Let $g(x) = f(x)-x,$ and you're (quickly) done ^^.

I will write the solutions with more details. Using the above subtitution we obtain that $g(x+g(y))=g(x)+y$ (1). For $x=0$ the last relation becomes $g(g(y))=g(0)+y$ From this we find that $g$ is surjective ,so also f is surjective so there exist u such that $f(u)=0$ .Setting in the first relation where x,y the u we obtain $f(f(u))=2f(u)$ so $f(0)=0$ so $g(g(y))=y$ (2) .Now put in (1) where y the g(y) to have $g(x+y)=g(x)+g(y)$ (3) due to (2) and (3) holds for all $x,y\in Z$ due to the surjectivity of g. So $g(x)=cx$ where $c=g(1)$ so $f(x)=x+xc'$ . But subtituting this in the first relation gives c'=-1 or c'=1 so the fuctions are $f(x)=2x$ and $f(x)=0$

http://www.mathlinks.ro/Forum/viewtopic.php?p=521359#p521359

silouan wrote: I From this we find that $g$ is surjective ,so also f is surjective I don't think it is true

silouan wrote: $ g(g(y)) = g(0) + y$ From this we find that $ g$ is surjective ,so also f is surjective so there exist u such that $ f(u) = 0$ .Setting in the first relation where x,y the u we obtain $ f(f(u)) = 2f(u)$ so $ f(0) = 0$ so $ g(g(y)) = y$ From $ g(g(y))=g(0)+y$ we find that $ g$ is injective,hence by taking $ x=0$ in (1):we obtain that $ g(0)=0\Rightarrow g(g(y))=y$.

For $ x=y$ we have $ f(f(x))=2f(x)$(1). For $ x=y+1$ $ =>$ $ f(1+f(y))=f(y+1)+f(y)$ and by induction obtain $ f(k+f(x))=f(x)+f(x+k)$(2).Now for $ x->f(x)$ and $ y->f(y)$ we have $ f(f(x)-f(y)+f(f(y)))=f(fx))+f(f(y))$ $ =>$ $ f(f(x)+f(y))=2f(x)+2f(y)$ (1) $ =>$ $ f(x)+f(f(y)+x)=2f(x)+2f(y)$ (2) $ =>$ $ f(x)+f(y)+f(x+y)=2f(x)+2f(y)$(2) $ =>$ $ f(x+y)=f(x)+f(y)$ $ =>$ $ f(x)=kx$, k-integer. Now we have $ k(x-y+f(y))=k(x+y)$ and we have $ k=0$ or $ k=2$ $ =>$ $ f(x)=0$ and $ f(x)=2x$.

For x=y, we get f(f(x))=2f(x). For $y \mapsto f(y)$, we have f(f(y)+x)=f(x)+2f(y) (*), and for $x=f(y), 2f(y), ...$ in (*), we get by induction that $f(nf(y))=2nf(y)$, $(\forall n\in \mathbb{N})(\forall y\in \mathbb{Z})$. Hence, for $x=2y,3y,...$ in the functional equation, and using (*), we obtain by induction f(ny)=nf(y), $(\forall n\in \mathbb{N})(\forall y\in \mathbb{Z})$. Taking y=1 and y=-1, we have $f(n)=nf(1) (\forall n\in \mathbb{N})$ and $f(n)=-nf(-1) (\forall n\in \mathbb{Z}- \mathbb{N})$ We deduce that f(n)=2n or f(n)=0 after reporting in the functional equation.

By substituting $ x=y $ we have $ f(f(x))=2f(x) $.Define $ a_1=f(x) $, $ a_n=f(a_{n-1})=2 \cdot a_{n-1} $,then $ a_n=a_1 \cdot 2^{n-1},n>1 $.If the function is not injective,let $ a,b $ be numbers such that $ a\neq b $ and $ f(a)=f(b)=c $.By substituting these values into the original equation we have $ f(x-a+c)=f(x)+c=f(x-b+c) $,therefore $ f(x) $ is periodic.However,it is an integer function,and since it is periodic it must be bounded.However $ a_n=a_1 \cdot 2^{k-1} $ is not bounded unless $ f(x)=0 $.Therefore the function is injective and we have $ f(x-y+f(y))=f(y-x+f(x))$ which is equivalent to $ x-y+f(y)=y-x+f(x)\Rightarrow f(x)-2x=f(y)-2y $,thus $ f(x)=2x+c $,where $ c $ is a constant.Plugging into the original equation we find that $ f(x)=2x $.

AlexanderMusatov wrote: By substituting $ x=y $ we have $ f(f(x))=2f(x) $.Define $ a_1=f(x) $, $ a_n=f(a_{n-1})=2 \cdot a_{n-1} $,then $ a_n=a_1 \cdot 2^{n-1},n>1 $.If the function is not injective,let $ a,b $ be numbers such that $ a\neq b $ and $ f(a)=f(b)=c $.By substituting these values into the original equation we have $ f(x-a+c)=f(x)+c=f(x-b+c) $,therefore $ f(x) $ is periodic.However,it is an integer function,and since it is periodic it must be bounded.However $ a_n=a_1 \cdot 2^{k-1} $ is not bounded unless $ f(x)=0 $.Therefore the function is injective and we have $ f(x-y+f(y))=f(y-x+f(x))$ which is equivalent to $ x-y+f(y)=y-x+f(x)\Rightarrow f(x)-2x=f(y)-2y $,thus $ f(x)=2x+c $,where $ c $ is a constant.Plugging into the original equation we find that $ f(x)=2x $. i know it's a stupid question but i still have so some questions about fct.equations...why can't we proove injectivity like this? suppose $f(f(x))=f(f(y))$ $P(x,x)$ we have $f(f(x))=2f(x)$ thus from our assumption $f(f(x))=2f(x)=2f(y)=f(f(y))$ so from $f(a)=f(b)$ follows $a=b$ with $a=f(x)$ , $b=f(y)$ ?????????????????

