i think your solution it's not all right

dangerousliri wrote: i think your solution it's not all right @@ Can you point out the line that isn't right? Thks

USJL wrote: If $x\in F$, then from (4) we get that $-x\in F$ or $x=1$. However, $1,-1\in F$, so $-x\in F\quad\forall x\in F..........(5)$. ??

USJL wrote: If $x\in F$, then from (4) we know that $(x-1)f(-x)=-x^2+x$, so $f(-x)=-x$ or $x=1$ i see it in detail and it's wrong here it is why as you said $f(x)-f(-x)=2x-x(f(x)+f(-x))$ tha is true but the moment you said if $x\in F$ than $(x-1)f(-x)-x^2+x$ so $f(-x)=-x$ or $x=1$ whiich also true now here it is the mistake since $1,-1\in F$ than $-x\in F$ $\forall x\in F$ this is wrong that's why becuase $f(f(x))=f(x)$ this mean $f(x)\in F$ $\forall x$ but than from $(4)$ we have $(f(x)-1)(f(-f(x))=-(f(x))^2+f(x)$ so $(f(x)-1)(f(-f(x))+f(x))=0$ this mean $f(x)=1$ or $f(-f(x))=-f(x)$ if you prove $f(x)=1$ only for $x=1$ than rest of your solution is right

Let $P(x,y)$ be the assertion $f(x+f(x+y))+f(xy)=x+f(x+y)+yf(x)$ $P(0,0)\Rightarrow f(f(0))=0$ $P(0,f(0))\Rightarrow 2f(0)=f(0)^2\Rightarrow f(0)=0$ or $2$ case 1.$f(0)=2\Rightarrow f(2)=0$ $P(x,1)\Rightarrow f(x+f(x+1))=x+f(x+1)$ $P(0,f(x+1)+x)\Rightarrow f(x+1)+x+2=f(x+1)+x+2(f(x+1)+x)$ $\Rightarrow f(x)=2-x\forall \in\mathbb{R}$ case 2.$f(0)=0$ $P(x,0)\Rightarrow f(x+f(x))=x+f(x)$ $P(x,1)\Rightarrow f(x+f(x+1))=x+1+f(x)$ $P(1,f(x+1)+x)\Rightarrow f(1+f(1+x+f(x+1)))+f(x+f(x+1))=1+f(x+1+f(x+1))+f(x+1)+x$ $\Rightarrow f(f(x)+x+1)=f(x)+x+1$ $P(x,-1)\Rightarrow f(x+f(x-1))+f(-x)=x+f(x-1)-f(x)$ $\Rightarrow -f(x)=f(-x)$ $P(x,-x)\Rightarrow f(x)+f(-x^2)=x-xf(x)$ $P(-x,x)\Rightarrow f(-x)+f(-x^2)=-x+x(-x)$ $\Rightarrow f(x)-f(-x)=2x-x(f(x)+f(-x))$ $\Rightarrow f(x)=x \forall x\in\mathbb{R}$ hence, $f(x)=2-x$ and $f(x)=x$ are solutions.

dangerousliri wrote: USJL wrote: If $x\in F$, then from (4) we know that $(x-1)f(-x)=-x^2+x$, so $f(-x)=-x$ or $x=1$ i see it in detail and it's wrong here it is why as you said $f(x)-f(-x)=2x-x(f(x)+f(-x))$ tha is true but the moment you said if $x\in F$ than $(x-1)f(-x)-x^2+x$ so $f(-x)=-x$ or $x=1$ whiich also true now here it is the mistake since $1,-1\in F$ than $-x\in F$ $\forall x\in F$ this is wrong that's why becuase $f(f(x))=f(x)$ this mean $f(x)\in F$ $\forall x$ but than from $(4)$ we have $(f(x)-1)(f(-f(x))=-(f(x))^2+f(x)$ so $(f(x)-1)(f(-f(x))+f(x))=0$ this mean $f(x)=1$ or $f(-f(x))=-f(x)$ if you prove $f(x)=1$ only for $x=1$ than rest of your solution is right Since $x\in F$, $f(x)=x$ by definition. So $x=1$ is for-sure true.

dangerousliri wrote: USJL wrote: If $x\in F$, then from (4) we know that $(x-1)f(-x)=-x^2+x$, so $f(-x)=-x$ or $x=1$ i see it in detail and it's wrong here it is why as you said $f(x)-f(-x)=2x-x(f(x)+f(-x))$ tha is true but the moment you said if $x\in F$ than $(x-1)f(-x)-x^2+x$ so $f(-x)=-x$ or $x=1$ whiich also true now here it is the mistake since $1,-1\in F$ than $-x\in F$ $\forall x\in F$ this is wrong that's why becuase $f(f(x))=f(x)$ this mean $f(x)\in F$ $\forall x$ but than from $(4)$ we have $(f(x)-1)(f(-f(x))=-(f(x))^2+f(x)$ so $(f(x)-1)(f(-f(x))+f(x))=0$ this mean $f(x)=1$ or $f(-f(x))=-f(x)$ if you prove $f(x)=1$ only for $x=1$ than rest of your solution is right When $f(x)=1$,then $-f(x)=-1$,and this already shows that $f(-f(x))=f(-1)=-1$

