momo1729 wrote: Find all functions $f:\mathbb{R}\rightarrow \mathbb{R}$ such that for all $x,y$ in $\mathbb{R}$, $xf(x+xy)=xf(x)+f(x^{2}).f(y)$ Let $P(x,y)$ be the assertion $xf(x+xy)=xf(x)+f(x^2)f(y)$ $P(0,0)$ $\implies$ $f(0)=0$ If $f(1)=0$, then $P(1,x-1)$ $\implies$ $f(x)=0$ which indeed is a solution Let us from now consider that $f(1)=a\ne 0$ If $a\ne 1$, $P(1,x)$ $\implies$ $f(x+1)=af(x)+a$ and we easily get $f(n)=a\frac{a^n-1}{a-1}$ $\forall n\in\mathbb N$ Plugging this expression in $P(m,n)$, we see that this is not a solution (rather ugly, I think). So $a=1$ and $P(1,x)$ $\implies$ $f(x+1)=f(x)+1$ and so $f(n)=n$ and $f(x+n)=f(x)+n$ $P(x,-1)$ $\implies$ $f(x^2)=xf(x)$ Plugging this in $P(x,y)$, we get $xf(x(y+1))=xf(x)(f(y)+1)=xf(x)f(y+1)$ And so $f(xy)=f(x)f(y)$ $P(x,y)$ becomes then $xf(x)f(y+1)=xf(x)+f(x)^2f(y)$ $\iff$ $xf(x)(f(y)+1)=xf(x)+f(x)^2f(y)$ And so, setting $y=1$ : $f(x)(f(x)-x)=0$ and so $\forall x$, either $f(x)=0$, either $f(x)=x$ But, if for some $x\ne 0$, we have $f(x)=0$, then $f(x+1)=f(x)+1$ implies $f(x+1)=1$ which is impossible since either $f(x+1)=x+1\ne 1$, either $f(x+1)=0\ne 1$ So $f(x)=x$ $\forall x$, which indeed is a solution. Hence the answer : $f(x)=0$ $\forall x$ $f(x)=x$ $\forall x$

$ f:\mathbb{R}\rightarrow\mathbb{R} $ $ xf(x+xy)=xf(x)+f(x^{2}).f(y) $ $y=-1=>0=xf(x)+f(x^2)f(-1)$ $x=-1=>f(-1)=f(1)f(-1)=> f(-1)=0$ or $f(1)=1$ 1) $f(-1)=0=>xf(x)=0=>f(x)=0$ where $x!=0$ if $x=0=>f(0)=0=>f(x)=0 \forall x\in\mathbb{R}$ 2) $f(1)=1=>x=1=>f(x+1)=f(x)+1$ $f(-1)=-1$ $y=-1=>xf(x)=f(x^2)=>xf(x+xy)=xf(x)(f(y)+1)=>f(x+xy)=f(x)f(y+1)=>y=x-1=>f^2(x=f(x^2)=>xf(x)=f^2(x)=>f(x)(f(x)-x)=0$ if there exists such nonzero $a,b$ that $f(a)=0=>x=a=>f(a+ay)=0 \forall y\in\mathbb{R}$ if there exists such nonzero $b$ hat $f(b)=b=>y=\frac {b-a} {a}=>f(b)=0$ which is condraction.) Thus the answer is $f(x)=0 \forall x \in\mathbb{R}$ $f(x)=x \forall x \in\mathbb{R}$

$P(0,0) \Rightarrow f(0)=0$ $P(x,-1) \Rightarrow xf(x)=-f(x^2)f(-1)$ $P(x,x-1) \Rightarrow xf(x^2)=xf(x)+f(x^2)f(x-1) \Rightarrow xf(x^2)=-f(x^2)f(-1)+f(x^2)f(x-1) \Rightarrow f(x^2)=0$ or $f(x-1)=x+c$ which is $f(x^2)=0$ or $f(x-1)=x-1$ proving that a mixed function does not occur is not hard

$\bigstar \color{green}{\textit{\textbf{ANS:}}}$ $f(x)=x \quad \textrm{and} \quad f(x)=0 \quad \forall x\in \mathbb{R}.$ $\blacklozenge \color{blue}{\textit{\textbf{Proof:}}}$ It's easy to see that these are indeed solutions to the given FE. Let $P(x,y)$ denote the given assertion, \[P(0,x): f(0)f(x)=0\]we see that $f(x)=0 \quad \forall x\in \mathbb{R}$ is a solution here. If $f$ is not a zero function, then $f(0)=0$. Then, \[P(-1,-1): f(-1)(f(1)-1)=0\]if $f(-1)=0$, then $P(x,-1)\implies xf(x)=0 \implies f(x)=0 \quad \forall x\in \mathbb{R}$ which is a contradiction. So, $f(1)=1$ here, and \[P(1,-1): f(-1)=-1\]which gives \[P(x,-1): f(x^2)=xf(x).\]Rewrite our original $P(x,y)$ as \[P(x,y): f(x+xy)=f(x)+f(x)f(y)\]and as \[P(1,x): f(x+1)=1+f(x),\]let $x=u, v=y+1$, \[f(uv)=f(x(1+y))=f(x)(1+f(y))=f(u)f(v)\]which implies $f$ is multiplicative. Also, let $x=a, xy=b$, $f(a+b)=f(x+xy)=f(x)+f(x)f(y)=f(x)+f(xy)=f(a)+f(b)$ which implies $f$ is also additive. All together, $f(x)=x \quad \forall x\in \mathbb{R}. \quad \blacksquare$

Let $P(x,y)$ the assertion of the given F.E. $P(0,x)$ gives $f(0)f(x)=0$ so if $f(0) \ne 0$ then $f$ is constant but $f$ constant means $f(x)=0$ contradiction!, therefore $f(0)=0$. $P(x,-1)$ gives $-f(-1)f(x^2)=xf(x)$, now if $f(1)=0$ then $P(1,x)$ again gives $f$ constant so $f(x)=0$ which is a solution, otherwise $f(1) \ne 0$ and then by $P(1,-1)$ we get $f(-1)=-1$ and therefore $f(x^2)=xf(x)$ which means for $x \ne 0$ that $f$ is odd, but since $f(0)=0$ we get that $f$ is odd everywhere. Now we can re-write $P(x,y)$ as $f(x+xy)=f(x)(1+f(y))$ for $x \ne 0$, also note from the previous equation we get $f(1)=1$ by replacing $x=-1$, and now $P(1,x)$ for $x \ne 0$ gives $f(x+1)=f(x)+1$ which by induction gives $f(x+n)=f(x)+n$ and $f(n)=n$ for all integers $n$. Now note that again our F.E. for $x,y \ne 0$ can be re-written as $f(xy)=f(x)f(y)$ so $f$ is multiplicative (even when $x$ or $y$ are $0$, clearly). But this means $f(x)^2=f(x^2)=xf(x)$ so $f(x)=x$ or $f(x)=0$, but note that if $f(c)=0$ for some $c \ne 0$ then $f(cx)=0$ so $f(x)=0$ everywhere, in the other case $f(x)=x$ for all reals $x$. Therefore $f(x)=x$ and $f(x)=0$ are the only solutions thus we are done .