dudeldai wrote: i know it's a stupid question but i still have so some questions about fct.equations...why can't we proove injectivity like this? suppose $f(f(x))=f(f(y))$ $P(x,x)$ we have $f(f(x))=2f(x)$ thus from our assumption $f(f(x))=2f(x)=2f(y)=f(f(y))$ so from $f(a)=f(b)$ follows $a=b$ with $a=f(x)$ , $b=f(y)$ ????????????????? $f: \mathbb{Z}\rightarrow\mathbb{Z}$ What you need to prove, in order to prove injectivity : $\forall (x,y) \in \mathbb{Z}^2, f(x)=f(y) \Rightarrow x=y$. What you have proven : $\forall (x,y) \in (f(\mathbb{Z}))^2, f(x)=f(y) \Rightarrow x=y$.

Varal7 wrote: dudeldai wrote: i know it's a stupid question but i still have so some questions about fct.equations...why can't we proove injectivity like this? suppose $f(f(x))=f(f(y))$ $P(x,x)$ we have $f(f(x))=2f(x)$ thus from our assumption $f(f(x))=2f(x)=2f(y)=f(f(y))$ so from $f(a)=f(b)$ follows $a=b$ with $a=f(x)$ , $b=f(y)$ ????????????????? $f: \mathbb{Z}\rightarrow\mathbb{Z}$ What you need to prove, in order to prove injectivity : $\forall (x,y) \in \mathbb{Z}^2, f(x)=f(y) \Rightarrow x=y$. What you have proven : $\forall (x,y) \in (f(\mathbb{Z}))^2, f(x)=f(y) \Rightarrow x=y$. yeah i told you it was a stupid question.. btw thanks

Let $p(x,y)\Rightarrow f(x-y+f(y))=f(x)+f(y)$. $p(x,x)\Rightarrow f(f(x))=2f(x).....(i)$ $p(x+y,y)\Rightarrow f(x+f(y))=f(x+y)+f(y).....(ii)$ $p(x,f(y))\Rightarrow f(x+f(y))=f(x)+2f(y)......(iii)$(By using $(i)$) So $f(x+y)+f(y)=f(x)+2f(y)\Rightarrow f(x+y)=f(x)+f(y)\; \; \forall x,y\in \mathbb {Z}$ Remember this function is from $\mathbb {Z}$ to $\mathbb {Z}$. So $f(x)=\zeta x\; \; \forall x\in \mathbb {Z}$ with an integer constant $\zeta $. By plunging in the main equation we find only $\zeta =0$ and $\zeta =2$ satisfies the equation.So all the functions are $f(x)=0$ and $f(x)=2x$.

From x=y we have f(f(x))=2f(x) Now when y=f(z), we got f(x-f(z)+f(f(z)))=f(x)+f(f(z)) -> f(x+f(z))=f(x)+2f(z) When x=x-z -> f(x-z+f(z))=f(x-z)+2f(z) -> f(x)+f(z)=f(x-z)+2f(z) -> f(x)=f(x-z)+f(z) Therefore f is an additive function -> f(x)=cx where c=const. From f(x-y+f(y))=f(x)+f(y) ----> c=0 or c=2

Let $f(0)=c$ $x=y$ gives $f(f(0))=2f(0)$ so $f(c)=2c$ If $x=c,y=c$ we get $f(2c)=2f(c)=4c$ If $x=c,y=0$ we get $f(2c)=3c$ So $c=0$ and $f(0)=0$ Now let $g(x) = f(x)-x$ so $g(0)=0$ and $g(x+g(y))=g(x)+y$ and we easily get that $g(x)$ is injective so $g(g(x))=x$ that gets $g(x)=x$ So $f(x)=2x$ for all $x$.

The only solutions to the equation are $f(x)=2x$ and $f(x)=0$. Let $P(x,y)$ be the given assertion. From $P(x,x)$, we get that $f(f(y))=2f(y)$. From $P(-f(0),0)$ we get that $f(0)=f(-f(0))+f(0) \implies f(-f(0))=0$. From $P(x,-f(0))$, we get that $f(x+f(0))=f(x)=f(x)+f(0)$. Thus we must have that $f(0)=0$. From $P(0,y)$ we get that $f(f(y)-y)=f(y)$. Now we must have that $f(x+f(y)-y)=f(x)+f(f(y)-y)$. From $P(x,f(t)-t)$, we get that $f(x+t)=f(x)+f(t)$. Thus we get that $f$ is additive on $\mathbb{Z}$, i.e. $f(x)=cx$. When we plug that in $f(f(y))=2f(y)$, we get that $c\in \{ 0,2\}$