A slight alternative: Substitute $f(x) = x - g(x)$. $\Rightarrow$ $\quad \quad \quad g(2x + y - g(x+y)) = g(xy) - y g(x)$ for all $x, y \in \mathbb{R}$. From here immediately follow two identities: $\quad \quad \quad \quad \quad \>\> (*) : g(2x + 1 - g(x+1)) = 0$ when $y = 1$; and $\quad \quad (**) : g(2x - 1 - g(x-1)) = g(-x) + g(x)$ when $y = - 1$... Let $a$ be some zero of $g(x)$, i.e. $g(a) = 0$. Then by setting $x = 0, y = a$ $\Rightarrow$ $g ( a - g(a) ) = 0 = (1 - a) g(0)$. Here there are two options: Case 1) $g(0) \neq 0$ : $\Rightarrow$ $g(x)$ has only one zero and it is $a = 1$. Hence, $(*)$ $\Rightarrow$ $g(x+1) = 2x$.

Thus, substitute $x=1$ in the general identity to get: $\qquad \qquad \qquad \qquad \qquad \qquad g(y + 2 - g(y + 1)) = g(y)$. Set in the latest $y = 2x + 1 - g(x + 1)$ and apply $(*), (**), (***)$ to get at the end $g(\tilde{x}) + g(- \tilde{x}) = 0$, where $\tilde{x} = x+2$. However, one has for $x + y = 0$ that $g(x) (1-x) = g (-x^2) = g(-x) (1 + x)$, so $g(x) \equiv 0$. Thus the two possible solutions are $f (x) = x - 2(x-1) = 2 - x$ (case 1) and $f(x) = x$ (case 2).

I think it is nice one in Imo2015, but so easy to solve. We have \[ (1) \quad f(f(y))+f(0)=f(y)+yf(0), \] \[ (2) \quad f(x+f(x))+f(0)=x+f(x), \quad \quad f(x+f(x+1))=x+f(x+1) \] and there exists fixed point($f(x)=x$). Since (1) we get that $f(0)=0$ or only $x=1$ is fixed point. We are done! Answers: $f(x)=x$ or $f(x)=2-x$

Why are we done?

Let $P(x,y)$ be the assertion $f(x+f(x+y))+f(xy)=x+f(x+y)+yf(x)$ $P(0,0) \implies f(f(0))=0$. let $f(0)=a \implies f(a)=0$. $P(x,1) \implies f(x+f(x+1))=x+f(x+1),\forall x \in \mathbb{R}.(*)$ $P(0,y) \implies f(f(y))=f(y)+a(y-1). (1)$ From $(1)$, let $y=x+f(x+1) \implies a(f(x+1)+x-1)=0 \Leftrightarrow a=0$ or $f(x)=2-x, \forall x \in \mathbb{R}$. it is easy to see that $f(x)=2-x$ is a function satisfying the conditions of problem. for $f(0)=0$, $P(x,-x) \implies f(x)+f(-x^2)=x-xf(x).(2)$ from $(2)$, let $x=-x \implies f(-x)+f(-x^2)=-x+xf(-x).(3)$ Suppose that $\exists b \neq 0: f(b)=f(-b)$ from $(2)(3)$ we have $f(b)=1=f(-b)$. From $(1)$ we have $f(1)=1$, from $(2) \implies f(-1)=-1$ $P(-b,0) \implies f(-b+1)=-b+1$. $P(1,-b) \implies f(-b+2)=1-2b$ $P(-1,2-b) \implies f(b-2)=-3$. From $(2),(3)$, this is a contradiction.Thus $f(x)=-f(-x),\forall x\neq0 \implies f(x)=x,\forall x\in \mathbb{R}$. So $f(x)=2-x,\forall x \in \mathbb{R}$ and $f(x)=x,\forall x \in \mathbb{R}$ are two functions satisfying the conditions of problem.

this problem is proposed by me

How did you get $f(b-2)=1$?

It seems that we can solve this functional equation by usual routine working, but I think it is not easy to write the complete solution out of our expectation. The problem is proposed by Dorlir Ahmeti, Albania.

