Mr.C wrote: find all functions from the reals to themselves. such that for every real $x,y$. $$f(y-f(x))=f(x)-2x+f(f(y))$$ Let $P(x,y)$ be the assertion $f(y-f(x))=f(x)-2x+f(f(y))$ Let $a=f(0)$ and $u=f(a)$ $P(0,a)$ $\implies$ $f(u)=0$ $P(u,x)$ $\implies$ $f(f(x))=f(x)+2u$ and so $f(f(f(x)))=f(f(x))+2u=f(x)+4u$ $P(x,f(x))$ $\implies$ $a=f(x)-2x+f(f(f(x))=2f(x)-2x+4u$ and so $f(x)=x+c$ for some $c=\frac a2-2u$ Plugging this back in original equation, we get $c=0$ And so $\boxed{f(x)=x\quad\forall x\in\mathbb R}$

$$P(x,y)\Rightarrow f(y-f(x))=f(x)-2x+f(f(y))$$$$P(x,f(x))\Rightarrow f(0)=f(x)-2x+f(f(f(x)))...(1)$$then $f$ is injective $$P(0,f(0))\Rightarrow f(f(f(0)))=0$$Then exist $c$ such that $f(c)=0$. In $(1)$ put $x=c$ then: $$ f(0)=-2c+f(f(0)) \Rightarrow f(2c+f(0))=0\Rightarrow 2c+f(0)=c\Rightarrow c=-f(0) $$But $$P(c,c)\Rightarrow 0=-2c+f(0)\Rightarrow c=\frac{f(0)}{2} $$then $f(0)=0$ $$ P(0,y)\Rightarrow f(y)=f(f(y)) \Rightarrow f(y)=y$$Then the only solution are $\boxed{f(x)=x \ \forall \ x\in\mathbb{R}}$

Mr.C wrote: find all functions from the reals to themselves. such that for every real $x,y$. $$f(y-f(x))=f(x)-2x+f(f(y))$$ My Solution Step 1. $P(0,f(0))\implies f(f(f(0)))=0$ Step 2. $P(f(f(0)),f(0))\implies f(f(0))=0$ Step 3. $0=f(f(0))\implies f(0)=f(f(f(0)))=0\implies f(0)=0$ Step 4. $P(0,x)\implies f(f(x))=f(x)$ Step 5. $P(x,f(x))\implies 0=f(x)+f(f(f(x)))-2x=f(x)+f(f(x))-2x=2f(x)-2x=0\implies f(x)=x$ $\boxed{f(x)=x\quad\forall x\in\mathbb R}$

Let $P(x, y)$ be the assertion $f(y-f(x))=f(x)-2x+f(f(y))$. By changing $x$ we can see that $f(y-f(x))-f(x) = f(f(y))-2x$ takes any value. $\forall a \exists p, q: f(p)-f(q) = a$ $P(f(x), x): f(0)=f(x)-2x+f(f(f(x)))$ Let $Q(x)$ be the assertion $f(0)=f(x)-2x+f(f(f(x)))$ $P(f(y), x): f(f(y)-f(x))=f(x)-2x+f(f(f(y))) = f(x)-2x+2y-f(y)+f(0)$ $P(f(y), x)+P(f(x), y): f(f(y)-f(x))+ f(f(x)-f(y))=2f(0)$ $\implies \forall a: f(-a)+f(a)=2f(0)$ $P(x, 0):f(-f(x))=f(x)-2x+f(f(0))\implies 2f(0)=f(f(x))+f(x)-2x+f(f(0))$ $P(x, 0)-P(f(x), 0)+Q(x):f(0)=4f(x)-4x$ $\implies f(0)=4f(0)-4\cdot 0\implies f(0)=0\implies \boxed{f(x)=x}$

