$P(x,y): f(x+f(x+y))=x+f(f(x)+y)$ $P(x,0):f(x+f(x))=x+f(f(x))$ $f(a)=0\Longrightarrow P(a,0):0=a+0\Longrightarrow a=0$ $P(x,-x): f(x)=x+f(f(x)-x)$ $f(x)-x=t\Longrightarrow f(t)=t$

ratatuy wrote: $P(x,y): f(x+f(x+y))=x+f(f(x)+y)$ $P(x,0):f(x+f(x))=x+f(f(x))$ $f(a)=0\Longrightarrow P(a,0):0=a+0\Longrightarrow a=0$ $P(x,-x): f(x)=x+f(f(x)-x)$ $f(x)-x=t\Longrightarrow f(x)=x$ This is so messed up, well first u have to prove that there exists an $a$ such that $f(a)=0$ now the replace after that doesn't give $a=0$ and finally the fact that $t$ is an identity doesn't mean $f(x)=x$

Plugging $y=-f(x)$, we see that $f$ is surjective. Let $g(x)+x=f(x)$, so equation rewrites as $g(2x+y+g(x+y))+g(x+y)=g(x)+g(x+y+g(x))$, and putting $y \to y-x$, we get $g(x+y+g(y))+g(y)=g(x)+g(y+g(x))$, call this $P(x,y)$. Note that for any $a,b \in \mathbb{R}$ such that $g(a)=g(b)$, $P(a,y)$ and $P(b,y)$ gives that $g(y+g(y)+a)=g(y+g(y)+b)$ for all $y \in \mathbb{R}$. $P(x,0) \implies g(x+g(0))=g(x)+g(g(x))-g(0)$. $P(0,x) \implies g(x+g(0))=g(x+g(x))+g(x)-g(0)$, comparing this with above, we get $g(g(x))=g(x+g(x))$ for all $x \in \mathbb{R}$. Now, letting $a=g(x) , b=x+g(x)$, since $g(a)=g(b)$, we get $g(y+g(y)+g(x))=g(y+g(y)+x+g(x))$ for all $x,y \in \mathbb{R}$. Since $x+g(x)=f(x)$ was surjective, we get $g(z+g(x))=g(z+x+g(x))$ for all $x,z \in \mathbb{R}$, and putting $z \to z-g(x)$ we get that $g(z)=g(z+x)$ for all $x,z \in \mathbb{R}$, so $g$ is constant and $f(x)=x+c$ for some constant $c$. It is easy to see that all such functions work.

The issue was a bit complicated. I hope I have not solved any carelessness in the process A bit hard problem Let $P(x,y) : f(x+f(x+y))=x+f(f(x)+y)$ . $P(x,0) : f(x+f(x))=x+f(f(x))$ . $P(x,f(y)) , P(y,f(x))$ : $f(x+f(x+f(y)))-x=f(y+f(y+f(x)))-y$ . Now put $y=0$ . then $f(x+f(x+f(0)))-x=f(f(f(x)))$ . ( ) $P(x,-f(x))$ : $f(x+f(x-f(x)))=x+f(0) $ then $f$ is surjective ! . ( ) Now we know $\exists \alpha $ : $f(\alpha ) = 0$ . Now by $P(\alpha , 0)$ we get $f(0)=0$ . Now we know $f(0)=0$ . By ( ) we get $f(f(x))=f(f(f(x)))$ . ( ) $P(x,-x)$ : $f(x)=x+f(f(x)-x) \to f(f(x)-x) = x-f(x) \to f(f(f(x)-x)) = f(x-f(x)) \to x+f(f(f(x)-x)) = f(x-f(x))+x$ Now we get $f(x+f(f(f(x)-x))) = f(f(x-f(x))+x)$ . By ( ) and ( ) we get $f(x+f(f(x)-x))=x$ . ( ) $P(0,x+f(f(x)-x)) : f(f(x))=f(x+f(f(x)-x)) = x$ . then we can see $f$ is injective . Now by ( ) we get $f(x)=x$. $\blacksquare$