Let $P(x,y)$ be the assertion $f(x+f(x+y))+f(xy)=x+f(x+y)+yf(x)$.$P(0,1)$ implies that $f(f(1))=f(1)$ and $P(0,f(1))$ implies that $f(f(f(1)))+f(0)=f(1)+f(1)\cdot f(0)$ i.e. $(f(1)-1)\cdot f(0)=0$(we used the fact that $f(f(1))=f(1)$).From here we have two cases: Case $1$:$f(0)=0$ $P(x,-x)$ implies $x=f(-x^2)+(x+1)f(x)$,from which it follows that $f(1)=1$ and $f(-1)=-1$.Now $P(-x,x-1)$ implies $f(-x-1)+f(x(1-x))=-x-1+(x-1)f(-x)=-x-1+f(-x^2)+x=f(-x^2)-1(1)$ and $P(x,1-x)$ implies $f(x+1)+f(x(1-x))=x+1+(1-x)f(x)=f(-x^2)+1+2f(x)(2)$.Substracting $(1)$ from $(2)$ we obtain $f(x+1)-f(-x-1)=2+2f(x)$,so $f(x+1)=f(-x-1)+2+2f(x)$. But $(x+2)f(x+1)+xf(-x-1)=2(x+1)$(this follows from the fact that $x+1=f(-(x+1)^2)+(x+2)f(x+1)$ and $-x-1=f(-(x+1)^2)-xf(-x-1)$),so $(x+2)f(x+1)+x(f(x+1)-2-2f(x))=2(x+1)$,from which it follows that $(x+1)f(x+1)=xf(x)+2x+1(3)$. $P(x,-x-1)$ implies that $f(x-1)+f(-x(x+1))=x-1-(x+1)f(x)(4)$ and $P(x+1,-x)$ implies that $f(x+2)+f(-x(x+1))=x+2-xf(x+1)(5)$.Substracting $(4)$ from $(5)$ we obtain $f(x+2)-f(x-1)=3+(x+1)f(x)-xf(x+1)(6)$.From $(3)$ it follows that $f(x+2)=\frac{(x+1)f(x+1)+2x+3}{x+2}=\frac{xf(x)+4x+4}{x+2}$,$f(x-1)=\frac{xf(x)-2x+1}{x-1}$ and $f(x+1)=\frac{xf(x)+2x+1}{x+1}$.Plugging those values back in $(6)$,after a bit of work,we will obtain $\fbox{f(x)=x}$,$\forall x\in\mathbb{R}$. Case $2$:$f(1)=1$,but $f(0)\neq 0$(if $f(0)=0$,then this will be the same as case $1$). $P(0,y)$ implies $f(f(y))=f(y)+f(0)(y-1)(7)$ and $P(x,1)$ implies $f(x+f(x+1))=x+f(x+1)$.Now from $(7)$ it follows that $f(x+f(x+1))+f(0)(x+f(x+1)-1)=f(f(x+f(x+1))=f(x+f(x+1))$,so $f(0)(x+f(x+1)-1)=0$.But $f(0)\neq 0$,forcing that $x+f(x+1)=1$.Therefore $f(x+1)=1-x=2-(x+1)$,from which it results that $\fbox{f(x)=2-x}$,$\forall x\in\mathbb{R}$.

I have a doubt Given the function is non constant and $f(x)=f(f(x))=f(f(f(x))).... $,also $ f(2 f(x))=2f(x)$ can we deduce that $f(x)=x$ ?

@Devesh14: No, e.g. it could be f(x)=x for x rational and f(x)=0 for x irrational.

But the function is non constant even for irrational $x$ because we have $P(x,0)\Rightarrow f(x+f(x))=x+f(x)$

If u want to lose braincells confusing urself with thinking about what to do on a Functional Equation and getting like 112131 replaces that make u go to nowhere before finding the right ones, then this problem is for u . Let $P(x,y)$ the assertion of the given F.E. $P(0,0)$ $$f(f(0))=0$$$P(0,f(0))$ $$f(0)^2=2f(0) \implies f(0)=0 \; \text{or} \; f(0)=2$$Case 1: $f(0)=0$ $P(0,x)$ $$f(x)=f(f(x))$$$P(x,0)$ $$f(x+f(x))=x+f(x)$$$P(-1,1)$ $$f(-1)=-1$$$P(1,-1)$ $$f(1)=1$$$P(x,1)$ $$f(x+f(x+1))=x+f(x+1)$$Claim 1: $f$ is injective at $0$ Proof: Assume that there exists $c$ such that $f(c)=0$ then by $P(c,0)$ u get $c=0$. Claim 2: $f$ is injective at $1$ Proof: Assume that there existed $f(d)=1$ then $P(d-1,1)$ gives $d=1$ Claim 3: If $f(a)=a$ then $f(-a)=-a$ Proof: By $P(a,-a)$ $$f(-a^2)=-a^2$$And now by $P(-a,a)$ $$a(1-a)=(a-1)f(-a) \implies a=1 \; \text{or} \; f(-a)=-a$$But since $f(1)=1$ and $f(-1)=-1$ we can say that $f(-a)=-a$ holds becuase $f$ is injective at $1$. Finishing: $P(-1,x+f(x-1))$ $$f(-x-f(x-1))=-x-f(x-1) \implies f(x+f(x-1))=x+f(x-1)$$$P(x,-1)$ $$f(-x)=-f(x) \implies f \; \text{odd}$$$P(x,-x)-P(-x,x)$ $$f(x)=x \; \forall x \in \mathbb R$$Case 2: $f(0)=2$ Which means that $f(2)=0$ and $P(0,x)$ gives $$f(f(x))=f(x)+2x-2$$$P(1,1)$ $$f(1)=1$$$P(x,1)$ $$f(x+f(x+1))=x+f(x+1)$$$P(0,x-1+f(x))$ $$f(x)=2-x \; \forall x \in \mathbb R$$Since we found all the solutionsl we are done