Mr.C wrote: Find all functions from $\mathbb{R}\mapsto\mathbb{R}$. such that for every real $x,y$. $$f(y-f(x))=f(x)-2x+f(f(y))$$ Claim 1:- $f$ is injective. Proof:- As usual let $P(x,y)$ denote the above FE. FTSOC suppose $f(a)=f(b)$ for some $a\ne b$ then $$P(a,x)\rightarrow f(x-f(a))=f(a)-2a+f(f(fx))=f(x-f(b))=f(b)-2b+f(f(x))\implies a=b$$Hence $f$ is injective. __________________________________________________________________________________________ Claim 2:- $f(0)=0$ Proof:- $P(x,f(x))\rightarrow f(0)=f(x)-2x+f(f(f(x))$ $(\bigstar)$. Now plugging $x=0$ in $(\bigstar)$ we get $f(f(f(0)))=0$. Now let $f(f(0))=u$. So $f(u)=0$. Then $P(u,f(0))\rightarrow f(f(0))=-2u+f(f(f(0)))=-2u\implies 3f(f(0))=0\implies f(f(0))=u=0$. Hence, $f(0)=0$. __________________________________________________________________________________________ Now come back to the problem. $P(0,x)\rightarrow f(x-f(0))=f(0)+f(f(x))\implies f(x)=f(f(x))\overset{\text{Injectivity}}{\implies} f(x)=x \forall x\in\mathbb{R}$. $\blacksquare$

Mr.C wrote: find all functions from the reals to themselves. such that for every real $x,y$. $$f(y-f(x))=f(x)-2x+f(f(y))$$ Okayish for a P1.Here's my solution:- The only function that work is $f\equiv \text{id}$.Its easy to verify this.Now we will prove this is the only.Let $f(0)=c$ and $f^3(x)=f(f(f(x)))$.Then Lemma : $f^3(x)=2x+c-f(x)$ for all $x$ Proof : $P(x,f(x))\implies f^3(x)=2x+c-f(x)$ concluding the lemma.$\square$ Claim : $f$ is injective Proof : $P(x,0)\implies f(-f(x))=f(x)-2x+f(c)$ and now the claim is obvious. Back to the original problem \[P(f(f(x)),y)\implies f(y-2x+f(x)-c)=f(y-f^3(x))=f^3(x)-2f(f(x))+f(f(y))=2x+c-f(x)-2f(f(x))+f(f(y))-\clubsuit\]Plugging $x=0$ in $\clubsuit$ we get $f(y)=f(f(y))-2f(c)$.Now plugging $y=f(y)$ in this we get \[f(f(y))=f^3(y)-2f(c)=2y+c-f(y)-2f(c)\Longleftrightarrow f(y)+2f(c)=2y+c-f(y)-2f(c)\]Thus $f(y)=y+\alpha$ for some $\alpha$.Plugging back into the original assertion we get $\alpha=0$.Thus $f(x)=x$ and we are done.$\blacksquare$

same as @Functional_equation but after step $4$ i just proved that ${f}$ is injective $\mathcal{PROOF}:$ suppose that $f(x_1)=f(x_2)$ then we should prove that $x_1=x_2$ \[P(x_1,y)\implies f(y-f(x_1))=f(x_1)-2x_1+f(f(y)) \thickspace (1)\]\[P(x_2,y)\implies f(y-f(x_2))=f(x_2)-2x_2+f(f(y)) \thickspace (2)\]the LH of the $(1),(2)$ are equal so $f(x_1)-2x_1+f(f(y))=f(x_2)-2x_2+f(f(y)) \implies x_1=x_2$ DONE.

I don't wanna miss this moment. Solution. We have, $f(x)=x~\forall x\in\mathbb{R}$ as our only desired solution. It's easy to check that it works. We procced with a claim to show this is the only one. Let $P(x,y)$ denote the equation. We claim that $f(0)=0$. Notice that by $P(0,f(0))$ we obtain $f(f(f(0)))=0$. Combining this with $P(f(f(0)),0$ to get $f(0)=-f(f(0)),$ so $f(-f(0))=0$. To conclude, combine $P(f(0),-f(0))$ with previous result to get \begin{align*} f(0) &=f(f(0))-f(0) \\ &=-f(0)-f(0)=-2f(0) \\ \end{align*}Hence $f(0)=0$ as desired. $\square$ Therefore, $P(0,y)$ gives $f(f(y))=f(y)$. Consequently, via $P(x,2f(x))$ we have $f$ is Injective, moreover, we have $2x=f(f(2f(x)))=f(2f(x))$. To finish, replace $x \to f(x)$ in last result to get \begin{align*} 2f(x) &=f(2f(f(x))) \\ &=f(2f(x))=2x \\ \end{align*}Hence $f(x)=x~\forall x\in \mathbb{R}$ as desired. $\blacksquare$

