As usual, let $P(x,y)$ denote $(y + 1)f(x) + f(xf(y) + f(x + y))= y$ for all $x,y\in \mathbb{R}$. $P(0,y)$ gives us $f(f(y))=y(1-f(0))-f(0)$ for all $y\in \mathbb{R}$. Now we've two possible cases of $1-f(0)$: 1. If $1-f(0)\neq 0$. We get that $f$ is injective. $P(x,-1)$ gives us $f(xf(-1)+f(x-1))=-1$ for all $x\in \mathbb{R}$. By injectivity, there exists constant $c$ that $f(x-1)+xf(-1)=c$ for all $x\in \mathbb{R}$, in other words, $f$ is a linear function. Plugging in the initial condition easily gives us only one possible solution from this case: $f(x)=-x$ for all $x\in \mathbb{R}$. 2. If $1-f(0)=0$. We get that $f(f(y)) =-f(0)$ for all $y\in \mathbb{R}$. Note that $1+f(0)=2\neq 0$. $P(f(x),y)$ gives us $(y+1)(-f(0))+f(f(x)f(y)+f(f(x)+y))=y$ for all $x,y\in \mathbb{R}$. Hence, $f(f(x)f(y)+f(f(x)+y))=y(1+f(0))+f(0)\implies -f(0)=f( y(1+f(0))+f(0))$ for all $x,y\in \mathbb{R}$. Since $1+f(0)\neq 0$, $y(1+f(0))+f(0)$ take values of all real number when $y$ varies in $\mathbb{R}$. Hence, $f$ is a constant function, but plugging in the initial condition gives no possible solution. From both cases, we get only one solution the problem: $f(x)=-x$ for all $x\in \mathbb{R}$.

My solution is almost the same as above,but a bit simpler. its obvious that $f$ cannot be constant,so we immediately get that it's surjective(because $y(1-f(x))-f(x)$ can take any real value when $x$ is constant and $f(x) \neq 1$) now,$P(0,y)$ implies that $$(1)f(f(y))=y(1-f(0))-f(0) $$if $f(0)=1$ then $f(f(y))=-1$ for all y,contracting the surjectivity;so $f$ is bijective. from surjectivity,we know that there is a real $a$ such that $f(a)=0$ so $P(a,0)$ implies that $f(af(0))=0$ but since $f(0) \neq 1$ ,from injectivity we conclude that $a=0$. plugging in equation (1) we reach that $f(f(x))=x$.now,$P(f(x),0)$ implies that $f(x)=-x$.We can easily see that it works. so the only solution is $f(x)=-x$.

i don't know how is it problem 4 in the TST because i feel it's too easy anyways...

Here is my solution for this problem Solution $(y + 1)f(x) + f(xf(y) + f(x + y)) = y$ Let $x = 0$, we have: $f(f(y)) = y(1 - f(0)) - f(0), \forall y \in \mathbb{R}$ - If $f(0) = 1$ then: $f(f(y)) = - 1, \forall y \in \mathbb{R}$ Let $x = y = 0$, we have: $f(f(0)) = - f(0)$ Let $x = 0$, $y = f(0)$, we have: $f(- f(0)) = - f^2(0)$ So: $f(1) = - 1$, $f(- 1) = - 1$ Let $x = - 1$, $y = 1$, we have: $f(2) = 3$ Then: $f(3) = - 1$ Let $x = 3$, $y = 0$, we have: $f(2) = 1$, conflict with $f(2) = 3$ - If $f(0) \ne 1$ then: $f$ is bijective So there exists $a \in \mathbb{R}$ which satisfies $f(a) = 0$ Let $x = a$, $y = 0$, we have: $f(af(0)) = 0$ Then: $af(0) = a$ Let $x = 0$, $y = a$, we have: $(a + 1)f(0) + f(0) = a$ Hence: $(a + 1)f(0) + f(0) = af(0) = a$ or $f(0) = a = 0$ So: $f(f(y)) = y, \forall y \in \mathbb{R}$ Let $y = 0$, we have: $f(f(x)) = - f(x), \forall x \in \mathbb{R}$ Then: $ - f(x) = x$ or $f(x) = - x, \forall x \in \mathbb{R}$ Retry, we see that $f(x) = - x$ satisfies the problem In conclusion, we have: $f(x) = - x, \forall x \in \mathbb{R}$

