Nice problem I can prove that $f$ is one-to-one (for non constant) and if $f(k)=0$ then $k=0$ but it seems both facts can't lead to the solution. If someone knows the proof that use them, please post it. As usual, let $P(x,y)$ denote $f(x + y + yf(x)) = f(x) + f(y) + xf(y)$. If $f$ is constant function, $f\equiv 0$ is the only solution. Easy to see that $f(x)\equiv x$ is also a solution. Now we'll look for other $f$ than these two solutions. $P(1,0)$ gives $f(0)=0$. $P(-1,y)$ gives $f(y-1+yf(-1))=f(-1)$ for all $y\in \mathbb{R}$. If $f(-1)+1\neq 0$, $y-1+yf(-1)=y(f(-1)+1)-1$ is surjective on $\mathbb{R}$, and so $f$ is constant function. Hence, $f(-1)=-1$. $P(x,-1)$ gives $f(x-1-f(x))=f(x)-1-x$ for all $x\in \mathbb{R}$. $P(x-1-f(x),1)$ gives $f(-1)=f(x)-1-x+(x-f(x))f(1)=-1$ for all $x\in \mathbb{R}$, and so $f(x)(1-f(1))=x(1-f(1))$ for all $x\in \mathbb{R}$. Setting $x=1$ in it gives $f(1)=1$. $P(1,y)$ gives $f(2y+1)=2f(y)+1$. $P(x-1-f(x),y)$ gives $f(x-1-f(x)+y(f(x)-x))=f(x)-x-1+f(y)(x-f(x))$ for all $x,y\in \mathbb{R}$. Setting $y=3$ gives $f(2f(x)-2x-1)=2x-2f(x)-1$ for all $x\in \mathbb{R}$. (Note that $f(3)=2f(1)+1=3$.) Since $f(2f(x)-2x-1)=2f(f(x)-x-1)+1$ for all $x\in \mathbb{R}$, combining these two equations gives $f(f(x)-x-1)=x-f(x)-1$ for all $x\in \mathbb{R}$. $P(f(x)-x-1,z)$ gives $f(f(x)-x-1+z(x-f(x))=x-f(x)-1+f(z)(f(x)-x)$ for all $x,z\in \mathbb{R}$. Since $f(x)-x-1+z(x-f(x))=x-f(x)-1+y(f(x)-x)\implies y+z=2$, we get that $f(x)-x-1+f(y)(x-f(x))=x-f(x)-1+f(z)(f(x)-x)$ for all $x,y,z\in \mathbb{R}$ that $y+z=2$. That is equivalent to $(x-f(x))(f(y)-f(z)-2)=0$. Since $f(x)\not\equiv x$, $f(y)+f(z)=2$ for all $y,z\in \mathbb{R}$ that $y+z=2$. Hence, $2=f(2t+1)+f(1-2t)=(2f(t)+1)+(2f(-t)+1)\implies f(t)+f(-t)=0$ for all $t\in \mathbb{R}$. $P(x,-y)$ and $P(x,y)$ give $f(x+y+yf(x))+f(x-y-yf(x))=2f(x)$ for all $x,y\in \mathbb{R}$. Suppose there exists $r\in \mathbb{R}$ that $f(r)=-1$. $P(r,r)$ gives $r=-1$. So, $f(x+z)+f(x-z)=2f(x)$ for all $x,z\in \mathbb{R}$ that $x\neq -1$. Note that when $x=-1$, we have $f(-1+z)+f(-1-z)=-f(-z+1)-f(z+1)=-2$, since $(-z+1)+(z+1)=2$. This gives $f(x+z)+f(x-z)=2f(x)$ for all $x,z\in \mathbb{R}$. In particular, $f(2x)+f(0)=f(2x)=2f(x)$ for all $x\in \mathbb{R}$. So, $f(x+z)+f(x-z)=f(2x)$ for all $x,z\in \mathbb{R}$, which is equivalent to $f(p)+f(q)=f(p+q)$ for all $p,q\in \mathbb{R}$. Plugging in $P(x,y)$ gives $f(yf(x))=xf(y)$ for all $x,y\in \mathbb{R}$. So, $f(f(x))=x$ and $f(x)f(y)=f(xy)$ for all $x,y\in \mathbb{R}$. The only non-constant $f$ that satisfies both Cauchy equation and $f(x)f(y)=f(xy)$ for all $x,y\in \mathbb{R}$ is $f(x)\equiv x$, which contradicts our assumption. Finally, the only two solutions are $f(x)=0$ for all $x\in \mathbb{R}$ and $f(x)=x$ for all $x\in \mathbb{R}$.