I guess this is troll solution. Let $P(x,y)$ denotes assertion of given functional equation. Observe that $P(x, f(x))$ gives us: $$ f(0) = f(x) - 2x + f(f(f(x))) \quad (1)$$This implies that $f$ is injective. Now $P(x, 2f(x))$ gives us: $$ f(f(x)) = f(x) - 2x + f(f(2f(x))) \quad (2) $$On another hand setting $x =0$ in $(1)$ yields to $f^3(0)= 0$, so there exists $a$ such that $f(a) =0$. Considering $P(a, x)$ gives us: $$ f(x) = -2a + f(f(x)) \implies f(f(x)) - f(x) = 2a $$Consequently $(2)$ becomes: $$ 2a + 2x = f(f(2f(x))) $$This implies $f$ is surjective. So since $f(x)$ varies through all real numbers we get that: $$ f(f(x)) -f(x) =2a \implies f(x) - x =2a $$Plugging back in original equation we get that $a=0$, therefore $f(x) = x$, which clearly works.

Let $P(x,y)$ be the assertion of $f(y-f(x))=f(x)-2x+f(f(y))$. $P(x,f(x))\implies f(0)=f(x)-2x+f(f(f(x)))\quad \diamondsuit$ $P(0,f(0))\implies f(0)=f(0)+f(f(f(0)))\implies f(f(f(0)))=0$ $P(0,0)\implies f(-f(0))=f(0)+f(f(0))$ $P(0,x)\implies f(x-f(0))=f(0)+f(f(x))$ $P(f(f(0)),y)\implies f(y)=-2f(f(0))+f(f(y))$ Plugging $y$ as $0$ into the last, we obtain that $f(0)+f(f(0))=0$, thus $f(-f(0))=0$. $P(f(0),-f(0))\implies f(0)=-2f(0)\implies f(0)=0$. Thus, $f(f(x))=f(x)$. And thus, by $\diamondsuit$, we get that $f(x)=x$. This does work. Answer. $\boxed{f(x)=x\forall x\in\mathbb{R}}$.

What does the title mean?

Mr.C wrote: find all functions from the reals to themselves. such that for every real $x,y$. $$f(y-f(x))=f(x)-2x+f(f(y))$$ A bit easy for Iran TST. Took a minute. just seeing if arrogance gets you lots of upvotes We claim that all solutions are of the form $\boxed{f(x)=x}$ for all reals $x$ and this definitely satisfies the condition and so it suffices to show that this is the only solution. Let $P(x, y)$ be the assertion. $P(x, f(x)) \implies f(0) = f(x)-2x + f(f(f(x)))$ and so $f$ is injective. Here, $x = 0 \implies f(f(f(0)))=0$and so if $x$ is the unique real number such that $f(x)=0$, then $P(x, x) \implies f(0)=2x$ and so $f(x) = f(f(f(0)))=f(f(2x))=0=f(0)+2x-f(2x) \implies f(2x)=4x$ and $f(4x)=f(x)$ which means that $x=0$. Now, $P(k, 0) \implies f(f(k))=f(k)$ and so $f(k)=k$ as desired. @below I was to write injective but for some reason wrote bijective, today is a bad day

@above Although not important in your solution, starting from $ P (x, f (x)) $ you only get injective, not bijective

\[P(x,f(x)) \implies f(0)=f(x)-2x+f(f(f(x))) \implies f(x)=2x+f(0)-f(f(f(x))) \cdots (i)\]\[P(x,x+f(x)) \implies 2x=f(f(x+f(x))) \implies f\text{ surjective.}\]Now,assume $f(c)=0$. Then, \[P(c,y) \implies f(y)=-2c+f(f(y)) \implies f(y)=-4c+f(f(f(y)))\]add $(i)$ to this equation to get,$f(x)=kx+c$ plugging this in we get $f(x)=x$

MiltonMath12 wrote: $ 2c+f(0)=c\Rightarrow c=f(0)$ @2above Your solution is wrong either.

I've already corrected it. Thanks for checking again, what do you think now?

$P(x,0)\implies$ injective. $f(x,x+f(x))\implies$ surjective. So, $f$ is a bijection. Let $f^{-1}$ be its inverse. Then, $P(f^{-1}(y-f(y)),y)\implies y-f(y)=2f^{-1}(y-f(y))$ or $$f(\frac{y-f(y)}{2})=y-f(y)$$So, $P(\frac{x-f(x)}{2},y)\implies f(y-x+f(x))=f(f(y))\implies y-x+f(x)=f(y)$. So, $f(x)=x+f(0)$ and plugging gives that $f(x)=x$, which clearly works.

