Let $P(x,y)$ be the assertion. $P(y,y)\implies f(y+f(y))=\pm(2y+f(0))^{(!)}\implies f(f(0))=\pm f(0)$ $P(x,0)\implies f(x+f(0))f(f(x))=(2x+f(-x))f(x)^{(!!)}$ Now,we will split into 2 cases: Case 1 $f(0)\neq 0$. If $f(f(0))=-f(0)$,then $(!)\implies f(0)=\pm (3f(0))\implies f(0)=0$,a contradiction. Hence $f(f(0))=f(0)$.Thus $(!)\implies f(2f(0))=\pm (3f(0))$ and then $(!!)\implies f(2f(0))f(f(f(0)))=(2f(0)+f(-f(0)))f(f(0))\implies f(2f(0))=2f(0)+f(-f(0))$ If $f(2f(0))=3f(0)$,then $f(-f(0))=f(0)$ and then $(!!)\implies f(0)f(f(-f(0)))=(-2f(0)+f(f(0)))f(-f(0))\implies f(0)=0$,which is again a contradiction.Hence $f(2f(0))=-3f(0)$ and thus $f(-f(0))=-5f(0)$. So $(!!)\implies f(0)f(f(-f(0)))=(-2f(0)+f(f(0)))f(-f(0))\implies f(-5f(0))=5f(0)$. Finally,$(!)\implies f(-5f(0)+f(-5f(0)))=\pm (-9f(0))\implies f(0)=\pm (9f(0))\implies f(0)=0$,a contradiction.Hence no solution in this case. Case 2 $f(0)=0$. If for some integer $c$,we have $f(c)=0$,then $P(c,c)\implies 0=4c^2\implies c=0$.Hence $f(c)=0\iff c=0$. Now,$(!)\implies f(y+f(y))=\pm 2y$ and $(!!)\implies f(x)f(f(x))=(2x+f(-x))f(x)\implies f(f(x))=2x+f(-x)\,\forall x\neq 0\implies f(f(x))=2x+f(-x)\,\forall x\in\mathbb{Z}^{(*)}$ $P(-f(y),y)\implies\forall y,f(y+f(y))=2f(y)\vee f(-f(y)-y)=-2y^{(-)}$ If for some $a\neq 0$,we have $f(a+f(a))=-2a$,then $(-),(*)\implies f(a)=-a\vee -2f(a)-2a+f(f(a)+a)=f(f(-f(a)-a))=f(f(f(a)+a))=2f(a)+2a+f(-f(a)-a)$ $\implies f(a)=-a\vee f(a)=-a\implies f(a)=-a\implies f(0)=-2a\implies a=0$,a contradiction. Hence $\boxed{f(y+f(y))=2y}$. Now $(-)\implies f(y)=y\vee [f(2y)=f(f(y+f(y)))=2y+2f(y)+f(-f(y)-y)=2f(y)$ $\wedge f(-2y)=f(f(-y-f(y)))=-2y-2f(y)+f(f(y)+y)=-2f(y)]$ But it is easy to see that if for some $a$,we have $f(a)=a$,then $f(2a)=2a$ and $f(-2a)=-2a$.Hence we have $\forall y,f(2y)=2f(y),f(-2y)=-2f(y)\implies \forall y,f(2y)=2f(y),f(y)=-f(-y)$. Hence $(*)$ become $f(f(x))=2x-f(x)$. $P(x,-x)\implies f(x-f(x))=\pm 2(x-f(x))$ If for some $x$,we have $f(x-f(x))\neq -2(x-f(x))$,then $x-f(x)\neq 0$ and $f(x-f(x))=2(x-f(x))$ So $(*) \implies f(2f(x-f(x)))=0\implies x-f(x)=0$,a contradiction. Hence $\boxed{f(x-f(x))=-2(x-f(x))}^{(+)}$ $P(f(y)+y,-y)\implies f(y)^2=(2f(y)+2y-f(2y+f(y)))(-2y+f(2y+f(y)))\implies \boxed{f(2y+f(y))=2y+f(y)}^{(++)}$ Now,it is easy to prove that $f(a)=a\implies f(na)=na\,\forall n\in\mathbb{Z}$ and $f(a)=-2a\implies f(na)=-2na\,\forall n\in\mathbb{Z}$ by induction. Hence there don't exist integer $a,b\neq 0$ such that $f(a)=a$ and $f(b)=-2b$. Finally,it is easy to see that $f(x)\equiv x$ is a solution.Now,if there exists $c$ such that $f(c)\neq c$,then from $(+)$ there must exists $d\neq 0$ such that $f(d)=-2d$.Hence $[f(k)=k\iff k=0]\implies 2y+f(y)=0\,\forall y\implies f(x)\equiv -2x$. Hence,all solutions are $\boxed{f(x)\equiv x}$ and $\boxed{f(x)\equiv -2x}$ which clearly satisfied the assertion.