@bryanguo Where have u proved that $f$ is injective in the case $f(0)=0$? @awsomeming327 @RedFlame2112 excuse me but i dont see where u have proved that $f(x+f(x-1))=x+f(x-1)$

MathLuis wrote: @bryanguo Where have u proved that $f$ is injective in the case $f(0)=0$? @awsomeming327 @RedFlame2112 excuse me but i dont see where u have proved that $f(x+f(x-1))=x+f(x-1)$ Darn it, it's been so long I had to take a while to figure out this $P(x,0)$ gives $x+f(x)$ fixed point and we still have $x-1+f(x)$ fixed point, and so $P(1,x-1+f(x))$ is $f(1+f(x+f(x)))+f(x-1+f(x))=1+f(x+f(x))+f(1)(x-1+f(x))$ so if we can somehow show that $f(1)=1$ then we're $f(1+x+f(x))=1+x+f(x)$ and we'd be done. That seems like a relatively easy task but I'm mentally burnt out so I'll try tomorrow I think I just substituted $x-1$ into $P(x-1,1)$ and somehow messed it up.

Let $P(x,y)$ denote the assertion. Then, $P(0,y)$ gives $f(f(y))+f(0)=f(y)+yf(0)$. Therefore, $y=0$ gives $f(f(0))=0$ and $y=f(0)$ gives $2f(0)=f(0)^2$. This implies $f(0)=0$ or $f(0)=2$. Case 1: $f(0)=2$ Then, $f(2)=0$ and $f(f(y))=f(y)+2y-2$. This implies $f$ is injective and $f(y)=y$ if and only if $y=1$. Now, $P(x,1)$ gives $f(x+f(x+1))=x+f(x+1)$, so $f(x+1)=1-x$. Therefore, $f(x)=2-x$. This works because both sides are equal to $y+2-xy$. Case 2: $f(0)=0$ Then, $f(f(y))=f(y)$. Now, $P(f(k),k-f(k))$ gives $$f(2f(k))+f(f(k)(k-f(k)))=2f(k)+(k-f(k))f(k)$$and $P(f(k),0)$ gives $$f(2f(k))=2f(k).$$This means that $f(f(k)(k-f(k)))=(k-f(k))f(k)$. Therefore, $P(k-f(k),f(k))$ gives $$f(k)+f(f(k)(k-f(k)))=f(k)+(k-f(k))f(k)=k-f(k)f(k-f(k)),$$so $$(k-f(k))(f(k)-1)=-f(k-f(k)).$$Therefore, if $f(a)-a=f(b)-b\neq0$, then $f(a)=f(b)$, so $a=b$. Since $P(1,-1)$ gives $f(1)+f(-1)=1-f(1)$ and $P(-1,1)$ gives $f(-1)+f(-1)=-1+f(-1)$, we get $f(-1)=-1$ and $f(1)=1$. Now, $P(1,y)$ gives $f(1+f(1+y))-(1+f(1+y))+f(y)-y=0$, so if $g(x)=f(x)-x$, then $g(y)=-g(1+f(1+y))$. If $g(y)\neq0$, then $g(y)=-g(1+f(1+y))=g(1+f(1+1+f(1+y)))$, so $y-1=f(f(y+1)+2)$. Therefore, $f(y-1)=y-1$. If $f(y+1)\neq y+1$, then $f(y)=y$, contradiction. Therefore, $f(y+1)=y+1$, so $f(y+3)=y-1$, which implies $f(y+2)=y+2$. However, $P(1,y+2)$ gives $f(y)-y+f(y+2)-(y+2)=0$, contradiction since $f(y+2)=y+2$ but $f(y)\neq y$. Therefore, we must have $f(y)=y$ for all $y$, which works since both sides are equal to $2x+y+xy$. Therefore, the only solutions are $\boxed{f(x)=x}$ and $\boxed{f(x)=2-x}$.