This problem was proposed by me.

socrates wrote: This problem was proposed by me. nice problem. would you kindly share the official solution with us?

sheripqr wrote: socrates wrote: This problem was proposed by me. nice problem. would you kindly share the official solution with us? See http://emc.mnm.hr/wp-content/uploads/2016/12/EMC_2016_Seniors_ENG_Solutions.pdf

ThE-dArK-lOrD wrote: I can prove that $f$ is one-to-one (for non constant) and if $f(k)=0$ then $k=0$ but seem that both facts can't lead to the solution, if someone know the prove that use it, please post it. I have a solution that use injectivity of $f(x)$,but it maybe have typo(s).Please tell me if something is wrong . Let $P(x,y)$ be the assertion. Clearly the only constant solution is $\boxed{f(x)\equiv 0}$,now we will assume that $f(x)$ is non-constant. $P(0,0)\implies f(0)=0$ If for some $g\neq 0,f(g)=0$,then $P(g,y)\implies f(g+y)=f(y)(g+1)\implies f(2g)=f(g)(g+1)=0\implies [P(2g,y)\implies f(y)(2g+1)=f(2g+y)=f(y)(g^2+2g+1)]\implies f(x)\equiv 0$,a contradiction since $f(x)$ is non-constant.Hence $f(x)=0\iff x=0$. $P(x,\frac{a-x}{f(x)+1})\implies f\left(\frac{a-x}{f(x)+1}\right)=\frac{f(a)-f(x)}{x+1}$ for all $x$ such that $f(x)\neq -1$. $f(h)=-1\implies [P(h,h)\implies h=-1]$,so if $a\neq b$ and $f(a)=f(b)$,then $f(a)=f(b)\neq -1\implies f\left(\frac{a-b}{f(b)+1}\right)=0\implies a=b$,a contradiction.Hence $f(x)$ is injective. $P(-1,1)\implies f(f(-1))=f(-1)\implies f(-1)=-1$. $P(-2,-2)\implies f(-4-2f(-2))=0\implies f(-2)=-2\implies [P(-2,y)\implies f(-2-y)=-2-f(y)]$. Let $f(1)=a$,then $P(1,-1)\implies f(-a)=a-2\implies [P(-a,1)\implies -1=2a-2-a^2]\implies a=1\implies f(1)=1\implies f(-3)=-2-f(1)=-3$. Define $g(x)=f(x-1)+1$,then $g(x)$ is injective and $P(x-1,y)\implies Q(x,y):g(x+yg(x))=g(x)+x(g(y+1)-1)$. $f(y)+f(-2-y)=-2\implies g(1+y)+g(-1-y)=0\implies g(-x)=-g(x)$. $g(-2)=-2,g(-1)=-1,g(0)=0,g(1)=1,g(2)=2$. $Q(2,y-1)\implies g(2y)=2g(y)$. If $g(c)=d$,then $Q(c,y)\implies g(c+dy)=d+c(g(y+1)-1)\implies g(c+d)=c+d\implies g(-c-d)=-c-d\implies [Q(c,-\frac{2c}{d}-1)\implies g\left(-\frac{2c}{d}\right)=-\frac{2d}{c}]$ $\implies g\left(-\frac{c}{d}\right)=-\frac{d}{c}\implies [Q(c,-\frac{c}{d}-1)\implies g(-d)=-c\implies g(d)=c$.Hence $g(g(x))=x$ and $f(f(x))=f(g(x+1)-1)=g(g(x+1))-1=x$,thus $f(x)$ and $g(x)$ are bijective. Assume that there exists a constant $n$ such that $f(n)\neq n$,let $m=n+1$,then $g(m)\neq m$,let $g(m)=k$,then we have $k\neq m,k\neq 0,g(k)=m$,so $Q(m,y)\implies g(m+yk)=k+m(g(y+1)-1)\implies g(2yk+m+k)=k+m(g(2y+2)-1)=k+m(2g(y+1)-1),g(2yk+2m)=2k+m(2g(y+1)-2)\implies