$ x = y = 0$ gives $ f(0)^2 = 0$, so $ f(0) = 0$. $ x = 1, y = -1$ gives $ f(1)f(-1) = -f(1)$. Either $ f(1) = 0$ or $ f(-1) = -1$. Suppose $ f(1) = 0$. Then $ x = 1$ gives $ f(1+y) = 0$, or $ f \equiv 0$. This is a solution. Now assume it away. $ f(-1) = -1$. $ y = -1$ gives $ f(x^2) = xf(x)$. Put $ x = -1$ in to get $ f(1) = 1$. $ x = 1$ gives $ f(y+1) = f(y)+1$. Substituting $ f(x^2)$ for $ xf(x)$ gives $ xf(x+xy) = xf(x) + xf(x)f(y)$. Let's only allow $ x$ to be nonzero, and then the equation reduces to $ f(x(y+1)) = f(x) + f(x)f(y)$, which reduces to $ f(xy) = f(x)f(y)$ when $ y$ is shifted by $ 1$. Put in $ x = y$ in this new equation to get $ f(x^2) = f(x)^2$, or $ xf(x) = f(x)^2$. So for all $ x$, $ f(x) = 0$ or $ f(x) = x$. If $ f(a) = 0$ for some $ a \neq 0$, then $ f(xa) = f(x)f(a) = 0$. $ xa$ can take any nonzero real, so $ f \equiv 0$ in this case. We end up with $ f(x) = x$ or $ f(x) = 0$ as our solutions.

MellowMelon wrote: $ x = y = 0$ gives $ f(0)^2 = 0$, so $ f(0) = 0$. $ x = 1, y = -1$ gives $ f(1)f(-1) = -f(1)$. Either $ f(1) = 0$ or $ f(-1) = -1$. Suppose $ f(1) = 0$. Then $ x = 1$ gives $ f(1+y) = 0$, or $ f \equiv 0$. This is a solution. Now assume it away. $ f(-1) = -1$. $ y = -1$ gives $ f(x^2) = xf(x)$. Put $ x = -1$ in to get $ f(1) = 1$. $ x = 1$ gives $ f(y+1) = f(y)+1$. Substituting $ f(x^2)$ for $ xf(x)$ gives $ xf(x+xy) = xf(x) + xf(x)f(y)$. Let's only allow $ x$ to be nonzero, and then the equation reduces to $ f(x(y+1)) = f(x) + f(x)f(y)$, which reduces to $ f(xy) = f(x)f(y)$ when $ y$ is shifted by $ 1$. Put in $ x = y$ in this new equation to get $ f(x^2) = f(x)^2$, or $ xf(x) = f(x)^2$. So for all $ x$, $ f(x) = 0$ or $ f(x) = x$. If $ f(a) = 0$ for some $ a \neq 0$, then $ f(xa) = f(x)f(a) = 0$. $ xa$ can take any nonzero real, so $ f \equiv 0$ in this case. We end up with $ f(x) = x$ or $ f(x) = 0$ as our solutions. Can you pleas explain " which reduces to $ f(xy) = f(x)f(y)$ when y is shifted by 1" I just cant figure out what you did there. Help please

$f(x(y+1)) = f(x) + f(x)f(y)$ is equivalent to $f(xy) = f(x) + f(x)f(y-1) = f(x)(1 + f(y-1) = f(x)f(y)$ when you replace $y+1$ with $y$.

Let $P(x,y)$ be the assertion into the problem statement. $P(0,y)\implies f(y)f(0)=0.$ Either $\boxed{f(x)=0}$ for all $x$ or we have $f(0)=0$. Assume that there is some $x$ such that $f(x)\neq 0$; thus, we have $f(0)=0$. $P(1,-1)\implies 0=f(1)+f(1)f(-1)\implies f(1)(1+f(-1))=0.$ Either $f(1)=0$ or $f(-1)=-1.$ Case 1: $f(1)=0$. $P(1,y)\implies f(y+1)=0$ but this doesn't satisfy the assumption that there exists some $x$ such that $f(x)\neq 0$. We already counted this solution before the assumption of $f(0)=0$. Case 2: $f(-1)=-1$ $P(-1,-1)\implies 0=1-f(1)$ so $f(1)=1$. Thus, $P(1,y)\implies f(y+1)=f(y)+1$. $P(x,-1)\implies xf(x)=f(x^2)$. Substituting these two identities into the problem yields $xf(x+xy)=xf(x)+f(x^2)f(y)=xf(x)(1+f(y))=xf(x)(f(y+1))$ so $$f(x+xy)=f(x)(f(y+1))$$given that $x\neq 0.$ Let $Q(x,y)$ be the assertion into this new function. $Q(x,x-1)\implies f(x^2)=(f(x))^2$. Hence, $(f(x))^2=f(x^2)=xf(x)$ so $$f(x)(f(x)-x)=0.$$Now, we've run into the Pointwise Values Trap. We claim that the only possible solution from this if $\boxed{f(x)=x}$ and pathological solutions do not exist. Suppose for the sake of contradiction that there exists distinct values $s$ and $t$ such that $f(s)=s$ and $f(t)=0$ for $s,t\neq 0$(since they still can be in the same function). Then, $f(s+st)=s+st$ or $f(s+st)=0$. We consider the revised function $$xf(x+xy)=xf(x)(f(y)+1).$$ Case 2.1: $f(s)=s, f(t)=0$, and $f(s+st)=s+st$ Let $x=s$ and $y=t$ in the revised function yields $sf(s+st)=sf(s)(f(t)+1)\implies s^2+s^2t=s^2$ or $s^2t=0$ so we must have either $s=0$ or $t=0$, contradiction. Case 2.2: $f(s)=s, f(t)=0$, and $f(s+st)=0$. Let $x=s$ and $y=t$ in the revised function yields $sf(s+st)=sf(s)(f(t)+1)\implies 0=s^2$ or $s=0$, contradiction. Hence, we have proven that $f(x)=x$ is the only solution that is not identically zero and our two solutions are $f(x)=x$ and $f(x)=0.$ $\blacksquare$