Obviously function is injective. $P(x,x+f(x))\implies$ function is surjective Let $c$ be such that $f(f(c))=0$ $P(0,c)\implies c=f(0)$ So $f(f(f(0)))=0$ $P(f(f(0)),f(0)\implies f(f(0))=0$.Therefore $f(0)=0$ $P(0,y)\implies f(x)=x $ which clearly works

The only solution is $f(x)=x$. It is easy to see that this is indeed a solution of the given equation. Now, we prove that this is the only solution. Let $P(x,y)$ denote the assertion in the problem. $P(0,f(0))$ gives \begin{align} f(f(0)-f(0))=& f(0)+f(f(f(0))) \\ \implies f(f(f(0)))=& 0. \end{align}Now, $P(f(f(0)), f(f(f(0))))$ gives \begin{align*} f(f(f(f(0)))-f(f(f(0))))=& f(f(f(0)))-2(f(f(0))+f(f(f(f(0)))) \\ \implies f(0)= & -2f(f(0))+f(0). \\ \implies f(f(0))=0. \end{align*}Plugging this in $(2)$ we get $f(0)=0$. Now, $P(0, x)$ gives us \begin{align*} f(x-f(0))=& f(0)+f(f(x)) \\ \implies f(f(x))=& f(x) \\ \implies f(f(f(x)))=& f(f(x))=f(x). \end{align*}Now, $P(x,f(x))$ gives \begin{align*} f(f(x)-f(x))= & f(x)-2x+f(f(f(x))) \\ \implies 0= & 2f(x)-2x \\ \implies f(x)=& x, \end{align*}and we are done.

Let $P(x,y)$ be the given assertion. $P(x,f(x))\implies f(0)=f(x)-2x+f(f(f(x))),$ thus f is injective. $P(0,f(0))\implies f(f(f(0)))=0.$ $P(f(f(0)),f(0))\implies f(f(0))=0.$ So $f(f(f(0)))=f(f(0))\implies f(0)=0.$ $P(0,y)\implies f(f(y))=f(y)\implies f(y)=y,$ thus the solution is the identity function only and it is easy to check that it satisfies. $\square$

Let $P(x,y)$ the assertion of the given F.E. Assume $f(a)=f(b)$ and now by $P(a,x)-P(b,x)$ $$-2a+2b=0 \implies a=b \implies f \; \text{injective}$$$P(x,x+f(x))$ $$f(f(x+f(x)))=2x \implies f \; \text{surjective}$$Now use that there exists $c$ such that $f(c)=0$ $P \left(\frac{f(f(x))}{2},x \right)$ $$f \left(\frac{f(f(x))}{2} \right)+\frac{f(f(x))}{2}=x$$From here in this ecuation set $x$ such that $f(f(x))=2c$ so u get that $c=x$ hence $f(0)=2c$. Now by $P(0,2c)$ $$f(f(2c))=0 \implies f(2c)=c$$$P(c,2c)$ $$3c=0 \implies c=0 \implies f(0)=0$$$P(0,x)$ $$f(x)=f(f(x)) \implies f(x)=x$$Thus we are done

Let $P(x,y) :f(y-f(x))=f(x)-2x+f(f(y))$ $P(x,f(x)) : f(0)=f(f(x))-2x+f(f(f(x))) $ then $f :$ is injective ( ) $P(0,f(0)) : f(0)=f(0)+f(f(f(0)))$ that meant $f(f(f(0))) = 0$ $P(x,f(f(0)) : f(x) = f(f(x))-2f(f(0))$ ( ) $P(f(f(f(0))) , f(f(0)) : f(0)=f(f(f(f(0))))+f(f(f(f(0)))) \Leftrightarrow 2f(0)=f(0) \Leftrightarrow f(0)=0$ Now if we put $2f(f(0))=2f(0)=0$ in ( ) then we know : $$f(x)=f(f(x))$$and ( ) then we get $f(x)=x$

Nice one! Solved with Sunaina !! We claim that the only solution is $f(x)=x$. It clearly satisfies given equation. Let $P(x,y)$ be the assertion in given equation. Assume possible $f(x_1)=f(x_2)$. $P(x_1, y)$ and $P(x_2,y)$ imply $x_1=x_2$, implying injectivity. $P(0,f(0)) : f(f(f(0)))=0$. $P(f(f(0)), f(f(0))) : f(0) = 2f(f(0))$. $P(f(0),f(0)) : 2f(0) = f(f(0))$. Above two equations together imply $$f(0)=0$$From $P(0,x)$ and injectivity, $$f(x)=f(f(x)) \Rightarrow x=f(x) $$Hence, $\boxed{f(x)=x}$ $\forall x$ as claimed, and we are done!