How did you get $ f(a)=-2a \to f(na)=-2na $?

First,assume that we have $f(a)=-2a$,then from $f(2y)=2f(y)$,we must have $f(2a)=-4a$ and from $f(x-f(x))=-2(x-f(x))$,we must have $f(3a)=-6a$. Now we will prove that for $n\ge 2$,if $f(ma)=-2ma$ for all $0\le m\le 2n-1$,then $f(ma)=-2ma$ for all $0\le m\le 2n+1$ First,we have $f(2na)=2f(na)=-4na$. Now,$P(a,(n+1)a)\implies f(-(2n+1)a)f((n-1)a)=(2a+f(na))(2(n+1)a+f(-na))$ $\implies -f((2n+1)a)(-2(n-1)a)=(2a-2na)((2n+2)a-(-2na))$ $\implies f((2n+1)a)=-(4n+2)a$ as desired.

My solution is almost same with 3above other than last steps: Let's assume that there exist $x\in Z $ s.t $f (x)\neq x $ and $f (x)\neq -2x $.So we proved that $f (2x+f (x))=2x+f (x) $. $P (x,2x) $ gives that $f (x+2f (x))=4x-f (x) $. $(*) $ $P (f (x),x+f (x)) $ $\implies $ $f (3x)=2x+f (x) $. Then $P (x,x+f (x)) $ yields that $f (3x)f (x+2f (x))=3f (x)(4x-f (x))$. So from $(*) $ we get that either $3f (x)=2x+f (x) $ which is contradiction to our assumption.So $f (x)=4x $ then $x+2f (x)=0$ $\implies $ $x=0$ contradiction. (Remember that we have $f$ is injective at zero). If there are $a,b\in Z $ with $a,b\neq 0$ and $f (a)=a $ and $f (b)=-2b $ we can easily get contradiction (This part is left to the reader as it is just boring casebashing.)

tenplusten wrote: $P (f (x),x+f (x)) $ $\implies $ $f (3x)=2x+f (x) $. Shouldn't it be $P(f(x),x+f(x))\implies f(f(x)+2x)f(x+f(x)+f(f(x)))=(2f(x)+f(x))(2x+2f(x)+f(-x))$ $\implies (f(x)+2x)f(3x)=3f(x)(2x+f(x))\implies f(3x)=3f(x)$?

Difficult and long problem.