Let $P(x,y)$ denote $f(x+f(x+y))+f(xy)=x+f(x+y)+yf(x)$. First, $P(0,0)$ and $P(0,f(0))$ implies $f(0)\in \{0,2\}$. Also, $P(x,1)$ implies $f(x+f(x+1))=x+f(x+1)$. Consider $f(0)=2$: then $P(0,x+f(x+1))$ implies $f(x+1)=1-x$, which works. Let's assume $f(0)=0$. $P(-1,1)$ and next $P(1,-1)$ imply $(f(1),f(-1))=(1,-1)$. $P(x+1,0)$ implies $f(x+1+f(x+1))=x+1+f(x+1)$ and so $P(1,x+f(x+1))$ implies $f(x+2+f(x+1))=x+2+f(x+1)$. Now $P(x+2,-1)$ implies $f(-(x+2))=-f(x+2)$, i.e. odd. Now comparing $P(1,-x)$ and $P(-1,x)$ implies $f\equiv \text{Id}$, which works.

The answer is $f \equiv 2 - x$ or $f \equiv x,$ $ \forall x \in \mathbb{R}.$ We can check these easily works. Now we prove they are the only ones. Let $P(x,y)$ denote the assertion. Then, $$P(x,1) : f(x + f(x+1)) = x + f(x+1). \qquad{(\star)}$$ Hence there exists a fixed point. Let this be $a$. Then, $$P(0,x) : f(f(x)) + f(0) = f(x) + xf(0).$$ Let $x = a$ then, $f(0) = af(0) \implies a = 1$ or $f(0) = 0.$ Thus we have two cases. When $a = 1$, we have $x + f(x+1) = 1$ $ \forall x \in \mathbb{R}.$ Then, $x \rightarrow x - 1$ gives $$x - 1 + f(x) = 1 \implies f(x) = 2 - x, \forall x \in \mathbb{R}.$$ When $f(0) = 0,$ we let $x = -1$ in $\star$, $$f(-1 + f(0)) = -1 + f(0) \implies f(-1) = -1.$$ Now, $P(1,x)$ gives, $$f(f(x+1)+1) + f(x) = 1 + f(x+1) + xf(1).$$ Let $x = -1,$ then $f(f(0) + 1) + f(-1) = 1 + f(0) - f(1)$. Hence $f(1) = 1$. Thus, the above equation becomes, $$f(f(x+1) + 1) + f(x) = 1 + f(x+1) + x.$$ Let $x \rightarrow x-1$ then $$f(f(x) + 1) + f(x-1) = f(x) + x. \qquad{(\star \star)}$$ Now we prove by induction on $x \in \mathbb{Z},$ such that $f(x) = x, \forall x\in \mathbb{Z}.$ The base case for positive integers is $f(1) = 1$. Then, assume $f(k) = k$, for all $x \leq k$. Let $x = k$ in $(\star \star)$, $$f(f(k) +1) + f(k-1) = f(k) + k \implies f(k+1) = k+1.$$ Similarly for negative integers, we can proceed the same way. Hence $f(x) = x, \forall x \in \mathbb{Z}.$ Then, fix $x \in \mathbb{Z}$ and let $y$ be any real number. Thus, $$P(x,y-x) : f(x + f(x)) + f(x(y-x)) = x + f(x) + (y-x)f(x) \implies 2x + f(x(y-x)) = 2x + (y-x)x \implies f(x(y-x)) = x(y-x).$$ Now, let $x=1$ and $y \rightarrow y+1,$ we get $f(y) = y, \forall y \in \mathbb{R}.$ Hence, $f(x) = x, \forall x \in \mathbb{R}.$ Completing the proof..

Solution is attached.

