$P(x,-1): f(xf(-1)-1)=0$. If $f(-1) \neq 0 \implies f \equiv 0 \implies $ Contradiction. Say $f(y_0)=0$. Thus $P(0,y_0): f(-y_0^2)=(y_0+1)f(x-y_0)$. If $y_0 \neq -1 \implies f \equiv c$ for constant $c$, hence $c \equiv 0$. $P(x,x+1): f(xf(x+1)-(x+1)^2)=0=f(-1) \implies xf(x+1)=x^2+2x \implies f(x+1)=x+2$, for $x \neq 0$. Thus $f(x)=x+1$ for $x \neq 1$. Using $P(x,1): f(xf(1)-1)=2f(x-1)$, and as $f(1) \neq 0$, for $x$, say $>\frac{100}{f(1)}$, we get $xf(1)=2x \implies f(1)=2$. Thus our 2 solutions are $f \equiv x+1$ and $ \equiv 0$. @below because we proved $f$ is injective at 0.

Why we have this: $\Rightarrow xf(x+1)={{x}^{2}}+2x$?

TuZo wrote: Why we have this: $\Rightarrow xf(x+1)={{x}^{2}}+2x$? Because he showed that $f(z)=0$ happens only for $z=-1$. So using $z=xf(x+1)-(x+1)^2$ we must have $z=-1$ and hence $xf(x+1)=x^2+2x$. Edit: Oh, I didn't see that OP already answered your query...

Well, this problem was proposed by me. I'm sorry if people felt bored about it and wasted their valuable time on what I thought to be a nice study in classical functional equation techniques, admittedly a rather easy one (but by no means too easy for the competition as the results clearly show). In any case, I can just encourage you to submit your own surely more interesting problem proposals so that future juries can choose their problems more wisely and there will hopefully be no more need for AoPS threads titled "Boring Problem XY".

Tintarn wrote: Well, this problem was proposed by me. I'm sorry if people felt bored about it and wasted their valuable time on what I thought to be a nice study in classical functional equation techniques, admittedly a rather easy one (but by no means too easy for the competition as the results clearly show). In any case, I can just encourage you to submit your own surely more interesting problem proposals so that future juries can choose their problems more wisely and there will hopefully be no more need for AoPS threads titled "Boring Problem XY". Well every FE is boring nowadays. Sorry if we hurt your feelings

@Tintarn, I like your problem. It fits the style of Baltic Way. Keep the good job done. I look forward to your next propositions on contests.

Hamel wrote: Well every FE is boring nowadays. I don't agree with this claim. I think that AoPS users dangerousliri and socrates have provided great counter-examples to it by proposing nice and innovative functional equations in recent years.

Tintarn wrote: Hamel wrote: Well every FE is boring nowadays. I don't agree with this claim. I think that AoPS users dangerousliri and socrates have provided great counter-examples to it by proposing nice and innovative functional equations in recent years. Actually what I mean by that is that ideas get revisited and transformed into new problems all the time. IMO Dangerousliri did a huge break proposing such a fe in IMO 2018!

Does anyone has diferent solution?

Let $P(x,y) : f(xf(y)-y^2)=(y+1)f(x-y)$ We only see when $ f(x) \neq 0$ $ P(x,-1) \implies f(-1)=0$ If $ f(c)=0 $, $P(x,c) \implies c=-1 $ $ P(y-1,y) \implies f((y-1)f(y)-y^2)=0 \implies f(y)=y+1 (y \neq 1) $ $ P(-1,-2) \implies f(1)=2 $, so the answers are $\boxed{f \equiv 0 , f(x)=x+1}$

I read the comment now and I'd like to add that if an individual (even if the individual is a very experienced and advanced in math,) finds something boring, this does not mean that the problem is not valuable, finding something interesting really depends on that person's level in mathematics and their interests. When we are talking about holding a competition or proposing to a competition or Olympiad what we must consider is the variety of students and their level and their interests. I personally don't like geometry and yet, I can't imagine an IMO without geometry. We must respect people who put their time and energy and study Olympiad math and at last share their results with us and let us enjoy the problem solving.