We claim that the only solutions are $$ \boxed {f(x)=x} \text{ } \forall x \in \mathbb Z \quad \text {and} \quad \boxed {f(x)=-2x} \text{ } \forall x \in \mathbb Z$$ It is easy to see that these work . We now claim that these are the only ones. Call the assertion as $P(x,y).$ Note that $$P(x,0) \implies f(f(x)) \cdot f(x+f(0))= f(x) \cdot (2x+f(0))$$$$P(x,x) \implies f(x+f(x)) = \pm(2x +f(0))$$ Claim$:=$ $f(0)=0$ Proof : We write $f(0)=c$ for convenience. Note that $P(0,0) \implies f(c) = \pm c$ Case 1 $:=$ $f(c)=-c$ Then , $P(-c,-c) \implies c^2=(3c)^2 \implies \boxed {c=0}$ Case 2 $:=$ $f(c)=c$ Then , $P(2c) \implies f(2c) =\pm 3c$ We again consider two cases $:=$ Case 2.1 $:=$ $f(2c)=3c$ Then we have $:=$ $$P(c,0) \implies f(2c)=2c+f(-c) \implies \boxed {f(-c)=c}$$$$P(-c,0) \implies f(f(-c))=-f(-c) \implies c=-c \implies \boxed {c=0}$$ Case 2.2 $:=$ $f(2c)=-3c$ Then we have $:=$ $$P(c,0) \implies f(2c)=2c+f(-c) \implies \boxed {f(-c)=-5c}$$$$P(-c,0) \implies c \cdot f(f(-c))=-5c(-2c+c)\implies c \cdot f(-5c)=5c^2 \implies \boxed {f(-5c)=5c}$$$$P(-5c,-5c) \implies c=f(0)= \pm (9c) \implies \boxed {c=0} $$ Hence the claim is proved . $\blacksquare$ We now note that $$P(x,x) \implies f(x+f(x))= \pm 2x$$The above equation implies that $f$ is injective at $0$ . Also this implies that $f(x)=-x \iff x=0$ Hence we have $:=$ $$P(x,0) \implies f(f(x))=2x+f(-x)$$ The above equation implies that if $f(f(x))=x$ then $f(f(x))=x \implies f(-x)=-x \implies f(x)=x$ Claim $:=$ $f(x+f(x))=2x$ holds $\forall x \in \mathbb Z$ Proof : FTSOC assume there exists $a\neq 0$ satisfying $f(a+f(a))=-2a$ . Note that we have $:=$ $$P(a,2a+f(a)) \implies\left( f( a + f(2a+f(a)\right) \cdot \left (f(2a+2f(a)) \right) =0$$ Since $f$ is injective at $0$ , we have the following cases $:=$ Case 1 $:=$ $f(a)=-a$ Then , by the remark made earlier , $a=0.$ Case 2 $:=$ $f(2a+f(a))=-a$ Write $2a+f(a)=b$ We have $:=$ $$\pm 2b = f(b+f(b))=f(a+f(a)) =-2a$$ We now consider two cases depending on the sign. Case 2.1 $:=$ $b=-a$ Then we have $b=-a \implies 2a+f(a)=-a \implies f(a)=-3a$ However we have $-a=f(2a+f(a))=f(2a-3a)=f(-a)$ Hence , $f(-a)=-a \implies f(a)=a=-3a \implies \boxed {a=0}$ Case 2.2 $:=$ $b=a$ Then we have $2a+f(a)=a \implies f(a)=-a \implies \boxed {a=0}$ Hence the claim is proved . $\blacksquare$ We now prove that $f$ is odd and injective. Claim$:=$ $f$ is odd . Proof : Note that $$P(-x,f(x)) \implies \left ( f(f(x))-x \right) \cdot \left ( f(x)+f(-x) \right) =0$$ Hence if $f(x)=-f(-x)$ then there is nothing to prove . If $f(f(x))=x$ then we have $f(-x)=-x$ and $f(x)=x$ by an earlier remark . Hence $f$ is odd . $\blacksquare$. Next we note that $f(x)+f(f(x))=f(x)+2x+f(-x)=2x$ This implies $f(2x)=f\left(f(x)+f(f(x)) \right) =2f(x)$ Claim$:=$ $f$ is injective . Proof : Assume otherwise and pick $a,b,c,d$ satisfying $ab \neq 0$ with $a\neq b$ satisfying $f(a)=f(b)=c$ and $b-a=d$ . $$P(a,b) \implies f(a+c) \cdot f(b+c) = (2a+f(d))(2b-f(d)) \implies f(d)=2d$$ This also means $f(2d)=4d$ This means that $2d= f(d)+f(f(d))= 2d+f(2d)=6d \implies d=0.