Filipjack wrote: Find the functions $f: \mathbb{R} \to \mathbb{R}$ that satisfy $$(f(x)-y)f(x+f(y))=f(x^2)-yf(y),$$for all real numbers $x$ and $y.$ The only constant solutions are $\boxed{\text{S1 : }f(x)=0\quad\forall x}$ and $\boxed{\text{S2 : }f(x)=1\quad\forall x}$ So let us from now look only for nonconstant solutions. Let $P(x,y)$ be the assertion $(f(x)-y)f(x+f(y))=f(x^2)-yf(y)$ Let $a=f(0)$ 1) $f(x)$ is injective

2) $f(0)=0$

3) $f(x)=x$ $\forall x$ $P(0,x)$ $\implies$ $-xf(f(x))=-xf(x)$ and so $f(f(x))=f(x)$ $\forall x\ne 0$ and so (injective) $f(x)=x$ $\forall x\ne 0$, still true when $x=0$ And solution $\boxed{\text{S3 : }f(x)=x\quad\forall x}$, which indeed fits

Nice FE! The answers are $f(x)=0$ for all $x\in \mathbb{R}$, $f(x)=1$ for all $x\in \mathbb{R}$ and $f(x)=x$ 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 start off with the following important claim. Claim : The function $f$ injective or constant. Proof : Say there exists $\alpha \neq \beta \in \mathbb{R}$ such that $f(\alpha)=f(\beta)=c$. Then, $P(x,\alpha)$ and $P(x,\beta)$ gives \begin{align*} (f(x)-\alpha)f(x+f(\alpha))&=f(x^2)-\alpha f(\alpha)\\ (f(x)-\beta)f(x+f(\beta))&=f(x^2)-\beta f(\beta) \end{align*}Taking the difference of these two equations gives, \begin{align*} (f(x)-\alpha)f(x+f(\alpha)) +\alpha f(\alpha) &= (f(x)-\beta)f(x+f(\beta))+\beta f(\beta)\\ f(x+c)(\beta - \alpha) &= c(\beta - \alpha) \end{align*}Thus, $\alpha = \beta$ or $f$ is constant. We can now quickly check that if $f$ is constant then, $f(x)=0$ for all $x\in \mathbb{R}$ or $f(x)=1$ for all $x\in \mathbb{R}$. In what follows we assume that $f$ is non-constant. Now, we look at $P(x,f(x))$ which yields, \[0=(f(x)-f(x))f(x+f(f(x)))=f(x^2)-f(x)f(f(x))\]So, \begin{align} f(x^2)&=f(x)f(f(x)) \end{align}for all $x\in \mathbb{R}$. Now, we can fix some values of $f$. Claim : $f(1)=1$ and $f(0)=0$. From this we have $f(1)=f(1)f(f(1))$. Thus, $f(1)=0$ or $f(f(1))=1$. Similarly, $f(0)=0$ or $f(f(0))=1$.Now, note that if $f(1)=0$ since $f$ is injective, $f(0)\neq 0$ and thus, $f(f(0))=1$. But then, note that, plugging $x\to f(0)$ into $(1)$, we have \[f(f(0)^2)=f(0)f(f(0))=f(0)\]which since $f$ is injective implies $f(0)^2-0$ which is a clear contradiction. Thus, $f(1)=1$. Further, this means, $f(f(0))\neq 1$ (since then using injectivity we must have $f(0)=1$ which is absurd). Thus, $f(0)=0$ as desired. Now we are almost there. Simply note that $P(0,x)$ gives \[-xf(f(x))=-xf(x)\]From which it follows that $f(f(x))=f(x)$ for all $x\neq 0$ and thus, since $f$ is injective we conclude that $f(x)=x$ for all $x\neq 0$ which then implies that all solutions are indeed of the claim forms and we are done.

Let $P(x,y)$ the assertion of the F.E., now that if constant $f=0,1$ work so we assume non-constant. Claim 1: $f$ is injective. Proof: Suppose $f(a)=f(b)$ then by $P(x,a)-P(x,b)$ we get $(b-a)f(x+f(a))=(b-a)f(a)$, as $f$ is non-constant from here it must hold that $b=a$ as desired. Claim 2: $f(0)=0$ Proof: Suppose not then $P(0,0)$ gives $f(0)=f(0)f(f(0))$ so $f(f(0))=1$ by injectivity we can't have $f(1+f(0))=1$ therefore by $P(1,0)$ we get $f(1)=0$. Now for $x \ne 0$ set $P(1,x)$ which gives $f(1+f(x))=f(x)$ therefore $f(x)=x-1$ which is false as then $f(0)=-1$ but $f(-1) \ne 1$, therefore $f(0)=0$ must happen to avoid the contradiction. Finishing: $P(x,0)$ gives $f(x)^2=f(x^2)$ while $P(x,f(x))$ gives $f(x^2)=f(x)f(f(x))$ therefore for $x \ne 0$ we have $f(x)=f(f(x))$ and by injectivity $f(x)=x$ for $x \ne 0$, as $f(0)=0$ we get $f(x)=x$ as desired. Therefore the only solutions to the functional equation are $f(x)=0,1,x$ thus we are done .

Let $P(x,y)$ denote $(f(x)-y)f(x+f(y))=f(x^2)-yf(y)$. We assume $f$ non-constant (constant solutions are $f \equiv 0,1$. Suppose $f(a)=f(b)$. \begin{align*} P(x,a)&: (f(x)-a)f(x+f(a))=f(x^2)-af(a) \\ P(x,b)&: (f(x)-b)f(x+f(b))=f(x^2)-bf(b) \\ &\implies a[f(x+f(a))-f(a)]=b[f(x+f(b))-f(b)] \implies a=b \\ \end{align*}Thus, $f$ is injective. $P(x,0): f(x)f(x+f(0)=f(x^2)$ $P(x,f(x)): f(x^2)=f(f(x))$ Hence, $f(x)=0$, which fails for more than one $x$ because of injective. or $f(f(x))=f(x+f(0)) \implies f(x)=x+f(0)$. Subbing into $P(x,y)$, we get that $f(0)[3x+2f(0)-1]=0$, so $f(0)=0$, so $\boxed{f(x)=x}$.