$\spadesuit \color{green}{\textit{\textbf{ANS:}}}$ $f(x)\equiv 0 \quad \textrm{and} \quad f(x)\equiv x+1$. $\clubsuit \color{red}{\textit{\textbf{Proof:}}}$ It's easy to see that these are solutions to the given FE. Let $P(x,y)$ be the given assertion, if $f(0)=0$, then we have \[P(x,0): f(x)\equiv 0.\]If $f(0)\ne 0$, then \[P(0,-1): f(-1)=0.\]So, there exists $k\in \mathbb{R}$ such that $f(k)=0$, then \[P(k-1, k): f(-k^2)=0\]and also, \[P(0,k): (k+1)f(-k)=0.\]If $k\ne -1$, then $f(-k)=0$ and \[P(-k,-k): (1-k)f(0)=0 \implies k=1\]and this contradicts the fact that $f(-1)=0$ too. Therefore, $k=-1$, and \[P(x-1,x):f((x-1)f(x)-x^2)=0 \implies f(x)\equiv \frac{x^2-1}{x-1}\equiv x+1. \quad \blacksquare\]

Let $P(x,y)$ be the assertion $f(xf(y)-y^2)=(y+1)f(x-y)$. $\boxed{f(x)=0}$ works and is the only constant solution. Assume now that $f$ is nonconstant. $P(0,-1)\Rightarrow f(-1)=0$ Let there be $k$ such that $f(k)=0$. $P(x+k,k)\Rightarrow f(-k^2)=(k+1)f(x)$ so if $k\ne-1$ then $f$ is constant, contradiction. Thus $f(k)=0\Leftrightarrow k=-1$. $P(x-1,x)\Rightarrow f((x-1)f(x)-x^2)=0\Rightarrow (x-1)f(x)-x^2=1$, and we get $f(x)=x+1\forall x\ne1$. Now we just need to find $f(1)$, which is: $P(-1,-2)\Rightarrow f(1)=2$ So $\boxed{f(x)=x+1}$, which works.

The only solutions are $f \equiv 0$ and $f(x)=x+1$. Set $y=-1$, then we have that $f(xf(-1)-1)=0$. Let $S=\{ x \mid f(x)=0 \}$. Assume that $card(S) \geq 2$. Then set $y \in S$, but so that $y \neq -1$. Then we must have that $f(-y^2)=(y+1)f(x-y)$, this implies that $f$ is a constant valued function. Easily we get that $f \equiv 0$. Now assume that $card(S)=1$, this means that $f$ is injective at $x=-1$. Set $x-y=-1$, thus we must have that $xf(x+1)-(x+1)^2=-1$ and if $x \neq 0$,we have that $f(x+1)=x+2 \implies f(x)=x+1$. Now set $x=y$, then we must have that $f(xf(x)-x^2)=(x+1)f(0)$, if $x \neq 0$, we have that $x^2+x-x^2+1=(x+1)f(0)$, thus we have that $f(0)=1$. Thus we have that, for every real $x$, $f(x)=x+1$.

MarkBcc168 wrote: Find all functions $f:\mathbb{R}\to\mathbb{R}$ such that $$f(xf(y)-y^2)=(y+1)f(x-y)$$holds for all $x,y\in\mathbb{R}$. Assume that $f$ is constant $(f(x) \equiv c)$ and then: $$c=(y+1)c \; \forall y \in \mathbb R \implies c=0 \implies f(x) \equiv 0$$Now assume that $f$ is not constant: Let $P(x,y)$ the assertion of the given Funcional Equation: $P(x.-1)$ $$f(xf(-1)-1)=0 \implies f(-1)=0$$Note that $f(-1)=0$ is the only cero of the function because: Assume that existe a cero $c \ne -1$ such that $f(c)=0$ then: $P(x+c,c)$ $$f(-c^2)=(c+1)f(x) \implies f(x)=\frac{f(-c^2)}{(c+1)}$$This means that $c=0$ plugging $x=-c^2$ and hence: $P(x,0)$ $$f(x)=0 \; \forall x \in \mathbb R$$Contradiction!!!. Now using that $f(-1)=0$ is the only cero: $P(x.x+1)$ $$f(xf(x+1)-(x+1)^2)=0 \implies xf(x+1)=x^2+2x \implies f(x+1)=x+2$$This means that $f(x)=x+1$ for all $x$ on reals and $x \ne 1$. Use that $f(x-1)=x$ for $x \ne 2$ and $P(x,1)$ to get: $$f(xf(1)-1)=2f(x-1) \implies xf(1)=2x \implies f(1)=2$$And hence the solutions are: $\boxed{f(x) \equiv 0 \; \forall x \in \mathbb R}$ $\boxed{f(x)=x+1 \; \forall x \in \mathbb R}$ Thus we are done

