=-y$ gives: $$f(-2y+f(y))=f(-y)$$But because $f$ is injective: $$-2y+f(y)=-y$$$$ \iff f(y)=y$$Hence $f(x)=x$ is the only solution.

Ln142 wrote: We can clearly see that $f$ is injective. Setting $x=-y$ gives: $$f(-2y+f(y))=f(-y)$$But because $f$ is injective: $$-2y+f(y)=-y$$$$ \iff f(y)=y$$Hence $f(x)=x$ is the only solution. could you please explain how can we prove that f is injective? Actually i am a beginner.

$$P(0,y): f(f(y))=y+f(0)$$$$\implies f(f(0))=f(0)$$$$P(x,f(0)): f(2x+f(0))=x+f(0)+f(x)=x+0+f(x)$$$$\implies f(0)=0$$$$\implies f(f(x))=x$$$$P(-y,y): f(-2y+f(y))=f(-y)$$$$\implies f(f(-2y+f(y)))=f(f(-y)) \implies f(y)=y; \forall y\in \mathbb{R}$$And it is easy to see that this is in fact a solution.

Ucchash wrote: could you please explain how can we prove that f is injective? Actually i am a beginner. You get $f(f(x))=x+f(0)$ by plugging $x=0$, so if $f(a)=f(b), f(f(a))=f(f(b)) \implies a+f(0)=b+f(0) \implies a=b$.

$P(x,x) \Longrightarrow f(2x+f(x))=2x+f(x)$ $P(0,x) \Longrightarrow f(f(x))=x+f(0)$ $P(x,f(x)) \Longrightarrow f(3x+f(0))=3x+f(0) \Longrightarrow f(x)=x$

Ucchash wrote: Ln142 wrote: We can clearly see that $f$ is injective. Setting $x=-y$ gives: $$f(-2y+f(y))=f(-y)$$But because $f$ is injective: $$-2y+f(y)=-y$$$$ \iff f(y)=y$$Hence $f(x)=x$ is the only solution. could you please explain how can we prove that f is injective? Actually i am a beginner. Let $a$, $b$ be real numbers such that $f(a)=f(b)$. We will show that this implies $a=b$. Substituting $y=a$ in the original functional equation, we get: $$ x+a+f(x)=f(2x+f(a))=f(2x+f(b))=x+b+f(x)$$Therefore $a=b$.

Let $P(x,y)$ be the assertion of $f (2x + f (y)) = x + y + f (x)$. $P(0,y)\implies f(f(y))=y+f(0)\implies f\text{ is bijective}$. $P(-x,x)\implies f(-2x+f(x))=f(-x)\implies f(x)-2x=-x\implies f(x)=x$ This indeed satisfies the given.

Let $P(x,y)$ denote the assertion. By $P(0,y)$ we have $f$ injective, moreover by $P(0,0)$ we have $f(0)=0,$ hence by combining $P(0,y)$ and $P(f(x),y)$ we have $2x=2f(x)$. Hence $f(x)=x~\forall x\in\mathbb{R}$.

Very easy functional equation.

Here's my adaptation, a small bit harder. Let $\mathbb H\subseteq\mathbb R$ be some nonempty set. Determine all functions $f:\mathbb H\to\mathbb H$, such that $2x+f(y)\in\mathbb H$ and $f (2x + f (y)) = x + y + f (x)$ for all $x,y \in \mathbb H$.

Let $P(x,y)$ denote the given assertion. $P(-f(x),x): f(-f(x))=x-f(x)+f(-f(x))\implies x-f(x)=0\implies \boxed{f(x)=x}$, which works.

After finding $f$ injective $P(x,-x)$