strong_boy wrote: The issue was a bit complicated. I hope I have not solved any carelessness in the process A bit hard problem Let $P(x,y) : f(x+f(x+y))=x+f(f(x)+y)$ . $P(x,0) : f(x+f(x))=x+f(f(x))$ . $P(x,f(y)) , P(y,f(x))$ : $f(x+f(x+f(y)))-x=f(y+f(y+f(x)))-y$ . Now put $y=0$ . then $f(x+f(x+f(0)))-x=f(f(f(x)))$ . ( ) $P(x,-f(x))$ : $f(x+f(x-f(x)))=x+f(0) $ then $f$ is surjective ! . ( ) Now we know $\exists \alpha $ : $f(\alpha ) = 0$ . Now by $P(\alpha , 0)$ we get $f(0)=0$ . Now we know $f(0)=0$ . By ( ) we get $f(f(x))=f(f(f(x)))$ . ( ) $P(x,-x)$ : $f(x)=x+f(f(x)-x) \to f(f(x)-x) = x-f(x) \to f(f(f(x)-x)) = f(x-f(x)) \to x+f(f(f(x)-x)) = f(x-f(x))+x$ Now we get $f(x+f(f(f(x)-x))) = f(f(x-f(x))+x)$ . By ( ) and ( ) we get $f(x+f(f(x)-x))=x$ . ( ) $P(0,x+f(f(x)-x)) : f(f(x))=f(x+f(f(x)-x)) = x$ . then we can see $f$ is injective . Now by ( ) we get $f(x)=x$. $\blacksquare$ $P(\alpha,0) \implies \alpha+f(0)=0.$

my solution Let $P(x,y) : f(x+f(x+y))=x+f(f(x)+y)$. $P(x,-f(x)) : f(x+f(x-f(x)))=x+f(0)$ $\Rightarrow$ $f$ is surjective. Let $f(0)=c$, because $f$ is subjective then $\exists b \in \mathbb R, \ f(b) = 0$. $P(b,0) : b+c=0 \Rightarrow b=-c,\ f(-c)=0$. $P(b,y) : f(b+f(y+b))=b+f(y)$ or $ f(f(y-c)-c)=f(y)-c $. $\Rightarrow f(f(y)-c)=f(y+c)-c \ \forall y \in \mathbb R$. $(1)$ $P(0,y) : f(f(y))=f(y+f(0))=f(y+c)$. $(2)$ From $(1)$ and $(2)$ we have $f(f(y)-c)=f(f(y))-c\ \forall y \in \mathbb R$. because $f$ is subjective then $f(y-c)=f(y)-c \Rightarrow f(y)+c=f(y+c)\ \forall y \in \mathbb R$. $(3)$ From $(2)$ and $(3)$ we have $f(f(y))=f(y)+c \Rightarrow f(y)=y+c\ \forall y \in \mathbb R$.

Clearly $f$ is surjective. Define $g(x):=f(x)-x$ and let $z=x+y.$ $\textbf{Claim:}$ The equation $$g(x+z+g(z))+g(z)=g(z+g(x))+g(x) \quad \forall x,z \in \mathbb R$$denoted by $E(x,z)$, is satisfied by constant functions only. $\textbf{Proof.}$ $E(x,x)$ yields $g(x+g(x))=g(2x+g(x)).$ Take $u$ such that $g(u)+u=-g(x)-x.$ Comparing $E(x+g(x),u)$ and $E(2x+g(x),u)$ we get $g(x)=g(0).$ $\blacksquare$

Another solution without shifting. Let $P(x,y)$ denote the assertion $f(x+f(x+y))=x+f(f(x)+y).$ Clearly $f$ is surjective. Take $f(a)=0$ then $P(a,0)$ gives $a=-f(0).$ By $P(0,x)$, $f(f(x))=f(x-a).$ Take $x_0$ such that $f(x_0)=x-a$, then $P(a,x_0-a)$ implies $f(f(x))=f(x)-a$ and so, $f(x')\equiv x'+c$, which works.