Let $P(x,y) :f(y-f(x))=f(x)-2x+f(f(y))$ Assume $\exists a,b$ such that $f(a)=f(b)$.Then $P(a,x)-P(b,x) \implies a=b \implies \text{ f is injective } (*)$ $P(0,f(0)) \implies f(f(f(0)))=0.$ Let $f(f(0))=k,$ Then $f(k)=0$ $P(k,k) \implies f(0)=2k$ $P(k,f(k)) \implies f(0)=-k.$ Hence $k=0 \implies f(0)=0$ $P(0,x) \implies f(x)=f(f(x)).$ Injectivity yields $\boxed{f(x)=x}$

The answer is $f(x) \equiv x$ which works. Denote $P(x,y)$ as the assertion $f(y-f(x))=f(x)-2x+f(f(y))$ Claim 1. $f$ is surjective Proof: From $P(x,x+f(x))$ we get \[f(x)=f(x)-2x+f(f(x+f(x)))\]$\Rightarrow$ \[f(f(x+f(x)))=2x\]and thus $f$ is surjective. Claim 2. $f(0)=0$ Proof: Now denote a real number $k$ such that $f(k)=0$. From $P(k,k)$ we get $f(0)=2k$. From $P(0,f(0))$ we get $f(0)=f(0)+f(f(f(0))) \Rightarrow f(f(f(0)))=0$. From $P(k,f(0))$ we get $f(f(0))=-2k$. Finally from $P(k,0)$ we get $f(0)=-2k+f(f(0)) \Rightarrow 2k=-4k \Rightarrow k=0$ and thus $k=0$ as desired. Since we proved $f(0)=0$, $P(0,x)$ gives us $f(f(x))=f(x) \Rightarrow f(f(f(x)))=f(f(x))$ and $P(x,f(x))$ gives us $0=f(x)-2x+f(f(f(x)))=2f(x)-2x$. So we get $2f(x)=2x \Rightarrow f(x)=x$ as desired. $\square$

Solved with cxyerl. Clearly $f$ is injective(compare $P(a,y)$ and $P(b,y)$ where $f(a)=f(b)$). $P(0,f(0))$ gives $f(f(f(0)))=0$. $P(f(f(0)),f(0))$ gives $f(f(0))=0$. So $f(f(f(0)))=f(f(0))$ so $f(0)=0$. Now $P(0,y)$ gives $f(y)=f(f(y))$ so $\boxed{f(y)=y}$ which works. $\blacksquare$

Let $P(x, y)$ denote the assertion. Them, $P(x, f(x))$ implies $f(x)+f(f(f(x))) = 2x+f(0)$. Putting $x = 0$ here, we see that $f(f(f(0))) = 0$. Plugging this equation back into the original we see that $f(y-f(x)) = f(f(y))+f(0)-f(f(f(x)))$, and let this assertion be $Q(x, y)$. Now, we have the key claim of the problem Claim: $f$ is bijective Proof: if $f(a) = f(b)$ then $P(a, y)$ and $P(b, y)$ together imply $a = b$ on comparison. Therefore $f$ is injective. For surjectivity, we use the cancellation trick, i.e. $P(x, x+f(x))$. This gives $f(f(x+f(x))) = 2x$. Now, $Q(f(0), y)$ implies $f(y-f(f(0))) = f(f(y))$, i.e., $f(y) = y-f(f(0))$. This means $f(y) = y+c$ and plugging back, we see $c = 0$, so $f(x) = x$ is the only solution.

My solution was wrong...

