$(x, y) = (0, 0)\implies f(0) = f(0)^2\implies f(0) = 0$ or $f(0) = 1$. If $f(0) = 0$, $y=0 \implies f(x) = 0, \forall x\in\mathbb{R}$ which is clearly not a solution. So, $f(0) = 1$. Letting $y=-x$, we get $f(x)f(-x) = 1 - x^2 = 0 \iff x = \pm 1$, i.e. $f(1)f(-1) = 0$. If $f(1) = 0$, first of all note that $x=0$ is the unique value of $x$ so that $f(x) = 1$, for $x_0$ so that $f(x_0) = 1$, plugging $(x, y) = (0, x_0)\implies f(x_0f(x_0)) = x_0^2 + f(x_0) \iff 1 = x_0^2 + 1\iff x_0 = 0$. Then, plugging for $x+y = 1$, we get two relations $$f(x) = (1-x)^2 + f(x)f(1-x),$$$$f(1-x) = x^2 + f(x)f(1-x).$$By the second relation, we get $f(1-x)(1-f(x)) = x^2$. We multiply the first relation with $1-f(x)\neq 0$ to obtain that $(f(x) - (1-x))^2 = 0 \iff f(x) = 1-x$ for $x\neq 0$. Clearly, by $f(0) = 1$, $f(x) = 1-x$ for all $x\in\mathbb{R}$ is a solution. If $f(-1) = 0$, we proceed entirely similarly to obtain the second solution $f(x) = 1 + x$ for all $x\in\mathbb{R}$.

After $f(x_0)=1\Leftrightarrow x_0=0$ we can also finish this way: by putting $(x,y)=(x-1,1)$ when $f(1)=0$ and $(x,y)=(x+1,-1)$ when $f(-1)=0$.

$P(x,-x)$ $f(x)f(-x)=1-x^2$ the rest is as similar as2014 BMO TST P4 (I think).

The only answers are $f(x)=1-x$,$f(x)=x+1$ and $f(x)=0$, for all real $x$ Let $P(x,y)$ be the given assertion. From $P(x,0)$ we get that $f(x)=f(x)f(0)$. Let's say that $f(0)=0$, then we have that $f(x)=0$. Now let's say that $f(0) \neq 0$. Then we must have that $f(0)=1$. From $P(x,-x)$ we get that $f(x)f(-x)=1-x^2$. Thus giving us that $f(1)f(-1)=0$. Giving us that either $f(1)=0$ or $f(-1)=0$. So now let's assume that $f(1)=0$. From $P(1,y)$ we have that $f(1+yf(y+1))=y^2=f(1-yf(1-y))$. From $P(x,1)$ we have that $f(x+f(x+1))=x$. So now let's $S$ be the set of numbers $x$ such that $f(x)=1$, or in notation $S= \left\{ x \mid f(x)=1 \right\}$ Now let's say that $y+1 \in S$. Then we have from $f(1+yf(y+1))=y^2$, that $y^2=1$. This implies that $\text{max}\{card(S)\} = 2$, since $y$ doesn't depend on $S$. Now assume that $y=1$, this implies that $2 \in S$. Now plug in $P(2-y,y)$ to get that $1-y^2=f(2-y)f(y)$. Now plug in $y=2$ to get that $-3\neq f(2)f(0)=1$, thus we get a contradiction. Meaning that $y=-1$ and that $card(S)=1$, this means that $f$ is injective at point $x=0$. Thus from $f(x+f(x+1))=1$ we get that $f(x+1)=-x \implies f(x)=1-x$. The second case when $f(-1)=0$ is similar to $f(1)=0$ and we get that $f(x)=x+1$ in that case.