I claim that all solutions are of the form $f(x)=x+c$ for some $c$, these can be shown to work. Let $P(x,y)$ be the assertion. $P(x,-f(x))$ gives that $f(x+f(x-f(x)))=x+f(0)$, thus $f$ is surjective. For any $c$ such that $f(c)=0$, $P(c,-f(c))$ gives that $0=c+f(0)$, so $f(x)=0$ has exactly $1$ solution. Assume that there are distinct $a$ and $b$ such that $f(a)=f(b)$. Let $y_1$ be such that $f(f(a)+y_1)=a-f(a)$, and let $y_2$ be such that $f(f(a)+y_2)=b-f(a)$. Note that $y_1$ and $y_2$ must be distinct. $P(f(a),y_1)$ and $P(f(a),y_2)$ give that $f(f(f(a))+y_1)=f(f(f(a))+y_2)=0$, giving a contradiction, so $f$ is injective. $P(0,y)$ gives $f(y)=y+f(0)$ as desired.

chenjiaqi give a good solution ,but it is hard to type ,i am sorry

Different solution Let $P(x,y) : f(x+f(x+y))=x+f(f(x)+y)$ $P(0,x) \implies f(f(x))=f(f(0)+x) \ \ \ (1)$ $P(x,-f(x)) \implies f(x+f(x-f(x)))=x+f(0) \implies \text{ f is surjective }$ So $\exists k$ such that $f(k)=0$ $P(k,0) \implies f(k+f(k))=k+f(f(k)+0) \implies k=-f(0)$ $P(-f(0) , f(0)+y) \implies f(f(y)-f(0))=f(f(0)+y)-f(0) \ \ \ (2)$ Using $(1)$ in $(2)$ and substituting $f(y)$ with $z$ (we can do it since f is surjective) yields $f(z-f(0))=f(z)-f(0)$ Substituting $z$ with $z+f(0)$ , using $(1)$ and again substituting $f(z)$ with $t$ yields $f(t)=t+f(0) \implies \boxed{f(x) = x + c} \implies \text { check , works! }$ so we are done

Let $g(x)=f(x)-x$. $P(x,y-x)\Rightarrow Q(x,y):g(x+y+g(y))+g(y)=g(g(x)+y)+g(x)$ $Q(x,-g(x))\Rightarrow g(x-g(x)+g(-g(x)))+x-g(x)+g(-g(x))=x+g(0)$ so $g(x)+x$ is surjective. Let $g(a)+a=0$: $Q(a,a)\Rightarrow a=-g(0)\Rightarrow g(-g(0))=g(0)$ $Q(-g(0),x)-Q(0,x)\Rightarrow g(-g(0)+x+g(x))=g(x+g(x))\Rightarrow g(x)=g(x-g(0))$ since $x+g(x)$ is surjective. Then $g(x+g(0))=g(x)$ as well. $Q(0,x)\Rightarrow g(x+g(x))=g(0)\Rightarrow g(x)=g(0)$ so $g$ is constant. Then $f$ is linear, testing gives $f(x)=x+c$.

jasperE3 wrote: Let $g(x)=f(x)-x$. $P(x,y-x)\Rightarrow Q(x,y):g(x+y+g(y))+g(y)=g(g(x)+y)+g(x)$ $Q(x,-g(x))\Rightarrow g(x-g(x)+g(-g(x)))+x-g(x)+g(-g(x))=x+g(0)$ so $g(x)+x$ is surjective. Let $g(a)+a=0$: $Q(a,a)\Rightarrow a=-g(0)\Rightarrow g(-g(0))=g(0)$ $Q(-g(0),x)-Q(0,x)\Rightarrow g(-g(0)+x+g(x))=g(x+g(x))\Rightarrow g(x)=g(x-g(0))$ since $x+g(x)$ is surjective. Then $g(x+g(0))=g(x)$ as well. $Q(0,x)\Rightarrow g(x+g(x))=g(0)\Rightarrow g(x)=g(0)$ so $g$ is constant. Then $f$ is linear, testing gives $f(x)=x$. i think that f(x)=x+c is the solution.

EDIT: I made a mistake.

Parsia-- wrote: Setting $y = -f(x)$ we find that $f$ is surjective. $$P(0,y) \rightarrow f(f(y)) = f(f(0)+y)) \rightarrow f(y) = y+c$$which works. $f$ is inyective?