Mr.C wrote: find all functions from the reals to themselves. such that for every real $x,y$. $$f(y-f(x))=f(x)-2x+f(f(y))$$ $\color{blue}\boxed{\textbf{Answer: }f\equiv x}$ $\color{blue}\boxed{\textbf{Proof:}}$ $\color{blue}\rule{24cm}{0.3pt}$ $$f(y-f(x))=f(x)-2x+f(f(y))...(\alpha)$$In $(\alpha) x\to 0, y\to f(0):$ $$\Rightarrow f(0)=f(0)+f(f(f(0)))$$$$\Rightarrow f(f(f(0)))=0$$$$\Rightarrow \exists a/f(a)=0$$In $(\alpha) x\to a:$ $$\Rightarrow f(y)+2a=f(f(y))...(\beta)$$Replacing $(\beta)$ in $(\alpha):$ $$\Rightarrow f(y-f(x))=f(x)-2x+f(y)+2a...(\theta)$$In $(\theta) y\to f(x):$ $$\Rightarrow f(0)=f(x)-2x+f(f(x))+2a$$By $(\beta):$ $$\Rightarrow f(0)=f(x)-2x+f(x)+2a+2a$$$$\Rightarrow f(x)=x-2a+\frac{f(0)}{2}$$$$\Rightarrow f(x)=x+c, c\text{ is constant}...(\omega)$$Replacing $(\omega)$ in $(\alpha):$ $$\Rightarrow y-x-c+c=x+c-2x+y+c+c$$$$\Rightarrow 0=3c$$$$\Rightarrow c=0$$$$\Rightarrow \boxed{f\equiv x}_\blacksquare$$$\color{blue}\rule{24cm}{0.3pt}$

let P(x,y) be the assertion $P(0,f(0)) = f(0) = f(0)+f(f(f(0)))$ $f(f(f(0))) = 0$ let $f(f(0)) = u,f(u) = 0$ $P(u,y) = f(y) = f(f(y))-2u$ $P(x,y) = f(y-f(x)) = f(x)-2x+f(y)+2u$ $P(x,f(x)+y) = f(y) = f(x)-2x+f(f(x)+y) +2u$ $P(x,f(x)+x) = f(x)= f(x)-2x+f(f(x)+x)+2u$ $f(f(x)+x) = 2x-2u$ thus this mean that f is surjective $P(x,2f(x)) =f(f(x)) = f(x)-2x+f(2f(x))+2u$ $f(x)+2u=f(x)-2x+f(2f(x))+2u$ $f(2f(x)) = 2x$ if there exist a,b so that $f(a) = f(b) = k$ but $a>b$ contradiction so implies injectivity First We get that $f(f(f(0))) = f(2f(0))$ $f(f(0)) = 2f(0)$ $P(x,0) = f(-f(x)) = f(x)-2x+2f(0)$ $P(f(0),0) = f(-f(f(0)))= 2f(0)=f(f(0))$ $-f(f(0)) = f(f(0))$ $f(f(0)) = 0 =2f(0)$ $P(y,0) = f(y) = f(f(y))$ $f(y) = y$

Mr.C wrote: find all functions from the reals to themselves. such that for every real $x,y$. $$f(y-f(x))=f(x)-2x+f(f(y))$$ Proposed by Amin Fathpour

Maybe a little cleaner proof ... Answer : $ f(x) = x $. Solution : Let $ P(x,y) $ denote the given assertion and let $ f(0) = c $ where $ c $ is a fixed number. $ P(0,c) \Rightarrow f(f(c)) = 0 $. (1) $ P(f(c),c) $ and (1) $ \Rightarrow f(c) = 0 $. (2) (1) , (2)$ \Rightarrow f(0) = c = 0 $. (3) $ P(0,x) $ and (3) $ \Rightarrow f(x) = f(f(x)) $. (4) $ P(x,f(x)) $ and (4) $ \Rightarrow 2f(x) = 2x \Rightarrow f(x) = x $. Which is clearly true. As a result, our proof is complete.

Mr.C wrote: find all functions from the reals to themselves. such that for every real $x,y$. $$f(y-f(x))=f(x)-2x+f(f(y))$$ Mr.C wrote: find all functions from the reals to themselves. such that for every real $x,y$. $$f(y-f(x))=f(x)-2x+f(f(y))$$ Claim1: function is surjective Prove: by putting x=a/2 and y= x+f(x) we have a= f(f(y)) notice a is replaceable with any real therefore for any real number such as a there is a real number such as a' so that f(a')=a and claim is proved Claim 2: for any real x we have f(x)=x+c Prove: by using claim1 we know there exists real such as b so that f(b)=0 by putting x=b in the main problem we have f(y)=f(f(y))-2b and now because of claim 1 we know for any real such as x we have f(x)=x+2b and claim is proved Claim3: for any real such as x we have f(x)=x Prove: by putting results of claim 2 in main problem we have b=0 therefore claim is obvious

The ans is f(x)=x, which works . Now we show it’s the only one. Let P(x,y) be the given assertion. P(x+f(x),x) gives that f is surjective then if f(u)=0 then P(u,y) gives f(f(y))=f(y)+2u. Since f is surjective, we have f(x)=x+2u for all reals x. Plugging x = u gives u=0, so f is the identity. (Typed on phone)