g(2yk+m+k)-g(2yk+2m)=m-k\implies f(2yk+m+k-1)-f(2yk+2m-1)=m-k\implies f(x+k)-f(x+m)=m-k$. $f\left(\frac{a-x}{f(x)+1}\right)=\frac{f(a)-f(x)}{x+1}\implies f\left(\frac{a+(k-m)(f(x)+1)-x}{f(x)+1}\right)=\frac{f(a+(k-m)(f(x)+1))-f(x)}{x+1}$ $\implies f(a+(k-m)(f(x)+1))-f(a)=(x+1)\left[f\left(\frac{a-x}{f(x)+1}+k-m\right)-f\left(\frac{a-x}{f(x)+1}\right)\right]=(x+1)(m-k)$ $\implies f(a+(k-m)(f(-f(x))+1))-f(a)=(-f(x)+1)(m-k)$. $P(x+k-m,y)\implies f(x+k-m+y+yf(x+k-m))=f(x+k-m)+f(y)+(x+k-m)f(y)\implies f(x+y+yf(x)+(-y+1)(k-m))=f(x)+f(y)+xf(y)+(-f(y)+1)(m-k)=f(x+y+yf(x))+(-f(y)+1)(m-k)=f(x+y+yf(x)+(f(-f(y))+1)(k-m))\implies f(-f(y))=-y\implies f(-y)=f(-f(f(y)))=-f(y)$. $P(x,y)+P(x,-y)\implies f(x+(f(x)+1)y)+f(x-(f(x)+1)y)=2f(x)=f(2x+1)-1$.(Because $P(1,x)\implies f(2x+1)=2f(x)+1$.) $\implies f(a)+f(b)=f(a+b+1)-1\forall a\neq -2-b \implies f(a)+f(b)=f(a+b+1)-1$.(Because if $a+b=-2$,then $f(a)+f(b)=-2=f((-2)+1)-1=f(a+b+1)-1$.) $\implies g(a+1)+g(b+1)=g(a+b+2)$.Hence $g(x)$ is addictive,so $g(x)+g(yg(x))=g(x+yg(x))=g(x)+x(g(y+1)-1)\implies g(yg(x))=x(g(y+1)-1)\implies g(y)=g(yg(1))=g(y+1)-1\implies f(y)=g(y+1)-1=g(y)=f(y-1)+1\implies f(a)+f(b)=f(a+b+1)-1=f(a+b)$.Hence $f(x)$ is addictive,so $f(x)+f(y)+f(yf(x))=f(x+y+yf(x))=f(x)+f(y)+xf(y)\implies f(yf(x))=xf(y)\implies f(xy)=f(yf(f(x)))=f(x)f(y)$.Hence $f(x)$ is multiplicative and addictive,thus $f(x)\equiv 0$ or $f(x)\equiv x$,but since $f(x)$ is non-constant,we must have $f(x)\equiv x$,a contradiction with the assumption $f(n)\neq n$.Hence the only non-constant solution is $\boxed{f(x)\equiv x}$ which clearly satisfied the condition.

ThE-dArK-lOrD wrote: Nice problem I can prove that $f$ is one-to-one (for non constant) and if $f(k)=0$ then $k=0$ but seem that both facts can't lead to the solution, if someone know the prove that use it, please post it. As usual, let $P(x,y)$ denote $f(x + y + yf(x)) = f(x) + f(y) + xf(y)$ If $f$ is constant function, we get that $f\equiv 0$ is the only solution, and easy to see that $f(x)\equiv x$ is a solution. Now we'll look for $f$ that is not constant and $\not\equiv x$ $P(1,0)$ give us $f(0)=0$ $P(-1,y)$ give us $f(y-1+yf(-1))=f(-1)$ for all $y\in \mathbb{R}$ If $f(-1)+1\neq 0$ we get that $y-1+yf(-1)=y(f(-1)+1)-1$ is surjective on $\mathbb{R}$ and so $f$ is constant function. So $f(-1)=-1$ $P(x,-1)$ give us $f(x-1-f(x))=f(x)-1-x$ for all $x\in \mathbb{R}$ $P(x-1-f(x),1)$ give us $f(x-1-f(x)+1+(f(x)-1-x))=(f(x)-1-x)+f(1)+(x-1-f(x))f(1)$ for all $x\in \mathbb{R}$ So $f(-1)=f(x)-1-x+(x-f(x))f(1)=-1$ for all $x\in \mathbb{R}$ which is $f(x)(1-f(1))=x(1-f(1))$ for all $x\in \mathbb{R}$ Set $x=1$ in it give us $f(1)=1$ $P(1,y)$ give us $f(2y+1)=2f(y)+1$, $P(x-1-f(x),y)$ give us $f(x-1-f(x)+y(f(x)-x))=f(x)-x-1+f(y)(x-f(x))$ for all $x,y\in \mathbb{R}$ Set $y=3$ give us $f(2f(x)-2x-1)=2x-2f(x)-1$ for all $x\in \mathbb{R}$ (note that $f(3)=2f(1)+1=3$) Since $f(2f(x)-2x-1)=2f(f(x)-x-1)+1$ for all $x\in \mathbb{R}$, combine this two equations give us $f(f(x)-x-1)=x-f(x)-1$ for all $x\in \mathbb{R}$ $P(f(x)-x-1,z)$ give us $f(f(x)-x-1+z(x-f(x))=x-f(x)-1+f(z)(f(x)-x)$ for all $x,z\in \mathbb{R}$ Since $f(x)-x-1+z(x-f(x))=x-f(x)-1+y(f(x)-x)\Leftarrow y+z=2$ We get that $f(x)-x-1+f(y)(x-f(x))=x-f(x)-1+f(z)(f(x)-x)$ for all $x,y,z\in \mathbb{R}$ such that $y+z=2$ It's equivalent to $(x-f(x))(f(y)-f(z)-2)=0$, so since $f(x)\not\equiv x$, we get $f(y)+f(z)=2$ for all $y,z\in \mathbb{R}$ such that $y+z=2$ So $2=f(2t+1)+f(1-2t)=(2f(t)+1)+(2f(-t)+1)\Rightarrow f(t)+f(-t)=0$ for all $t\in \mathbb{R}$ $P(x,-y)$ and $P(x,y)$ give us $f(x+y+yf(x))+f(x-y-yf(x))=2f(x)$ for all $x,y\in \mathbb{R}$ Suppose there exist $r\in \mathbb{R}$ such that $f(r)=-1$, $P(r,r)$ give us $r=-1$ So $f(x+z)+f(x-z)=2f(x)$ for all $x,z\in \mathbb{R}$ that $x\neq -1$ Note that when $x=-1$, we have $f(-1+z)+f(-1-z)=-f(-z+1)-f(z+1)=-2$ (since $(-z+1)+(z+1)=2$) This give us $f(x+z)+f(x-z)=2f(x)$ for all $x,z\in \mathbb{R}$, thus $f(2x)+f(0)=f(2x)=2f(x)$ for all $x\in \mathbb{R}$ So $f(x+z)+f(x-z)=f(2x)$ for all $x,z\in \mathbb{R}$, which is equivalent to $f(p)+f(q)=f(p+q)$ for all $p,q\in \mathbb{R}$ Plug in $P(x,y)$ give us $f(yf(x))=xf(y)$ for all $x,y\in \mathbb{R}$, so $f(f(x))=x$ and $f(x)f(y)=f(xy)$ for all $x,y\in \mathbb{R}$ Then since non-constant $f$ satisfy Cauchy equation and $f(x)f(y)=f(xy)$ for all $x,y\in \mathbb{R}$, we get that $f(x)\equiv x$, contradict our assumption that $f(x)\not\equiv x$ Finally, the only two solutions are $f(x)=0$ for all $x\in \mathbb{R}$ and $f(x)=x$ for all $x\in \mathbb{R}$ you use cauchy equation to conclude however the function is not continuous !!!! so you have to prove it before

you can also try to find another particularity like the increasing... Anyway you can't conclude directly

neba42 wrote: you can also try to find another particularity like the increasing... Anyway you can't conclude directly In fact the only function $f$ satifies that : $f(x+y)=f(x)+f(y)$ and $f(xy)=f(x)f(y)$ is $f(x)=x$ that is because : from the second relation we can get $f(x^2)=f^2(x)\ge 0$ so $\forall x\ge 0 f(x)\ge 0$. Now set $y>0$ in the first relation to find that f is increasing from here we can use cauchy equation.