This works out really nicely. Let $P(x,y)$ be the assertion $f\left(x+yf(x+y)\right)=y^2+f(x)f(y)$. $P(x,0)\Rightarrow f(x)=f(x)f(0)$ If $f(0)\ne0$ then $f(x)=0$, which isn't a solution, so we have $f(0)=0$. $P(1,-1)\Rightarrow 0=f(1)f(-1)$ Case 1: $f(1)=0$ $P(1-y,y)\Rightarrow f(1-y)=y^2+f(1-y)f(y)$ Setting $y\mapsto1-y$ and multiplying by $-1$: $$-f(y)=-1+2y-y^2-f(y)f(1-y)$$So $f(1-y)=f(y)+2y-1$. Plugging this back into the first equation, we have $f(y)^2+2(y-1)f(y)+(y-1)^2=0$, which is $\left(f(y)+y-1\right)^2=0$, so $\boxed{f(x)=-x+1}$, which is a solution. Case 2 is done similarly. Case 2: $f(-1)=0$ $P(-1-y,y)\Rightarrow f(-1-y)=y^2+f(-1-y)f(y)$ Subtracting when $y\mapsto-1-y$, $f(-1-y)=f(y)-2y-1$. Plugging this back into the first equation gives: $$f(y)^2-2(y+1)f(y)+(y+1)^2=0$$$$\Rightarrow\boxed{f(x)=x+1}$$

I think my solution is a bit different. Anyways another great problem by Dorlir Sir/Ma'am Let $P(x,y)$ denote the assertion in the given problem. $P(0,0) \implies f(0) = f(0)^2 \implies f(0) \in \{0,1\}$ . Suppose $f(0) = 0$. Then $P(x,0) \implies f(x) = f(x)f(0) = 0$ so $f \equiv 0$ but this doesn't work. So $f(0) = 1$. $P(1,-1) \implies f(1) = 0 \text{ or } f(-1) = 0$. First we would prove injectivity at $f(x) = 1$. Let $v$ be such that $f(v) = 1$. (There exists such a $v$ since $f(0) = 1$). Now $P(0,v) \implies v = 0$ So we have proven that $f(x) = 1 \iff x = 0$ Case 1: $f(1) = 0$ Now $P(x,1) \implies f(x+f(x+1)) = 1 = f(0)$. Due to injectivity at $f(y) = 1$ we have that $x +f(x+1) = 0 \implies f(x+1) = -x \implies \boxed{f(x) = 1-x}$ Case 2: $f(-1) = 0$ Now $P(x,-1) \implies f(x-f(x-1)) = 1 = f(0)$. Due to injectivity at $f(y) = 1$ we have that $x - f(x-1) = 0 \implies x = f(x-1) \implies \boxed{f(x) = x + 1}$

Answer: $f(x) = 1 + x \quad \text{or} \quad f(x) = 1 - x$ Proof: Let $P(x,y)$ denote the assertion. $P(x,0) \to f(x) = f(x) \cdot f(0) \implies f(0) = 1$ $P(1,-1) \to f(1 - 1) = 1 + f(1) \cdot f(-1) \implies f(1)\cdot f(-1) = 0$ $\implies f(1) = 0 \quad \text{or} \quad f(-1) = 0$ Case 1: $f(1) = 0$ $P(x,1) \to f(x + f(x+1)) = 1$ $RHS = \text{constant} \implies x + f(x+1) = c$ for some constant $c$ $\implies x-1 + f(x) = c$ $\implies f(x) = c + 1 - x$ Plugging this back in our original equation gives $c = 0$ $\implies \boxed{f(x) = 1 - x}$ Case 2: $f(-1) = 0$ $P(x,-1) \to f(x - f(x-1)) = 1$ $RHS = \text{constant} \implies x - f(x-1) = c$ for some constant $c$ $\implies x+1 - f(x) = c$ $\implies f(x) = 1+x - c$ Plugging this back in our original equation gives $c = 0$ $\implies \boxed{f(x) = 1 + x}$ @below since RHS is a constant doesn't that imply that the input inside the function is also a constant? do we really need injectivity for that? @below thanks i get it now sorry my bad ill edit the sol soon / trash it out Thanks @EpicNumberTheory for pointing out the mistake, got to learn a lesson today

MrOreoJuice wrote: Case 1: $f(1) = 0$ $P(x,1) \to f(x + f(x+1)) = 1$ $RHS = \text{constant} \implies x + f(x+1) = c$ for some constant $c$ Case 2: $f(-1) = 0$ $P(x,-1) \to f(x - f(x-1)) = 1$ $RHS = \text{constant} \implies x - f(x-1) = c$ for some constant $c$ $\implies \boxed{f(x) = 1 + x}$ I think you can't conclude this. You need injectivity at $f(x) = 1$ for this type of conclusion which you haven't proven (didn't mention it either). @above. What if $f(2) = 1$ and $f(3) = 1$. Now what ? This way your solution at here is also wrong

