fasttrust_12-mn wrote: Let \( \mathbb{R} \) denote the set of real numbers. Find all functions \( f: \mathbb{R} \to \mathbb{R} \) such that \[ f(x^2) - y f(y) = f(x+y)(f(x) - y) \] for all real numbers \( x \) and \( y \). Let $P(x,y)$ be the assertion $f(x^2)-yf(y)=f(x+y)(f(x)-y)$ $P(0,x)$ $\implies$ $f(0)=f(0)f(x)$ and so : Either $f(0)\ne 0$ and so $\boxed{\text{S1 : }f(x)=1\quad\forall x}$, which indeed fits Either $f(0)=0$ and then : Subtracting $P(x,-x)$ from $P(-x,x)$, we get $x(f(x)+f(-x))=0$ and so $f(-x)=-f(x)$ $\forall x\ne 0$, still true when $x=0$ $P(x,0)$ $\implies$ $f(x^2)=f(x)^2$ $P(x,-x)$ $\implies$ $f(x)^2-xf(x)=0$ and so $\forall x$, either $f(x)=0$, either $f(x)=x$ If $f(x)=0$ and $f(y)=y$ for some $x,y\ne 0$ : $f(x^2)=f(x)^2=0$ $P(x,y)$ $\implies$ $f(x+y)=y$, impossible since $LHS$ is either $0$, either $x+y$, while $x,y\ne 0$ And so : Either $\boxed{\text{S2 : }f(x)=0\quad\forall x}$, which indeed fits Either $\boxed{\text{S3 : }f(x)=x\quad\forall x}$, which indeed fits

Let $P(x, y)$ denote the assertion. $P(x, 0)$ gives us $f(x^2) = f(x)^2$; in particular, $f(0) \in \{ 0, 1 \}$. Case 1: $f(0) = 1$. $$P(0, x) \implies 1 - yf(y) = f(y)(1-y) $$$$\implies \boxed{f(y) = 1} \forall y.$$Case 2: $f(0) = 0$. $$P(-y, y) \implies f(y)^2 - yf(y) = 0$$$$\implies f(y) \in \{ 0, y \} \forall y.$$Now assume $\exists a, b \ne 0$ so that $f(a) = 0, f(b) = b$. Then $$P(a, b) \implies -b^2 = -bf(a+b)$$$$\implies f(a+b) = b,$$impossible. So the pointwise trap is removed, and we get three solutions, all of which can be checked to work; $$\boxed{f(x) = 0 \forall x}, \boxed{f(x) = 1 \forall x}, \boxed{f(x) = x \forall x}.$$$\square$

Putting $x=0$ we get $f(0)=f(y)f(0)$ and so $f(y)=1$ for all $y$ (this solution works indeed) or $f(0)=0$ which we shall now assume. Putting $y=0$ we find $f(x^2)=f(x)^2$ and putting $x=-y$ we get $f(y)^2=yf(y)$, i.e. $f(y)\in \{0,y\}$ for all $y$. The pointwise trap is easy to handle.