$ Hence we arrive at the desired contradiction. Hence $f$ is injective . $\blacksquare$ Next we consider the follow cases. Case 1 : $f(\mathbb Z)$ contains an odd integer. We claim that $f(x)=x$ is the only solution in this case . Proof : We claim that there exists a odd integer $k$ satisfying $f(k)= \text { odd} $ . Note that we have $f(a)=2b+1$ by the condition. If $a$ is odd take $k=a$ . Otherwise note that since $f(f(a))= 2a-f(a) \equiv 1 \pmod 2 $ then we can just take $k=f(a)$ . Now by $P(x,x+k)$ we have $:=$ $$f(x+f(x+k)) \cdot f(x+f(x)+k) \equiv (2x+f(k)) \cdot (2x-f(k)) \equiv 1 \pmod 2$$Hence $$ f(x+f(x)+k) \equiv 1 \pmod 2 \iff f(f(x)+f(f(x))+k) \equiv 1 \pmod 2 \iff f(2x+k) \equiv 1 \pmod 2$$This means that $f( \text {odd} ) = \text { odd}$ This combined with $f(2x)=2f(x)$ implies that $:=$ $$\nu_2 (f(x)) = \nu_2 (x) $$ Next we have $$P \left( x, \left( \frac {x-f(x)}{2} \right) \right) \implies f \left( x + f \left ( \frac {x-f(x)}{2} \right ) \right)=2x-f(x) = f(f(x)) \iff x+ f \left ( \frac {x-f(x)}{2} \right )= f(x)$$ However this means that $\nu_2\{f(f(x)-x)\} = 1 + \nu_2\{f(x)-x\}$ . Hence $\nu_2\{f(x)-x\} = \infty$ and hence $\boxed {f(x)=x}$ Hence the claim is proven .$\blacksquare$ Case 2 : $f(\mathbb Z)$ does not contain an odd integer. We claim that $f(x)=-2x$ is the only solution in this case . Define $f(x)=-2g(x)$ .Note that $g$ satisfies the following properties $:=$ $2g(g(x))=g(x)+x $ $ g(2g(x)-x) = x$ $g(2x)=2g(x)$ $g$ is odd and injective. Note that the condition $g(2x)=2g(x)$ implies $\nu_2 (g(x)) \geq \nu_2 (x) $ Note that $P(x,y)$ rearranges to $:=$ $$g(2g(x)-y) \cdot g(2g(y)-x) = (x+g(x-y) \cdot y+g(y-x)$$ Set $y \mapsto g(x)$ to get $:=$ $$\text { RHS} =\left( x + g(x-g(x)) \right) \cdot \left( g(x) - g(x-g(x)) \right) \overset {\text {AM-GM}}{\leq} \left ( \frac {x+g(x)}{2} \right)^2 = (g(g(x))^2 = g(g(x)) \cdot g(2g(x)-x) = \text { LHS}$$ Since equality holds in all the estimates , we have $$ x + g(x-g(x)) = g(x) + g(g(x)-x) \iff g(x-g(x)) = \frac {g(x)-x}{2}$$ Hence we have $\nu_2 \{g(x-g(x))\} = \nu_2 \{x-g(x)\} -1 $ However since we proved $\nu_2 (g(x)) \geq \nu_2 (x) $ , this means that $\nu_2 \{(x-g(x)\} = \infty$ . This means $g(x)=x$ and $\boxed {f(x)=-2x}.$ Having exhausted all cases , we are done $\blacksquare$