Easy but very beautiful! Medjl wrote: Find all functions $f : \mathbb{R} \rightarrow \mathbb{R}$ such that $$(y + 1)f(x) + f(xf(y) + f(x + y))= y$$for all $x, y \in \mathbb{R}$. Let $P(x:y)$ denote the above FE. Rewrite the given FE as $f(xf(y)+f(x+y))=y(1-f(x))-f(x)\implies f$ is surjective. Claim:- $f(0)\ne 1$ FTSOC Assume $f(0)=1$. Due to the surjectivity of $f$ we get that there $\exists= k$ such that $f(k)=0$. So, $P(0:k)\implies (k+1)f(0)+f(0)=k\implies (k+2)f(0)=k\implies \frac{k}{k+2}=1$. But there deos not exist such $k$! Hence, $f(0)\ne 1$ Now suppose there $\exists a,b$ such that $f(a)=f(b)$ and $a\ne b$. \begin{align*} P(0:a)\implies (a+1)f(0)+f(f(a))=a \end{align*}\begin{align*} P(0:b)\implies (b+1)f(0)+f(f(b))=b\end{align*}So, comparing both the above equations we get $(a+1)f(0)-a=(b+1)f(0)-b\implies a(f(0)-1)=b(f(0)-1)$. But from Claim we get that $f(0)\ne 1$. Hence, $a=b\implies f$ in injective. Now $P(k:0)\implies f(kf(0))=0=f(k)\implies k=kf(0)\implies f(0)=0$. Now $P(0:x)\implies f(f(x))=x\implies f$ is involutive and $P(x:0)\implies f(x)=-f(f(x)=-x\implies\boxed{f(x)=-x}$ is the only solution which clearly satisfies the given equation. $\qquad\blacksquare$

I hope I didn't fakesolved it. Btw that was fun! Solution : Let $P(x,y)$ denotes the assertion. We divide the solution in two cases, as we will see now: Case : If $f(0)=1,$ then $P(0,y):f(f(y))=-1,$ taking both sides $f,$ we have $f(-1)=-1$. Now $$P(1,-1)\implies -1=f(f(-1)+f(0))=f(-1+1)=f(0)=1$$Contradiction. $\square$ Case : If $f(0) \neq 1,$ then easy to see that, through $P(0,y),$ we have injectivity of $f$. Now combining $P(x+1,-1)$ and $P(0,-1)$ we have $$f(xf(-1)+f(-1)+f(x))=-1=f(f(-1))$$So $f(x)=-xf(-1)$. On substituting this in original $P(x,y)$ we get $f(x)=-x$ as our only desired solution. $\qquad \blacksquare$

Ans: $f(x)=-x.$ Proof: It's easy to see that the function above is indeed a solution. Let $P(x,y)$ denote the given assertion, we have \[P(0,x): f(f(x))=(1-f(0))x-f(0).\]We now divide into two cases: Case 1: $f(0)=1$, we can see that $f(f(x))=-1$, and in particular, $f(f(0))=f(1)=-1$. Also, \[P(f(1),0): f(f(1)-1)=f(-2)=1\]which we then take \[P(-2,0): f(-1)=-1\]to make \[P(1.-1): f(0)=-1\]a contradiction. Case 2: $f(0)\ne 1$, we can see that $f$ is bijective in this case. Therefore, \[P(0,-1): f(f(-1))=-1\]and \[P(x+1,-1): f((x+1)f(-1)+f(x))=-1\]imply $f(x)=-f(-1)x$. Upon checking, $f(x)\equiv -x$ is the only solution. $\quad \blacksquare$