Proposed by dkshield and sebasluna17 Be: $P(x,y): f(x+f(x+y))+f(xy)=x+f(x+y)+yf(x)$ $P(0,0): f(f(0))+f(0)=f(0)+0\rightarrow f(f(0))=0$ $P(0,f(0)): f(f(f(0)))+f(0)=f(f(0))+f(0)^2\rightarrow f(0)^2=2(f(0))\rightarrow f(0)=2, f(0)=0$ If $f(0)=2$ $P(0,y): f(f(y))+f(0)=f(y)+2y$ $$f(f(y))+2 = f(y)+2y$$ $f(a)=f(b)=k$ $P(0,a): f(f(a))+f(0)=f(a)+2b$ $P(0,b): f(f(b))+f(0)=f(b)+2a$ $f(k)+f(0)=k+2b=k+2a\Rightarrow a=b$ f is inyective $P(0,1): f(f(1))+2=f(1)+2\rightarrow f(f(1))=f(1)\Rightarrow f(1)=1$ $P(x,1): f(x+f(x+1))+f(x)=x+f(x+1)+f(x)\rightarrow f(x+f(x+1))=x+f(x+1)$ suppose there $\exists d/f(d)=d$ $P(0,d): f(f(d))+f(0)=f(d)+2d$ $d+2=3d \rightarrow d=1$ Now $x+f(x+1)=1\rightarrow f(1+x)=1-x \rightarrow \boxed{f(x)=2-x}$ If $f(0)=0$ the heaviest case. Let $m$ be a fixed point if and only if $f(m)=m$ $P(x,0): f(x+f(x))=x+f(x)\rightarrow x+f(x):$ fixed point $P(x,1): f(x+f(x+1))=x+f(x+1)\rightarrow x+f(x)-1(\equiv x+f(x+1)):$ fixed point $x=0: f(x+f(x)-1)=x+f(x)-1$ $f(-1)=-1$ $P(1,-1): f(1+f(0))+f(-1)=1+f(0)-f(1)$ $f(1)-1=1-f(1)\rightarrow 2f(1)=2\rightarrow f(1)=1$ what if $f$ were odd? $P(x,-x): f(-x)+f(-x^2)=-x+xf(-x)$ $P(-x,x):f(x)+f(-x^2)=x-xf(x)$ $$f(x)-f(-x)=2x-xf(x)-xf(-x)=2x\rightarrow \boxed{f(x)=x}$$The problem is reduced to proving that $f$ is odd. $P(x,-1):f(x+f(x-1))+f(-x)=x +f(x-1)-f(x)$, then $f$ is odd $\iff$ $x+f(x-1)\equiv x+f(x)+1(x\rightarrow x+1)$ is a fixed point. Lemma: If $m$ and $m-1$ are fixed points then so is $m+1$. Proof: $P(1,x-1): f(1+f(x))+f(x-1)=1+f(x)+x-1$ $f(x+1)+x-1=1+x+x-1$ $f(x+1)=x+1:$ fixed point$\square$ By the lemma and since $x+f(x)-1$, $x+f(x)$ are fixed points then $x+f(x)+1$ is a fixed point then $f$ is odd, as desired.

terrible awful problem. after doing it i believe life is ultimately meaningless. sorry mr ahmeti. The answer is $f(x)=x$ and $f(x)=2-x$, which clearly work. Let $P(x,y)$ denote the assertion. From $P(0,x)$ we find that $f(f(x))=f(x)+(x-1)f(0)$. By substituting $x=0$ we find that $f(f(0))=0$, and then by substituting $x=f(0)$ we find that $f(0)=(f(0)-1)f(0)$, hence $f(0)=0$ or $f(0)=2$. If $f(0)=2$, then $f(2)=0$, so from $P(1,1)$ we obtain $f(1)=1$. Now let $S$ be the set of fixed points, i.e. $a \in S \iff f(a)=a$. From $f(f(x))=f(x)+(x-1)f(0)$, it follows that $\mathrm{Img}(f) \cap S=\{1\}$. On the other hand, from $P(x-1,1)$ we find that $f(x-1+f(x))=x-1+f(x)$, so $x-1+f(x) \in S$. On the other hand, since all fixed points are in the range of $f$, it follows that we must have $x-1+f(x)=1 \implies f(x)=2-x$ for all $x$. If $f(0)=0$, then from $P(x,0)$ we find that $x+f(x) \in S$, and from $P(x-1,1)$ we find that $x-1+f(x) \in S$. Then by $P(1,x-1+f(x))$ we obtain $$f(1+f(x+f(x)))+f(x-1+f(x))=x-1+f(x)+f(x+f(x))+x-1+f(x) \implies f(x+1+f(x))=x+1+f(x).$$Then by $P(x+1,-1)$, we have $f(-(x+1))=-f(x+1)$ for all $x$, hence $f$ is odd. Finally, by comparing $P(x,-x)$ and $P(-x,x)$, we readily obtain $$(x+1)f(x)+(x-1)f(-x)=2x \implies 2f(x)=2x \implies f(x)=x,$$so we're done. $\blacksquare$

$f(x+f(x+y))+f(xy)=x+f(x+y)+yf(x)$ plug $y=0$ $f(x+f(x))+f(0)=x+f(x)$ $a$ plug $x=0$ $f(f(y))+f(0)=f(y)+yf(0)$ $b$ plug $x=y=0$ $f(f(0))=0$ plug $y=f(0)$ into $b$ $2f(0)=f(0)^2$ thus, $f(0)=0$ or $f(0)=2$, and either $f(0)=0$ or $f(2)=0$ in case $f(0)=0$, we get $f(x)=x$ as our only solution in case $f(2)=0$, we get $f(x)+x=2 \to f(x)=2-x$ as our only solution therefore, $f(x)=x$ or $f(x)=2-x$