The only solutions are $\boxed{f(x) = x}$ and $\boxed{f(x) = -2x}$. These work. Let $P(x,y)$ denote the given assertion. Claim: $f(0) = 0$. Proof: Let $f(0) = a$. Assume $a\ne 0$. $P(0,0): f(a) = \pm a$. $P(x,0): f(f(x)) f(x+a) =f(x) (2x + f(-x)) $. $P(x,x): f(f(x) + x) = \pm (2x + a)$. If $f(a) = -a$, then $P(a,a)$ gives $a^2 = 9a^2$, so $a=0$, contradiction. So $f(a) = a$. $P(a,a): f(2a) = \pm 3a$. $P(a,0): f(2a) = 2a + f(-a)$. If $f(2a) = 3a$, then $f(-a) = a$. $P(-a,0)$ gives $a^2 = -a^2\implies a=0$, contradiction. So $f(2a) = -3a$. This implies $f(-a) = -5a$. $P(-a,0): f(-5a) \cdot a = -f(-a)\cdot a$, so $f(-5a) = 5a$. $P(-5a,-5a): a^2 = (9a)^2$, so $a=0$, contradiction. $\square$ Now we may rewrite our equations: $P(x,x): f(f(x) + x) = \pm 2x$. So if $f(x) = 0$, then $2x= 0\implies x=0$. $P(x,0): f(f(x)) = 2x + f(-x)$ for any $x\ne 0$, and of course it also is true for $x=0$. Claim: $f(f(x) + x) = 2x$ for all integers $x$. Proof: Suppose there existed $k\ne 0$ with $f(f(k) + k) = -2k$. $P(k,f(k) + 2k): f(k + f(f(k) + 2k)) f(2f(k) + 2k) = 0$. If $f(k) = -k$, then $-2k = 0$, which is impossible. So $k + f(f(k) + 2k) = 0\implies f(f(k) + 2k) = -k$. $P(f(k) + 2k,f(k) + 2k): -2k = \pm 2(f(k) + 2k)$, so $f(k) + 2k\in \{-k,k\}$. We already know $f(k) = -k$ is impossible, so $f(k) = -3k$. Thus, $f(-2k) = -2k$ and $f(-k) = -k$. $P(-k,0): f(k) = k$, so $k=-3k$, contradiction. $\square$ $P(f(x),-x): f(f(x) + f(-x)) f(f(f(x)) - x) =0$, so either $f(f(x)) = x$ or $f(x) = -f(-x)$. If $f(f(x)) = x$, then by $P(x,0)$ we get $f(-x) = -x$, and by $P(-x,0)$, we get $f(x) = x$. Therefore, $f(x) = -f(-x)$ for each $x$, which implies $f$ is odd. We can rewrite $P(x,0)$ as $f(f(x)) = 2x - f(x)$, which implies $f$ is injective. By $P(x,0)$ and $P(f(x),f(x))$, we have $f(2x) = f(f(f(x)) + f(x)) = 2f(x)$. Notice that $f(x) = 2x\implies f(x) = f(f(x) +x)\implies x=0$. $P(x,-x): f(x-f(x))f(f(x) - x) = (2x- 2f(x)) (2f(x) - 2x)$, so $f(x-f(x)) = \pm 2(x-f(x)) $ If $f(x) = x$, then $f(x-f(x)) = 2(x-f(x)) = -2(x-f(x))$. If $f(x)\ne x$, then since $f(x-f(x)) \ne 2f(x-f(x))$,\[f(x-f(x)) =- 2(x-f(x))\]holds true. Claim: $f(f(x) + 2x) = f(x) + 2x$ for all integers $x$. Suppose there existed $t$ such that this was not true. Let $f(f(t) + 2t) = c$. $P(t+f(t),-t): f(t)^2 = (2t + 2f(t) - c)(-2t + c)$. Expanding this gives\[-4t^2 + 4tc - 4tf(t) + 2f(t)c - c^2 = f(t)^2\]Thus, $-c^2 + (4t + 2f(t))c = f(t)^2 + 4tf(t) + 4t^2 = (f(t) + 2t)^2$. We can rewrite this as\[(c - (f(t) + 2t))^2 =0,\]so $c = f(t) + 2t$, contradiction. $\square$ Claim: $\nu_2(f(x)) \in \{\nu_2(x), \nu_2(x) + 1\}$ for all integers $x\ne 0$. Proof: Since $f(2x) = 2f(x)$ for each, $x$, we clearly have $\nu_2(f(x))\ge \nu_2(x)$. Suppose there existed $x$ for which $\nu_2(f(x)) - \nu_2(x) > 1$. Then for some odd integer $k$, $4\mid f(k)$. By $P(k,0)$, we have $f(f(k)) \equiv 2\pmod 4$, which is not possible as $4$ obviously divides $f(f(k))$. $\square$ Claim: Either $\nu_2(f(x)) = \nu_2(x)$ for all $x\ne 0$, or $\nu_2(f(x))= \nu_2(x) + 1$ for all $x\ne 0$. Proof: Suppose not. Then there exist odd integers $a,b$ such that $f(a)$ is odd, and $\nu_2(f(b)) = 1$. $P(b, b+a): f(b + f(a+b)) f(a+b + f(b)) = (2b + f(a))(2a + 2b - f(a))$. Notice that the RHS is odd, while the LHS is even (because $a+b+f(b)$ is even), contradiction. $\square$ Case 1: $\nu_2(f(x)) = \nu_2(x)$ for all $x\ne 0$. Then $f(x-f(x)) = -2(x-f(x))$ implies that $x-f(x) = 0$, so $f(x) = x$. Case 2: $\nu_2(f(x)) = \nu_2(x) + 1$ for all $x\ne 0$. Then $f(f(x) + 2x) = f(x) + 2x$ implies that $f(x) + 2x = 0$, so $f(x) = -2x$.