Let $P(x,y)$ be the assertion $(y+1)f(x)+f(xf(y)+f(x+y))=y$. Note that there is only one linear solution, $\boxed{f(x)=-x}$. Assume that $f$ is nonlinear. $P(0,x)\Rightarrow x(f(0)-1)+f(0)+f(f(x))=0$ If $f(0)\ne1$ then $f$ is injective, so $P(x,-1)\Rightarrow f(xf(-1)+f(x-1))=-1$, thus $f$ is linear. Now we have $f(0)=1$. $P(0,x)\Rightarrow f(f(x))=-1$ $P\left(f(x),\frac{y-1}2\right)\Rightarrow f(f(x)f(\frac{y-1}2)+f(f(x)+\frac{y-1}2))=y$, so $f$ is surjective. Surjectivity gives $f(x)=-1\forall x$ from $f(f(x))=-1$, contradiction with $f(0)=1$. There are no more solutions. $\square$

Basically Just substituting x with 0 instantly gives us that $f$ bijective.Thereafter we assume $k$ to be a root and then substitute y with k and directly get that $f$ is linear. Then just comparing coefficients does the job. We get $f(x) =-x$.

dame dame

As usual let $P(x,y)$ denote the given assertion. $f(\text{something}) = y[1 - f(x)] - f(x)$ Clearly $f \equiv 1$ does not work, fix $x=c$ such that $f(c) \neq 1$ which implies surjectivity. $P(0,y): f(f(y)) = y[1 - f(0)] - f(0)$ and if $f(0)=1$ hence $f(f(y)) = -1$ which is clearly not surjective, hence $f(0) \neq 1$ and $f$ is bijective. $P(0,-1): f(f(-1)) = -1$ $P(1,-1): f(f(-1) + f(0)) = -1 = f(f(-1)) \implies f(-1) + f(0) = f(-1) \implies f(0)=0$. $P(0,y): f(f(y)) = y$ $P(x , 0): f(x) + f(f(x)) = 0 \implies f(x) + x = 0 \implies f(x) = -x$ which obviously works.

Solved with Rama1728 Claim 1: $f$ is surjective. Proof: Since $f(x)=1$ is not a solution, there exists $c$ such that $f(c) \ne 1$ then by $P \left(c, \frac{x+f(c)}{1-f(c)} \right)$ $$f \left(cf \left(\frac{x+f(c)}{1-f(c)} \right)+f \left(c+\frac{x+f(c)}{1-f(c)} \right) \right)=x \implies f \; \text{surjective}$$Claim 2: $f$ is bijective. Proof: Becuase of Claim 1: we need to show that $f$ is injective. By $P(0,x)$ $$f(f(x))=(1-f(0))x-f(0) \implies f \; \text{injective if} \; f(0) \ne 1$$Now if $f(0)=1$ then $f(f(x))=-1$ which is not posible, hence proven. Finishing the problem: We now have to show that $f(x)=-x$ with all the claims. By $P(x,-1)$ we can easly get $f$ linear becuase of Claim 2 so let $f(x)=ax+b$. We will compute $b$ to make the things easier so $f(b)=0$ and $f(0)=b$, and now we use $P(b,0)$ to get $b=0$. Now we replace $f(x)=ax$ on the F.E. with $y=0$ and $x \ne 0$ $$ax+a^2x=0 \implies a=-1$$Hence $f(x)=-x$ is the unique sol as desired.