Yes it's true i didn't see that

Beautiful problem by @socrates! Here's my solution: Let $g(x)=x-f(x)$. Then the given condition translates to $$P(x,y):=g(x+y+xy-yg(x))+yg(x)=g(x)+g(y)+xg(y)$$Note that $g \equiv 0$ is the only constant solution. So from now on assume $g$ is non-constant. $P(0,0)$ easily gives $g(0)=0$. If $g(-1) \neq 0$, then $$P \left(-1,\frac{1-y}{g(-1)} \right) \Rightarrow g(y)=y+(g(-1)-1)$$It is easy to check that the only linear function which works is the identity function. So from now on also assume $g(-1)=0$. Then $$P(x,-1) \Rightarrow g(g(x)-1)=2g(x) \cdots ( \spadesuit )$$Let $g(1)=a$. Using $( \spadesuit )$ we have $g(a-1)=2a$. Then $$P(a-1,1) \Rightarrow g(-1)+2a=2a+a+(a-1)a \Rightarrow a=0$$where we use $g(-1)=0$. Now, let $g(-2)=k$. Then $$P(-2,-2) \Rightarrow g(2g(-2))-2g(-2)=0 \Rightarrow g(2k)=2k$$Then we have $$P(2k,y) \Rightarrow g(y+2k)=(2k+1)g(y)+2k(1-y)$$In particular, note that putting $y=2k$ gives $g(4k)=4k$. Also, substituting $y \rightarrow y+2k$, we have $$g(y+4k)=(2k+1)g(y+2k)+2k(1-y-2k)=(2k+1)^2g(y)+2k(2k+1)(1-y)-2ky+2k(1-2k)=(2k+1)^2g(y)-4k(k+1)y+4k$$Also, we have $$P(4k,y) \Rightarrow g(y+4k)=(1+4k)g(y)+4k(1-y)$$Comparing the above two equations, we get that $$(2k+1)^2g(y)-4k(k+1)y+4k=(1+4k)g(y)+4k(1-y) \Rightarrow 4k^2g(y)=4k^2y$$Putting $y=-1$ then gives $k=0$. This gives $$P(-2,y) \Rightarrow g(-2-y)=-g(y)$$Letting $x \rightarrow -2-x$ in $( \spadesuit )$ then yields $g(-g(x)-1)=-2g(x)$. Using $( \spadesuit )$ and the previous two relations, we get $$P(g(x)-1,y) \Rightarrow g(g(x)-1-yg(x))=2(1-y)g(x)+g(x)g(y)$$$$\text{ and }$$$$P(-g(x)-1,2-y) \Rightarrow g(-g(x)-1+(2-y)g(x))=-2(y-1)g(x)-g(x)g(2-y)$$Comparing the above two equations, and using the assumption that $g$ is not identically zero, we get $$g(x)g(y)=-g(x)g(2-y) \Rightarrow g(y)+g(2-y)=0 \text{ } \forall y \in \mathbb{R} \cdots ( \clubsuit )$$Now, since $g(1)=0$ (proved earlier), so $P(1,y)$ gives that $g(2y+1)=2g(y)$. But, by $( \clubsuit )$, we have $$g(2y+1)+g(1-2y)=0 \Rightarrow 2g(y)+2g(-y)=0 \Rightarrow g \text{ is an odd function}$$Now, we show that there are no real solutions to $g(a)=a+1$ except for $a=-1$, before proceeding further. Assume $g(a)=a+1$. Then, using $( \spadesuit )$, we get $$2g(a)=g(g(a)-1)=g((a+1)-1)=g(a) \Rightarrow g(a)=0 \Rightarrow a=-1$$Then for $x \neq -1$, we have $$P \left(x,\frac{y}{x+1-g(x)} \right) \Rightarrow g(x+y)+\frac{yg(x)}{x+1-g(x)}=g(x)+(x+1)g \left (\frac{y}{x+1-g(x)} \right)$$Also, by the tranformation $y \mapsto -y$, and using the fact that $g$ is odd, we get $$g(x-y)-\frac{yg(x)}{x+1-g(x)}=g(x)-(x+1)g \left (\frac{y}{x+1-g(x)} \right)$$Adding the previous two equalities, we have $$g(x+y)+g(x-y)=2g(x) \text{ } \forall x \neq -1$$Putting $y=x$ in the above equality gives $g(2x)=2g(x)$. Then substituting $a=x-y$ and $b=x+y$, we get that $$g(a)+g(b)=g(a+b) \text{ } \forall a+b \neq -2$$But, we have already shown the result for $a+b=-2$. So we get that $g$ is an additive function. Using additivity on $P(x,y)$ and $P(y,x)$ we get $$g(xy)+yg(x)=xg(y)+g(yg(x)) \text{ and } g(xy)+xg(y)=yg(x)+g(xg(y))$$Adding these two equations forces $g(xg(y))+g(yg(x))=2g(xy)$. Then $y=1$ gives $g(g(x))=2g(x)$. From this, we have $$P(g(x),y) \Rightarrow g(yg(x))+2yg(x)=g(x)g(y)+g(2yg(x)) \Rightarrow g(x)g(y)+g(yg(x))=2yg(x)$$where the last equality follows from $g(2x)=2g(x)$. Adding this to $P(x,y)$ (in the form given above) gives $g(xy)+g(x)g(y)=xg(y)+yg(x)$. Substituting back $g(x)=x-f(x)$, we get that $$xy-f(xy)+(x-f(x))(y-f(y))=x(y-f(y))+y(x-f(x)) \Rightarrow f(xy)=f(x)f(y)$$Also, we have $$f(x+y)=x+y-g(x+y)=x+y-g(x)-g(y)=f(x)+f(y)$$Thus, $f$ is both additive and multiplicative, and so $f$ must be either constant or the identity. Both these solutions contradict with our assumptions on $g$. Thus, the only answers are $f \equiv 0$ and $f=\text{id}$, which clearly work. $\blacksquare$

Let $P(x,y)$ denote $f(x+y+yf(x))=f(x)+f(y)+xf(y)$. The linear solutions are $f\equiv \text{Id}$ and $f\equiv 0$, assume $f$ is non linear. Firstly since non constant, $P(-1,x)$ implies $f(-1)=-1$. And $P(x,-1)$ implies $f(x-f(x)-1)=f(x)-x-1$. Since non linear, $P(x-f(x)-1,1)$ implies $f(1)=1$. Now $f(3)=3$, and comparing $P(1,f(x)-x-1)$ with $P(x-f(x)-1,3)$ yields $f(f(x)-x-1)=x-f(x)-1$. Now comparing $P(x-f(x)-1,1+y)$ and $P(f(x)-x-1,1-y)$ yields $f(1+y)=2-f(1-y)$. Comparing $P(1,x)$ and $P(1,-x)$ yields $f(x)=-f(-x)$. Comparing $P(x,y)$ and $P(x,-y)$ implies $f(x+z)+f(x-z)=2f(x)$ (for $z=yf(x)+y$) and so $f(2x)=2f(x)$, whence $f$ is additive. Now $P(x,1)$ implies $f(f(x))=x$ and $P(f(x),y)$ implies $f$ is multiplicative. So $f\equiv \text{Id}$, contradiction.

$x=0:$ $f(y+yf(0))=f(0)+f(y)$ $y=0:$ $f(x)=f(x)+f(0)+xf(0) \Rightarrow f(0)=0$ by taking $x\neq -1$. $x=-1:$ $f(-1+y+yf(-1))=f(-1)$ If $f(-1)\neq -1$, then $-1+y+yf(-1)$ is linear in $y$ with non-zero leading coefficient, so it includes all real numbers. So $f(r)=f(-1)$ for all reals r. So $f(x)=c$ for some $c$, checking gives $c=0$. So $f(x)=0$ is a solution. Henceforth assume $f(-1)=-1$. $y=-1:$ $f(x-1-f(x))=f(x)-1-x$ $x\rightarrow x-1-f(x)$, $f([x-1-f(x)]+y+y[f(x)-1-x])=[f(x)-1-x]+f(y)+[x-1-f(x)]f(y)$ $$\Rightarrow f(x-1-f(x)+yf(x)-xy)=f(x)-1-x+xf(y)-f(x)f(y)$$ Suppose $f(x)\neq x$ for some $x$, let $t=f(x)-x$, then $f(-1-t+yt)=t-1-tf(y)$ for all y. Take $y=-1$, then $f(-1-2t)=2t-1$ Take $y=1$, $f(-1)=t-1-tf(1)$, but $f(-1)=-1$, so $t-tf(1)=0 \Rightarrow f(1)=1$, because $t\neq 0$. $x=1:$ $f(1+y+yf(1))=f(1)+f(y)+f(y) \Rightarrow f(2y+1)=2f(y)+1 \Rightarrow f(2y-1)=2f(y-1)+1$ Now, note that: $$f(-1-t+yt)=t-1-tf(y)$$$$\Rightarrow f(-1+(y-1)t)=-1-t[f(y)-1]$$Replace $y\rightarrow 2y-1$, then $$f(-1+(2y-2)t)=-1-t[f(2y-1)-1]$$$$\Rightarrow 2f(-1+(y-1)t)+1=-1-t[2f(y-1)]$$$$\Rightarrow f(-1+(y-1)t)=-1-tf(y-1)$$ So $f(y-1)=f(y)-1 \Rightarrow f(y+1)=f(y)+1$ for all $y$. [Assuming some $t=f(x)-x$ is non-zero] $x\rightarrow x+1:$ $f(x+y+yf(x)+[1+y])=f(x)+f(y)+xf(y)+[1+f(y)]$ $y\rightarrow y+1:$ $f(x+y+yf(x)+[1+f(x)])=f(x)+f(y)+xf(y)+[1+x]$ Replace $y=f(x)$ in both equations above implies $f(f(x))=x$. Also $f(-1-t+yt)=t-1-tf(y) \Rightarrow f((y-1)t)=t(1-f(y)) \Rightarrow f(yt)=t(1-f(y+1))=-tf(y)$ So $f(-t)=-tf(-1)=t$. This implies $f(t-1)=-t-1, f(-t-1)=t-1$. $x=-t-1:$ $$f((-t-1)+y-(-t-1)y)=(t-1)+f(y)+(-t-1)f(y) \Rightarrow f(-1-t+ty)=t-1-tf(y)$$[This is what was done before] $x=t-1:$ $$f((t-1)+y-(t-1)y)=(-t-1)+f(y)+(t-1)f(y) \Rightarrow f(-1+t-ty)=-t-1+tf(y)$$ Following the same idea as first equation that lead to $f(yt)=-tf(y)$, $f(-1+t-yt)=-t-1+tf(y) \Rightarrow f((y-1)(-t))=(-t)(1-f(y)) \Rightarrow f(y(-t))=(-t)(1-f(y+1))=tf(y)$ So $f(-yt)=tf(y)=-[-tf(y)]=-f(yt)$, so $f$ is odd. Consider $f(x+y+yf(x))=f(x)+f(y)+xf(y)$ Replace $y\rightarrow -y$, $f(x-y-yf(x))=f(x)-f(y)-xf(y)$ If $x\neq 1$, let $y+yf(x)=z$, then $f(x+z)+f(x-z)=2f(x)$ for all $x$, $z$, because $f(x)+1\neq 0$ unless $x=-1$ by injectivity. [because $f(f(x))=x$] For $x=-1$, then $f(z-1)+f(-z-1)=f(z)-1+f(-z)-1=[f(z)+f(-z)]-2=-2=2f(-1)$. So $f(x+z)+f(x-z)=2f(x)=f(2x)$ for all $x$, $z$, so $f$ is additive. Then $f(x+y+yf(x))=f(x)+f(y)+xf(y) \Rightarrow f(x)+f(y)+f(yf(x))=f(x)+f(y)+xf(y)$ $\Rightarrow f(yf(x))=xf(y) \Rightarrow f(xy)=f(x)f(y)$ [because $f(f(x))=x$] So $f$ multiplicative and additive, hence $f(x)=x$ for all $x$. Contradiction with $t=f(x)-x\neq 0$ exists for some $x$. So only solutions are $f(x)=0$ and $f(x)=x$. QED

Alternatively, one can exploit $f(f(x))=x$ more by: $f(yt)=-tf(y) \Rightarrow f(-tf(y))=yt \Rightarrow f(-ty)=f(y)t$ This immediately implies $f$ is odd. So the substitution $x=t-1$ and $x=-t-1$ wouldn't be neccesary. Afterthoughts: - Actually $f(f(x))=x$ was not even needed, it was only for proving $f$ is odd. There ought to be easier ways to prove this. - $f(2y+1)=2f(y)+1 \Rightarrow f(y+1)=f(y)+1$ part was neat. - Key claim is still $f$ being odd, that is to consider $f(yt)$ and $f(-y(-t))$ and compare them, swapping the roles of $t$ and $-t$. - The rest are substitutions according to results that was obtained in the step immediately before, nothing too out of the ordinary.