\begin{align*} (x,y) &\rightarrow (0,0) \Rightarrow f(f(0)) = 0 \\ (x,y) &\rightarrow (0,f(0)) \Rightarrow f(0) = 0 \text{ or } f(0) = 2 \\ \end{align*} For \( f(0) = 2 \), valid \( f(2) = 0 \) and \( f(1) = 1 \): \[ (x,y) \rightarrow (0,x) \Rightarrow f(f(x)) + 2 = f(x) + 2x \quad (1) \] \[ (x,y) \rightarrow (x-1,1) \Rightarrow f(x + f(x) - 1) = x + f(x) - 1 \quad (2) \] In \((1)\), \( (x) \rightarrow (x + f(x) - 1) \) gives \( f(x) = -x + 2 \), which verifies \( f(0) = 0 \). For \( f(0) = 0 \), we have: \[ (x,y) \rightarrow (0,x) \Rightarrow f(f(x)) = f(x) \quad (3) \] \[ (x,y) \rightarrow (x,0) \Rightarrow f(x + f(x)) = x + f(x) \] \[ (x,y) \rightarrow (x,-x) \Rightarrow f(x) + f(-x^2) = x - xf(x) \quad (4) \] \[ (x,y) \rightarrow (-x,x) \Rightarrow f(-x) + f(-x^2) = -x + xf(-x) \quad (5) \] By removing \((4),(5)\), we have: \[ f(x) - f(-x) = 2x - x(f(x) + f(-x)) \quad (6) \] \[ (x,y) \rightarrow (x,f(x)-x) \Rightarrow f(x(f(x)-x)) = (f(x)-x)(f(x)) \quad (7) \] \[ (x,y) \rightarrow (-x,x-f(x)) \Rightarrow f(-x(x-f(x))) = (x-f(x))(f(-x)) \quad (8) \] From \((7), (8)\) in conjunction with \((6)\), we have: \[ -f(x)(f(x)-x) = (f(-x))(f(x)-x) \] So \( f(x) = x \) or \( f(x) = -f(-x) \). If \( f(x) = -f(-x) \) and replacing in \((6)\) gives \( f(x) = x \). The unique solutions are \( f(x) = x \) and \( f(x) = 2 - x \).

$P(0,0) : f(f(0))=0$. $P(0,f(0)) : 2f(0)= f(0)^2 \implies f(0) \in \{0,2\}$. case 1: $f(0)=2.$ $P(x,1) : f(x + f(x+1))= x + f(x+1) $ $P(0,x + f(x+1)) \implies \boxed{f \equiv 2-x}$ case 2: $f(0)=0.$ Define the set $S = \{x : f(x)=x , x \in \mathbb{R}\}$. $P(-1,1): -1 \in S$. $P(1,-1): 1 \in S$. $P(0,y): f(y) \in S$. $P(f(y),0) : 2f(y) \in S$. $P(f(y)-1,1): 2f(y)-1 \in S$. $P(1,2f(y)-1): 2f(y)+1 \in S$. $P(f(y)+1,-1) : f(f(y)+1)= -f(-(f(y)+1))$. $P(f(y)+1,-(f(y)+1))+P(-(f(y)+1),f(y)+1): f(f(y)+1)= f(y)+1$. $P(1,y) \implies \boxed{f \equiv x}$. and we are done.

Solved with (carried by) vsamc.

that $f(x) = x$ and $f(x) = 2-x$ are the only solutions, both of which can be plugged in and checked to work easily. Let $P(x,y)$ denote the assertion in the problem. Taking $P(0,0)$ gives us $f^2 (0) = 0$. Taking $P(0,y)$ gives us \[f^2 (y) + f(0) = f(y)+ yf(0).\]In particular, taking $y=1$ gives us $f^2 (1) = f(1)$. Also, if $f(0) \neq 0$, the above equation implies that $f$ is injective. So, we split into cases on whether $f(0) = 0$ or not. Assume that $f(0) \neq 0$: Claim: We have that $f(0) = 2$ and $f(2) = 0$. Proof: Since $f^2 (1) = f(1)$, it follows by the injectivity that $f(1) = 1$. Now, taking $P(1,0)$ gives us $f(2) + f(0) =2$. Now, taking $P(f(0), f(2))$ gives us \begin{align*} f(f(0) + f(2)) + f(f(0)f(2)) &= f(0) + f(2) + f(2) \cdot 0 \\ f(2) + f(f(0)f(2)) &= f(0) + f(2) \\ f(0)f(2) &= 0, \end{align*}where we use our results from item 1 to simplify $f(f(0))$ as $0$, as well as the injectivity of $f$. Since we are assuming $f(0) \neq 0$ in this case, it follows that $f(2) = 0$ and that $f(0) = 2$. Claim: $f(x) = 2-x$. Proof: Suppose $C$ is a fixed point of $f$. Then, taking $P(0,C)$ gives us \[f(f(C)) = f(C) + 2C - 2,\]and solving for $C$ gives us $C = 1$ as $f(f(C)) = f(C) = C$. Now, taking $P(x,1)$ gives us \[ f(x+f(x+1))= x+f(x+1),\]so $x+f(x+1)$ is a fixed point, implying \begin{align*} x+f(x+1) &= 1 \\ f(x+1) &= 1-x \\ f(x) &= 2-x. \end{align*} Now, we assume the other case, that $f(0) = 0$: Claim: We have that $f(1) = -1$ and $f(-1) = 1$. Proof: Taking $P(x,-x)$, we get that \[f(x) + f(-x^2) = x - xf(x).\]Setting $x = -1$ gives us $f(-1) = -1$, and setting $x=1$ gives us $f(1) = 1$. Claim: If $C$ and $C+1$ are fixed points, then $C+2$ is a fixed point. Proof: Taking $P(1,C)$ gives us \begin{align*} f(1 + f(C+1)) + f(C) &= 1 + f(1+C) + Cf(1) \\ f(C+2) &= C+2. \end{align*} Taking $P(x-1,1)$ gives us \[ f(x-1 + f(x)) = x-1 + f(x),\]so $x-1 + f(x)$ is a fixed point of $f$ for all $x\in \mathbb{R}$. Taking $P(x, 0)$ gives us \[ f(x+f(x)) = x + f(x),\]so $x+f(x)$ is a fixed point of $f$ for all $x\in \mathbb{R}$. Claim: $f(x)$ is odd. Proof: By the claim in part (b), $x+f(x)+1$ is a fixed point of $f$ for all $x\in \mathbb{R}$. Now, note that $P(x+1, -1)$ gives us \[f(x+f(x)+1) + f(-x-1) = x+1 + f(x) - f(x+1).\]Since $x+f(x)+1$ is a fixed point of $f$, it follows that $f(-x-1) = -f(x+1)$ for all $x\in \mathbb{R}$, so it follows that $f$ is odd. Taking $P(x, -x)$ and $P(-x,x)$ gives us the two equations \[\begin{cases}f(x) + f(-x^2) &= x - xf(x) \\ f(-x) + f(-x^2) &= -x + xf(-x).\end{cases}\]Subtracting these two equations gives us \[2f(x) = 2x \implies f(x) = x,\]so we are done. With all cases exhausted, we are finished!