Let the assertion $P(x,y)=f(x+f(x+y))=x+f(f(x)+y)$ $P(0,x)$ yields: $f(f(x))=f(f(0)+x)$ (1) Since $f$ is surjective, $\exists \alpha, \text{such that}, f(\alpha)=0$ $P(\alpha,0)$ yields: $f(\alpha)=\alpha +f(0)\Longrightarrow \alpha=-f(0)$ $P(\alpha,x)$ yields: $f(f(x-f(0))-f(0))=f(x)-f(0)$, let $Q(x)$ denote $P(\alpha,x)$ $Q(y+f(0))$ yields: $f(f(y)-f(0))=f(y+f(0))-f(0)$ $\therefore f(f(y)-f(0)=f(f(y))-f(0)$ from (1) Furthermore let $f(y)=z\Longrightarrow f(z-f(0))=f(z)-f(0)$, thus let $z=k+f(0)$. This implies that: $f(k)=k+f(0)\Longrightarrow f(x)=x+c$ $\blacksquare$. And we are done!

f(x + f(x+y)) = x + f(f(x) + y)(1) (1) x --> 0 => f(f(y)) = f(y + f(0))(2) (1) y--> -f(x) => f : surjective There is exist some c such that f(c) = 0 (1) x--> c, y --> 0 => f(c) = c + f(0) => c = -f(0) (1) y --> -x+c => f(x) = x + f(f(x) - x +c ) => f(x) - x = f(f(x) - x +c) Let a = f(x) - x + c => f(a) = a+f(0) (1) x--> a, y --> y - a => f(a + f(y)) = a + f(f(a) + y - a) => f(a + f(y)) = a + f(y + f(0)) => f(a + f(y)) = a + f(f(y)) By (2) we can replace f(y) by z => f(z + a) = a + f(z)(3) (1) y --> y - x + c => f(x + f(y + c)) = x + f(f(x) + y - x +c) and a = f(x) - x +c => f(x + f(y+c)) = x + f(y) + f(x) - x + c => f(x + f(y - f(0))) = f(x) + f(y) + c, replace y by y + f(0) => f(x + f(y)) = f(x) + f(y + f(0)) + c => f(x + f(y)) = f(x) + f(y) + c By (2) we can replace f(y) by y => f(x + y) = f(x) + f(y) + c then g(x) = f(x) - f(0) => g(x+y) = g(x) + g(y) (1) => g(x + g(x + y) + f(0)) + f(0) = x + f(0) + g(g(x) + y + f(0)) => g(x + g(x + y)+ f(0)) - g(g(x) + y + f(0)) = x => g(x - y + g(y)) = x => put y = a => g(x) = x then f(x) = x+f(0) - is answer

Hardish. Good FE spam. Set $y=-f(x)$ to get surjectivity. Then, take $y=0$ to get $f(f(x))+x=f(x+f(x))$ (#). Then, rewrite the initial condition to $f(x+f(y))=x+f(f(x)+y-x)$ (1), by substituting $y$ by $y-x$. Put now $y=0$ here to get $$f(f(x))=f(f(x)-x)+x$$(*). Now, let for any $x$, $f(x)-x=t$. Let $a_t$ be the number (by surjectivity) that makes $f(a_t)=t$. Putting $y=a_t$ on (1), we get that $f(f(x))=x+f(t+a_t)$. But by (*) we may substitute $f(f(x))$ in the last expression to get $f(t)=f(t+a_t)$. But by (#), putting $x=a_t$, we get that $f(t+a_t)=a_t+f(f(a_t))=a_t+f(t)$. This implies, with the equality above, $a_t=0$ and thus $f(x)=x+f(0)$. The motivation behind this short solution is just trying everything and seeing how much $f(x)-x$ appears when you plug nice stuff.

Answer is $f(x)=x+c$ for any real constant $c$. Note that plugging $y=-f(x)$ yields the surjectivity of $f$. Let $f(x)-x=g(x)$. \[g(x+y)+2x+y+g(g(x+y)+2x+y)=x+y+x+g(x)+g(g(x)+x+y)\]Pick $x,y-x$ to get \[g(y)+g(g(y)+x+y)=g(x)+g(g(x)+y)\]$x=y$ gives $g(g(x)+2x)=g(g(x)+x)$ or $g(f(x)+x)=g(f(x))$. Comparing $P(f(x),y)$ with $P(f(x)+x,y)$ implies \[g(f(y)+f(x))=g(f(y)+f(x)+x)\]Since $f$ is surjective, we can replace $y-f(x)$ with $f(y)$. So $g(y)=g(y+x)$ which yields $g$ is constant. If $g\equiv c,$ then $f(x)=x+c$ as desired.$\blacksquare$