Nice Let $P(x,y)$ be the assertion. Let $f(0) = c$. \[P(x,x) : f(x+f(x))^2 = (2x+c)^2 \implies f(x+f(x)) = \pm(2x+c) \tag{A}\]\[P(x,0) : f(x+c)f(f(x)) = (2x+f(-x))f(x) \tag{B}\]Claim $: c = 0$.

Note that now $A$ and $B$ would be $:$ \[f(x+f(x)) = \pm 2x\]\[f(f(x)) = 2x + f(-x)\]Note that if for some $t$ we have $f(t) = 0$ then letting $x = t$ in $A$ we would get $t = 0$. Also if there exists $t$ such that $f(t) = -t$ then $x = t$ in $A$ would give $t = 0$ and if there exists $t$ such that $f(t) = f(-t)$ then comparing $B$ for $x = t$ and $x = -t$ we would get $t = 0$. Claim $: f(x+f(x)) = 2x$.

Claim $: f(2x) = 2f(x)$.

Claim $: f(-x) = -f(x)$.

Claim $: f(x-f(x)) = -(2x-2f(x))$.

Claim $:$ There doesn't exist $a,b \neq 0$ such $f(a) = a$ and $f(b) = -2b$.

Now if for all $x \in \mathbb{Z}$ we have $f(x) = x$, it suits the equation and is an answer so now assume not so there exists $t$ such that $f(t) \neq t$. we have that $f(t-f(t)) = -2(t-f(t))$ and since $f(t) \neq t$ we have that there exists $s = t-f(t) \neq 0$ such that $f(s) = -2s$ so there can't be any $k$ such that $f(k) = k$ from the last claim but $:$ $P(f(x)+x,-x) : f(x)^2 = (2x+2f(x)-f(2x+f(x)))(-2x+f(2x+f(x))) \implies f(2x+f(x)) = 2x + f(x)$ so we must have $2x + f(x) = 0 \implies f(x) = -2x$ which also suits the equation. So Answers are $f(x) = x$ and $f(x) = -2x$.