Amazing one @dangerousliri !! Let $P(x,y)$ the assertion to the given F.E. and note that $f$ is not constant. $P(x,0)$ $$f(x)=f(x)f(0) \implies f(0)=1$$Note that $f$ is injective at $1$ becuase if there exists $c$ such that $f(c)=1$ then by $P(0,c)$ $$1=c^2+1 \implies c=0$$$P(x,-x)$ $$1-x^2=f(x)f(-x) \overset{x=1}{\implies} f(1)=0 \; \text{or} \; f(-1)=0$$If $f(1)=0$ then by $P(x-1,1)$ $$f(x-1+f(x))=1 \implies f(x)+x-1=0 \implies f(x)=1-x$$If $f(-1)=0$ then by $P(x+1,-1)$ $$f(x+1-f(x))=1 \implies -f(x)+x+1=0 \implies f(x)=x+1$$Thus we are done

The solutions are $f(x)=1-x$ and $f(x)=x+1$, which both work. Let $P(x,y)$ denote the given assertion. $P(0,0): f(0)=f(0)^2$, so $f(0)=0$ or $f(0)=1$. Assume that $f(0)=0$ $P(x,0): f(x)=0$ for all $x\in \mathbb{R}$, which is clearly not a solution. So $f(0)=1$. $P(x,-x): 1-x^2=f(x)f(-x)$ so either $f(1)=0$ or $f(-1)=0$ or both. $P(0,x): f(xf(x))=x^2+f(x)$ Also note that if $f(x)=0$, we must have $x=1$ or $x=-1$. If $f(x)=1$, then from $P(0,x)$, we get $1=x^2+1\implies x=0$. Case 1: $f(1)=f(-1)=0$. $P(1,-2): f(1)=4$, a contradiction. Case 2: $f(1)=0$ and $f(-1)\ne0$. $P(x-1,1): f(x-1+f(x))=1, x-1+f(x)=0\implies \boxed{f(x)=1-x}$. Case 3: $f(1)\ne0$ and $f(-1)=0$. $P(1-x,-1): f(1-x-f(-x))=1$, so $1-x-f(-x)=0\implies f(-x)=1-x$, so $\boxed{f(x)=x+1}$.

Ooooh, I'm reviving an age old thread. Very standard. The answers are $f(x) = 1-x$ for all $x \in \mathbb{R}$ and $f(x)=x+1$ for all $x\in \mathbb{R}$. It’s easy to see that these functions satisfy the given equation. We now show these are the only solutions. We denote by $P(x,y)$ the assertion that $f\left(x+yf(x+y)\right)=y^2+f(x)f(y)$. We start off with the following observation. Claim :The function $f$ is injective at $1$, i.e $f(x)=1$ if and only if $x=0$. Proof : Note that $P(0,0)$ gives $f(0)^2 = f(0)$ which implies $f(0) \in \{0,1\}$. If $f(0)=0$ we have by $P(x,0)$ that $f(x)=f(x)f(0)=0$ for all real numbers $x$. But this is clearly impossible upon substitution. Thus, $f(0)=1$. Now, say there exists a real number $\alpha \neq 0$ such that $f(\alpha)=1$. $P(0,\alpha)$ gives, \begin{align*} f(\alpha f(0)) &= \alpha ^2 + f(0)f(\alpha)\\ f(\alpha) &= \alpha ^2 + 1 \\ \alpha ^2 &= 0 \end{align*}which is a clear contradiction. Thus, there exists no such $\alpha $ proving the claim. Now, $P(1,-1)$ implies \[1=f(0)=f(1-f(0))=1+f(1)f(-1)\]Thus, $f(1)f(-1)=0$ so $f(1)=0$ or $f(-1)=0$. We examine these two cases separately. Case 1 : $f(1)=0$. Here, $P(x,1)$ yields, \[f(x+f(x+1))=f(x)f(1)+1 = 1\]which due to previous claim implies $x+f(x+1)=0$. Thus, considering $x=t-1$ we have that $f(t)=1-t$ for all real numbers $t$. Case 2 : $f(-1)=0$. Here, $P(x,-1)$ yields, \[f(x-f(x-1))=f(x)f(-1)+1 =1 \]which due to the previous claim implies $x-f(x-1)=0$. Thus, considering $x=t+1$ we have that $f(t)=t+1$ for all real numbers $t$. Thus, all solutions to the given equation are as claimed and we are done.