$P(x,-1): f(xf(-1)-1)=0$. If $f(-1)\ne 0$, then we can vary $x$ and get $f\equiv 0$, which implies $f(-1)=0$, a contradiction. So $f(-1)=0$. Case 1: There exists $k\ne -1$ with $f(k)=0$. $P(x,k): f(-k^2)=(k+1)f(x-k)$. Setting $x=2k$ gives $f(-k^2)=0$. If there exists $a$ with $f(a)\ne 0$, $P(x,k+a)$ gives $f(-k^2)=(k+1)f(a)\ne 0$, a contradiction. So $\boxed{f\equiv 0}$ in this case. Case 2: $f$ is injective at $0$. $P(x,x+1): f(xf(x+1)-(x+1)^2)=0$. So \[xf(x+1)-(x+1)^2=-1\implies xf(x+1)=(x+1)^2-1=x^2+2x\] If $x\ne 0$, then $f(x+1)=x+2$. This implies $f(x)=x+1$ for $x\ne 1$. So to show that $\boxed{f(x)=x+1}$ for all $x$, we need to show $f(1)=2$. $P(1,1): f(f(1)-1)=2f(0)=2$. If $f(1)\ne 2$, then $f(1)-1\ne 1$, so $f(f(1)-1)=f(1)$. But this implies $f(1)=2$. Thus, $f(1)=2$, as desired.

$f(x)=0$ is a solution, let's consider the other cases. $P(x,x-k)$, where $f(k)\neq 0$ gives $f$ surjective. Let $f(c)=0$, $P(x,c)\Longrightarrow f(-c^2)=(c+1)f(x-c)$, LHS is constant so $c=-1$. $P(x,x+1)\Longrightarrow xf(x+1)=x^2-2x$, so $f(x)=x+1$ for all $x\neq 1$ $P(-1,-2)\Longrightarrow f(1)=2$, so $f(x)=x+1$ and $f(x)=0$ are the only solutions.

Actually this question needs only $f(-1)=0$ is unique And $P(x, 0)$+$P(-1,0)$+$P(x,-f(0))$ $\implies{f(0)=1}$ and $f(x) =0$ Then $P(x, x+1)$ $\implies{f(x) =x+1} $

Let $P(x, y)$ be the assertion. $P(x, x)$ gives $f(xf(x) - x^2) = 0$. So we know there exists some $\alpha$ such that $f(\alpha) = 0$. $P(0, -\alpha)$ gives $f(-\alpha^2) = 0$. $P(x, \alpha)$ gives: $$0 = (\alpha + 1)f(x - \alpha)$$Either $\alpha = -1$ or $f(x - \alpha) = 0$. Substitute $x$ as $n - \alpha$ and we get $f \equiv 0$ if $f(x - \alpha) = 0$. So we have $\alpha = -1$, which means "injectivity" if $f(x) = 0$. $P(x, x + 1)$ yields: $$f(xf(x + 1) - x^2) = f(-1) \implies xf(x + 1) = x^2 + 2x \implies f(x + 1) = x + 2 \implies f(x) = x + 1$$We still need to show $f(1) = 2$ (as we can't divide by 0) which is trivial by $P(x + 1, x)$ (where $ x\neq - 1$ which gives: $$2x + 2 = f(1)(x + 1) \implies f(1) = 2$$Hence, we are done.

Let $P(x,y)$ denote the assertion of the original FE Clearly $\boxed{\text{S1: }f(x)=0 \quad \forall x}$ is a solution (which indeed fit since $f(xf(y)-y^2)=0=(y+1)f(x-y)$) Now Suppose $\exists t \in \mathbb{R}$ such that $f(t)=0$ $P(t+y,y)$ $\implies$ $f(…)=(y+1)f(t)$,right hand side can be any real number, so $f$ is surjective $P(0,-1)$ $\implies$ $f(-1)=0$. Suppose $\exists t \in \mathbb{R}\wedge t \neq -1$ such that $f(t)=0$ $\because f$ surjective $\therefore$ $\exists\alpha$, $\beta \in \mathbb{R}$ such that $f(\alpha)=118$, $f(\beta)=1209$ $P(\alpha ,t)$ $\implies$ $\frac{f(-t^2)}{t+1}=118$ while $P(\beta ,t)$ $\implies$ $\frac{f(-t^2)}{t+1}=1209$, so $118=1209$, contradiction! $\therefore f(x)=0 \iff x=-1$ For $y \neq 1$, $P(y-1,y)$ $\implies$ $f((y-1)f(y)-y^2)=0$ $\implies$ $(y-1)f(y)-y^2=-1$ $\implies$ $f(y)=y+1$ $\forall y \neq 1$ On the other hand, $P(1119,1118)$ $\implies$ $1119^2-1118^2+1=1119f(1)$ $\implies$ $f(1)=2$ So we obtain the only other solution $\boxed{\text{S2: }f(x)=x+1 \quad \forall x}$ (which indeed fit since $f(xf(y)-y^2)=x(y+1)-y^2+1=-y^2+xy+x+1=(y+1)(x-y+1)=(y+1)f(x-y)$)