all the solutions are beautiful but i found an easier solution

Let $P(x,y)$ the assertion: $P(0,0)$ $f(0)^2=0 \implies f(0)=0$ $P(1,-1)$ $-f(-1)f(1)=f(1) \implies$ $f(-1)=-1$ or $f(1)=0$ If $f(-1)=0$: $P(1,x-1)$ $f(x)=0$ for all $x \in \mathbb R$ If $f(-1)=-1$ $P(x,-1)$ $xf(x)=f(x^2)$. $P(x+1,x)$ $(x+1)f((x+1)^2)=f((x+1)^2)(1+f(x)) \implies x+1=1+f(x) \implies f(x)=x$ for all $x \in \mathbb R$. Or also implies $f((x+1)^2)=0$ but we found $f(x)=0$ is another solution soo.. $f(x)=x$ or $f(x)=0$ for all $x \in \mathbb R$

MathLuis wrote: $(x+1)f((x+1)^2)=f((x+1)^2)(1+f(x)) \implies x+1=1+f(x) \implies f(x)=x$ for all $x \in \mathbb R$ No. $(x+1)f((x+1)^2)=f((x+1)^2)(1+f(x)) $ implies : $\forall x\in\mathbb R$, either $f((x+1)^2)=0$, either $f(x)=x$

orl wrote: Determine all functions $ f: \mathbb{R} \mapsto \mathbb{R}$ such that \[ x f(x + xy) = x f(x) + f \left( x^2 \right) f(y) \quad \forall x,y \in \mathbb{R}.\] Can anyone check this proof? The only solutions are $f\equiv \text{Id}$ and $f\equiv 0$ which clearly works.We will now prove that indeed these are the only solutions. Let $P(x,y)$ denote the assertion $x f(x + xy) = x f(x) + f \left( x^2 \right) f(y)$.$\qquad$. We proceed in the following steps:- Note that $P(0,0)$ gives $f(0)=0$.From $P(x,-1)$ we get $f(-1)f(x^2)+xf(x)=0$. Now split into two cases; If $f(-1)=0$ then we get one of the solutions $f\equiv 0$.Otherwise Compare $P(-1,x)$ and $P(-1,-x)$ to see $xf(x)=-xf(-x)$ which implies $f$ is odd. Now $x=1$ in the above para easily gives $f(1)=1$.Thus $f(-1)=-1$ and hence $f(x^2)=xf(x)$ and $P(1,y)$ gives $f(y+1)=f(y)+1$.To finish up $$P(x,x-1)\implies x^2f(x)=xf(x^2)=xf(x)+f(x^2)f(x-1)=xf(x)+f(x^2)f(x-1)=xf(x)+xf(x)(f(x)-1) \implies f(x)=x \forall x\neq 0 $$but since $f(0)=0$ hence we conclude $f(x)=x$ for all $x\in \mathbb{R}$.$\square$

Pluto1708 wrote: $x^2f(x)=......=xf(x)+xf(x)(f(x)-1) \implies f(x)=x \forall x\neq 0 $ No $x^2f(x)=xf(x)^2$ implies $\forall x\ne 0$, either $f(x)=0$, either $f(x)=x$

pco wrote: MathLuis wrote: $(x+1)f((x+1)^2)=f((x+1)^2)(1+f(x)) \implies x+1=1+f(x) \implies f(x)=x$ for all $x \in \mathbb R$ No. $(x+1)f((x+1)^2)=f((x+1)^2)(1+f(x)) $ implies : $\forall x\in\mathbb R$, either $f((x+1)^2)=0$, either $f(x)=x$ Yes @pco i made a mistake in that part i have done yet the case when $f(x)=0$ soo that doesnt make a problem i think

i will correct it right now

$xf(x+xy)=xf(x)+f(x^2)f(y)$ $P(0,0)$ $\implies$ $f(0)=0$ $P(-1,-1)$ $\implies$ $f(-1)=0$ or $f(1)=1$ If $f(-1)=0$ then $P(x,-1)$ $\implies$ $f(x)=0$ which is a solution. If $f(1)=1$ $P(1,x)$ $\implies$ $f(x+1)=1+f(x)$ $P(x,x-1)$ $xf(x^2)=xf(x)+f(x^2)f(x-1)=xf(x)+f(x^2)(f(x)-1)$ $xf(x^2)=xf(x)+f(x^2)f(x)-f(x^2)$ $f(x^2)=\frac{xf(x)}{x+1-f(x)}$ it is obvious that $f(x) \ne x+1$ Let's substitute this with original equation $xf(x+xy)=xf(x)+\frac{xf(x)f(y)}{x+1-f(x)}$ Call it's assertion $Q(x,y)$ $Q(1,-1)$ $\implies$ $f(-1)=-1$ $Q(-1,x)$ $\implies$ $f(x)=-f(-x)$ $Q(-x,-1)$ $\implies$ $x+1-f(x)=1$ so that, $f(x)=x$ We have pointwise trap which is easy to get contradicition. Sols:$f(x)=0,x$