Let $P(x,y)$ denote the given assertion. $P(0,x): xf(0)+f(0)+f(f(x))=x\implies f(f(x))=x(1-f(0))+f(0)$. Claim: $f$ is surjective. Proof: Note that $f(xf(y)+f(x+y))=y-(y+1)f(x)=y(1-f(x))-f(x)$. Clearly $f(x)\ne 1$ for some $x$ since $f\equiv 1$ is not a solution. So varying $y$ gives that $y(1-f(x))-f(x)$ can take any real value. $\blacksquare$ Claim: $f(0)\ne 1$. Proof: Suppose FTSOC $f(0)=1$. Then $f(f(x))=1$ for all $x$. If $f$ was surjective, then $f\equiv 1$, which contradicts surjectivity. Claim: $f$ is injective. Proof: Suppose $f(a)=f(b)$. Note that $f(f(a))=a(1-f(0))+f(0)$ and $f(f(b))=b(1-f(0))+f(0)$. So $a(1-f(0))=b(1-f(0))$. Dividing both sides by $1-f(0)$ gives $a=b$. $\blacksquare$ So $f$ is bijective. $P(x+1,-1): f((x+1)f(-1)+f(x))=-1$. Since $f$ bijective, let $c$ be the unique value such that $f(c)=-1$. Then $f(x)+(x+1)f(-1)=c\implies f(x)=-xf(-1)+c-f(-1)$, so $f$ is linear. Claim: $f(0)=0$. Proof: $P(0,0): f(f(0))=f(0)$. Since $f$ is injective, $f(0)=0$ $\blacksquare$ So $P(0,x)$ gives $f(f(x))=x$ and we also have $f(x)=ax$ for some constant $a$ (since $f$ is linear and $f(0)=0$). Thus, $a^2x=x\implies a=1$ or $a=-1$. Noting that $f(x)=x$ is not a solution, we have $\boxed{f(x)=-x}$, which works.

Denote the assertion by $P(x,y).$ Note that $f$ is surjective. And thus $f(0)\neq 1$ since otherwise $P(0,x)$ gives $f(f(x))=-1,$ absurd. In particular $f$ is bijective. Take $f(x_0)=0$ then $P(x_0,0)$ implies $x_0=0.$ Now $P(0,x)$ implies $f(f(x))=x$ and then $P(x,0)$ gives $f(x)=-x,$ which fits.

Let $P(x,y)$ denote the assertion. I claim that the only solutions are $f(x)=-x$. It is easy to check that these indeed work; now we prove that these are the only ones. Firstly, we deal with the case if $f(0)=1$. So now we assume that $f(0)=1$. Now,\[P(x,0)\implies f(x)+f(x+f(x))=0.\tag{1}\]Also,\[P(0,y)\implies f(f(y))=-1.\tag{2}\]Now using $(1)$ and $(2)$, we get,\[P(f(x),x)\implies f(f(x)^2-f(x))=2x+1.\tag{3}\]Clearly, we must have that $f(x)$ is surjective. Now if we have $f(a)=f(b)$, then setting $x=a$ and $x=b$ separately in $(3)$ we get that $a=b$. So this gives us that $f$ is bijective. Now pick some $m\not=n$. Now using the surjective property, we get some $k$, $l$ such that $f(k)=m$ and $f(l)=n$. Now putting $k$ and $l$ in $(2)$, we get that $f(m)=f(n)=-1$ which is a contradiction to the fact that $f$ was injective. So we must have $f(0)\not=1$. Now we get that,\[P(0,y)\implies f(f(y))=y(1-f(0))-f(0).\tag{i}\]This gives us that $f$ is surjective as $f(0)\not=0$. Now if $f(a)=f(b)$ for some $a$, $b$; then by plugging $a$ and $b$ in $(\text{i})$, we get that $a=b$. So $f$ is injective too; and thus is bijective. Now we must have some $k$, such that $f(k)=0$. Now using $f(a)=0$, $f(0)\not=1$ and the injectivity of $f$, we get that, \begin{align*} P(a,0)&\implies f(af(0))=0=f(a)\\ &\implies af(0)=a\\ &\implies a=0 .\end{align*}So $f(0)=0$. Now, \[P(x,0)\implies f(x)+f(f(x))=0.\tag{ii}\]Also,\[P(0,y)\implies f(f(y))=y.\tag{iii}\]So from $(\text{ii})$ and $(\text{iii})$, we thus finally have,\[f(x)=-f(f(x))=-x,\]and we are done.