Let $P(x,y)$ be the given assertion. We have \begin{align*} P(0,0) & \Rightarrow f(f(0))=0 \\ P(0,f(0)) & \Rightarrow 2f(0)=f(0)^2 \Rightarrow f(0)\in \{0,2\}\end{align*}First let's suppose $f(0)=0$. Then $P(0,x)$ gives us $f(f(x))=f(x)$. And, \begin{align*} P(-1,1) & \Rightarrow 2f(-1)=-1+f(-1) \Rightarrow f(-1)=-1 \\ P(1,-1) & \Rightarrow f(1)-1=1-f(1) \Rightarrow f(1)=1\end{align*}So, $$P(1,y-1)\Rightarrow f(1+f(y))+f(y-1)=f(y)+y \qquad (1)$$putting $y=f(y)$ in $(1)$ we get $$f(1+f(y))+f(f(y)-1)=2f(y).$$Comparing this and $(1)$ we get $$f(f(y)-1)-f(y-1)=f(y)-y \qquad (2)$$But also $$P(-1,y+1) \Rightarrow f(f(y)-1)+f(-y-1)=f(y)-y-2 \qquad (3)$$So from $(2)$ and $(3)$ we get $$f(-y-1)+f(y-1)=-2 \implies 2+f(x+2)=-f(-x-4) \qquad (4)$$for all $x$. Note that \begin{align*} P(f(x),-f(x)) & \Rightarrow f(-f(x)^2)=-f(x)^2 \\ P(-f(x),f(x)) & \Rightarrow f(f(-x))+f(-f(x)^2)=-f(x)+f(x)f(-f(x)) \\ & \Rightarrow f(-f(x))(1-f(x))=f(x)^2-f(x)\end{align*}so it follows that $f(x)=1$ or $f(-f(x))=-f(x)$ for all $x$. When the former is true, the latter is true as well. Thus regardless, $f(-f(x))=-f(x)$ for all $x$. Now we'll use this and $(4)$ to end this. $$P(2,x) \Rightarrow f(-f(-x-4))+f(2x)=2+f(x+2)+xf(2) \Rightarrow -f(-x-4)+f(2x)=-f(-x-4)+xf(2) \Rightarrow f(2x)=xf(2) \qquad (5)$$for all $x$. Putting $x=1/2$ in $(5)$ we get $f(2)=2$, thus $(5)$ implies that $f(x)=x$ for all real $x$. This function works, indeed. So now we shall assume that $f(0)=2$. Then $f(2)=0$ and $$P(0,x)\implies f(f(x))+2=f(x)+2x$$note that this implies $f$ is injective and $$P(x,0)\implies f(x+f(x))=x+f(x)-2 \qquad (6)$$Let $t$ be such that $f(t)=t-2$. Then $P(2,t-2)$ gives us $f(2t-4)=2=f(0)$, so by injectivity $t=2$. Thus $f(t)=t-2\iff t=2$. Thus $(6)$ implies that $x+f(x)=2$, i.e. $f(x)=2-x$ for all $x$. This clearly works, too.